An Undergraduate Introduction to Financial Mathematics [3 ed.] 9814407445, 9789814407441

This textbook provides an introduction to financial mathematics and financial engineering for undergraduate students who

4,123 476 10MB

English Pages 486 [481] Year 2012

Report DMCA / Copyright

DOWNLOAD FILE

Polecaj historie

An Undergraduate Introduction to Financial Mathematics [3 ed.]
 9814407445, 9789814407441

Table of contents :
Undergraduate_Introduction_To_Financial_Mathematic..._----_(Intro)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(About_the_Author)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(Contents)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(1._The_Theory_of_Interest)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(2._Discrete_Probability)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(3._Normal_Random_Variables_and_Probability)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(4._The_Arbitrage_Theorem)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(5._Random_Walks_and_Brownian_Motion)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(6._Forwards_and_Futures)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(7._Options)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(8._Solution_of_the_Black-Scholes_Equation)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(9._Derivatives_of_Black-Scholes_Option_Prices)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(10._Hedging)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(11._Extensions_of_the_Black-Scholes_Model)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(12._Optimizing_Portfolios)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(13._American_Options)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(Appendix_A_Sample_Stock_Market_Data)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(Appendix_B_Solutions_to_Chapter_Exercises)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(Index)
Undergraduate_Introduction_To_Financial_Mathematic..._----_(Bibliography)

Citation preview

10:56:00.

An Undergraduate Introduction to Financial Mathematics Third Edition

10:56:00.

8495.9789814407441-tp.indd 1

14/6/12 11:38 AM

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

This page intentionally left blank

10:56:00.

BC8495/

An Undergraduate Introduction to Financial Mathematics Third Edition

J Robert Buchanan Millersville University, USA

World Scientific 10:56:00.

8495.9789814407441-tp.indd 2

14/6/12 11:38 AM

Published by World Scientific Publishing Co. Pte. Ltd. 5 Toh Tuck Link, Singapore 596224 USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress Cataloging-in-Publication Data Buchanan, J. Robert. An undergraduate introduction to financial mathematics / by J. Robert Buchanan. -- 3rd ed. p. cm. Includes bibliographical references and index. ISBN 978-9814407441 1. Business mathematics. I. Title. HF5691.B875 2012 330.01'513--dc23 2012020686

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library.

Copyright © 2012 by World Scientific Publishing Co. Pte. Ltd. All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means, electronic or mechanical, including photocopying, recording or any information storage and retrieval system now known or to be invented, without written permission from the Publisher.

For photocopying of material in this volume, please pay a copying fee through the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to photocopy is not required from the publisher.

In-house Editor: Juliet Lee Ley Chin

Printed in Singapore.

10:56:00.

Juliet - An Undergraduate Introduction.pmd

1

6/14/2012, 11:40 AM

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

For my wife, Monika.

10:56:00.

v

BC8495/Chp. 0

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

This page intentionally left blank

10:56:00.

vi

BC8495/Chp. 0

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 0

Preface

This third edition of An Undergraduate Introduction to Financial Mathematics extends significantly the material found in previous editions. A new chapter on extensions of the Black-Scholes formula for option pricing has been added. This chapter treats the effects of discrete and continuous dividends on option prices and hedging. The other returning chapters have had new material added to them as well, in an effort to make them more complete and comprehensive and to help them stand alone as introductions to the mathematics necessary to establish the remaining results found throughout the textbook. The chapter on arbitrage and linear programming has been improved in terms of serving as an introduction to linear programming and in terms of serving as a standalone justification of the Arbitrage Theorem. The chapter on Brownian motion and random processes includes much more background information and material on stochastic differential equations than in previous editions. It should now serve as a high-level overview of stochastic calculus for anyone contemplating a graduate-level course in the subject. The number of end-of-chapter exercises has been significantly increased as well in order to make this textbook more useful for self-study and for classroom use. The notation of the text has been streamlined and re-worked in order to bring the notation into closer agreement with other sources, to make the notation more consistent from chapter to chapter, and to make it easier for the reader to follow. It remains the author’s hope that this text is an accurate, accessible introduction for undergraduates to the mathematics of options and derivatives. The prerequisite mathematical background (multivariable calculus) has been kept the same as in previous editions. In order to produce a more accurate and readable new edition of this textbook, I enlisted the help of some of my colleagues and associates at Millersville University. They

10:56:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

viii

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 0

An Undergraduate Introduction to Financial Mathematics

deserve a great deal of thanks for helping me root out misspellings, grammatically mangled sentences, and other imperfections in a draft of the third edition. In alphabetical order they are: Ms. Monika Buchanan, Dr. Antonia Cardwell, Dr. Ximena Catepill´an, Dr. James Fenwick, Dr. Noel Heitmann, Dr. Bruce Ikenaga, Dr. Kevin Robinson, Dr. Delray Schultz, Dr. Zhoude Shao, Dr. Janet White, and Dr. Michael Wismer. Both the author and the text benefit from interacting with readers of the book and students using it for a class. If a reader has corrections or suggestions to share with me, or to check the latest list of errata, please consult the links found at the web site: http://banach.millersville.edu/∼bob/book/ J. Robert Buchanan Wyomissing, PA, USA May 25, 2012

10:56:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 0

Preface to the Second Edition

This second edition of An Undergraduate Introduction to Financial Mathematics extends significantly the material found in the first edition. Owners of the first edition will find the second contains corrections and clarifications of the contents of the first as well as additional examples, exercises, and two entirely new chapters. As carefully as I proof-read the manuscript of the first edition, an embarrassingly large number of typos and garbled sentences managed to pass through my filter. Fortunately several of the readers of the first edition took the time to compile and send to me a list of errors and other suggestions. The improvements in the second edition are a result of the set of corrections and comments made by the readers. Two individuals stand out for the volume of suggestions and help they gave, Prof. M.M. Chawla and Prof. Josef Dick. To the ten chapters of the first edition have been added two more. The first addition is on the topic of “Forwards and Futures” and constitutes the new Chapter 6. This topic allows the reader to exercise their newly obtained knowledge of Brownian motion, stochastic processes, and arbitrage at an earlier stage of the book than in the first edition. Previously these various threads were woven together in the chapters on options and solving the Black-Scholes equation. The earlier application of these topics may help the reader to gain greater mastery and to feel more comfortable using these tools. Chapter 6 also includes a discussion of the practice of “Marking to Market” for futures. This is provided as a preview of the process of hedging for portfolios of securities and options which appears in a later chapter. The second addition is on the topic of “American Options”. In the first edition of the text, American options were mentioned and briefly described mainly to give the reader a sense of the broad array of financial instruments found in the world of investment and risk management. In the second edition,

10:56:00.

May 25, 2012

x

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 0

An Undergraduate Introduction to Financial Mathematics

properties of American options are more fully explored and an elementary algorithm for pricing a type of American option is explained. This material forms the new Chapter 12 of the second edition. Chapters 6–10 of the first edition are now Chapters 7–11 of the second. The chapters returning in the second edition from the first edition should not disappoint the reader as they have been corrected, expanded, and polished. New examples, exercises, and higher quality graphics appear in the returning chapters. Since the appearance of the first edition, I have taught a course for undergraduates using the first edition as the textbook. I appreciate the comments of the students I faced in the classroom and those of the students at other institutions who emailed me. Student Catherine Albright from the fall semester 2007 read the first edition with a careful eye and brought to my attention numerous typographical errors. It remains the author’s hope that this text is an accurate, accessible introduction for undergraduates to the mathematics of options and derivatives. The prerequisite mathematical background (multivariable calculus) has been kept the same as in the first edition. If a reader has corrections or suggestions to share with me, or to check the latest list of errata, please consult the links found at the web site: http://banach.millersville.edu/∼bob/book/ J. Robert Buchanan Wyomissing, PA, USA August 12, 2008

10:56:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 0

Preface to the First Edition

This book is intended for an audience with an undergraduate level of exposure to calculus through elementary multivariable calculus. The book assumes no background on the part of the reader in probability or statistics. One of my objectives in writing this book was to create a readable, reasonably self-contained introduction to financial mathematics for people wanting to learn some of the basics of option pricing and hedging. My desire to write such a book grew out of the need to find an accessible book for undergraduate mathematics majors on the topic of financial mathematics. I have taught such a course now three times and this book grew out of my lecture notes and reading for the course. New titles in financial mathematics appear constantly, so in the time it took me to compose this book there may have appeared several superior works on the subject. Knowing the amount of work required to produce this book, I stand in awe of authors such as those. This book consists of ten chapters which are intended to be read in order, though the well-prepared reader may be able to skip the first several with no loss of understanding in what comes later. The first chapter is on interest and its role in finance. Both discretely compounded and continuously compounded interest are treated there. The book begins with the theory of interest because this topic is unlikely to scare off any reader no matter how long it has been since they have done any formal mathematics. The second and third chapters provide an introduction to the concepts of probability and statistics which will be used throughout the remainder of the book. Chapter Two deals with discrete random variables and emphasizes the use of the binomial random variable. Chapter Three introduces continuous random variables and emphasizes the similarities and differences between discrete and continuous random variables. The nor-

10:56:00.

May 25, 2012

xii

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 0

An Undergraduate Introduction to Financial Mathematics

mal random variable and the closely related lognormal random variable are introduced and explored in the latter chapter. In the fourth chapter the concept of arbitrage is introduced. For readers already well versed in calculus, probability, and statistics, this is the first material which may be unfamiliar to them. The assumption that financial calculations are carried out in an “arbitrage free” setting pervades the remainder of the book. The lack of arbitrage opportunities in financial transactions ensures that it is not possible to make a risk free profit. This chapter includes a discussion of the result from linear algebra and operations research known as the Duality Theorem of Linear Programming. The fifth chapter introduces the reader to the concepts of random walks and Brownian motion. The random walk underlies the mathematical model of the value of securities such as stocks and other financial instruments whose values are derived from securities. The choice of material to present and the method of presentation is difficult in this chapter due to the complexities and subtleties of stochastic processes. I have attempted to introduce stochastic processes in an intuitive manner and by connecting elementary stochastic models of some processes to their corresponding deterministic counterparts. Itˆ o’s Lemma is introduced and an elementary proof of this result is given based on the multivariable form of Taylor’s Theorem. Readers whose interest is piqued by material in Chapter Five should consult the bibliography for references to more comprehensive and detailed discussions of stochastic calculus. Chapter Six introduces the topic of options. Both European and American style options are discussed though the emphasis is on European options. Properties of options such as the Put-Call Parity formula are presented and justified. In this chapter we also derive the partial differential equation and boundary conditions used to price European call and put options. This derivation makes use of the earlier material on arbitrage, stochastic processes and the Put-Call Parity formula. The seventh chapter develops the solution to the Black-Scholes PDE. There are several different methods commonly used to derive the solution to the PDE and students benefit from different aspects of each derivation. The method I choose to solve the PDE involves the use of the Fourier Transform. Thus this chapter begins with a brief discussion of the Fourier and Inverse Fourier Transforms and their properties. Most three- or foursemester elementary calculus courses include at least an optional section on the Fourier Transform, thus students will have the calculus background necessary to follow this discussion. It also provides exposure to the Fourier

10:56:00.

June 25, 2012

13:21 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Preface to the First Edition

BC8495/Chp. 0

xiii

Transform for students who will be later taking a course in PDEs and more importantly exposure for students who will not take such a course. After completing this derivation of the Black-Scholes option pricing formula students should also seek out other derivations in the literature for the purposes of comparison. Chapter Eight introduces some of the commonly discussed partial derivatives of the Black-Scholes option pricing formula. These partial derivatives help the reader to understand the sensitivity of option prices to movements in the underlying security’s value, the risk-free interest rate, and the volatility of the underlying security’s value. The collection of partial derivatives introduced in this chapter is commonly referred to as “the Greeks” by many financial practitioners. The Greeks are used in the ninth chapter on hedging strategies for portfolios. Hedging strategies are used to protect the value of a portfolio against movements in the underlying security’s value, the risk-free interest rate, and the volatility of the underlying security’s value. Mathematically the hedging strategies remove some of the low order terms from the Black-Scholes option pricing formula making it less sensitive to changes in the variables upon which it depends. Chapter Nine will discuss and illustrate several examples of hedging strategies. Chapter Ten extends the ideas introduced in Chapter Nine by modeling the effects of correlated movements in the values of investments. The tenth chapter discusses several different notions of optimality in selecting portfolios of investments. Some of the classical models of portfolio selection are introduced in this chapter including the Capital Assets Pricing Model (CAPM) and the Minimum Variance Portfolio. It is the author’s hope that students will find this book a useful introduction to financial mathematics and a springboard to further study in this area. Writing this book has been hard, but intellectually rewarding work. During the summer of 2005 a draft version of this manuscript was used by the author to teach a course in financial mathematics. The author is indebted to the students of that class for finding numerous typographical errors in that earlier version which were corrected before the camera ready copy was sent to the publisher. The author wishes to thank Jill Bachstadt, Jason Buck, Mark Elicker, Kelly Flynn, Jennifer Gomulka, Nicole Hundley, Alicia Kasif, Stephen Kluth, Patrick McDevitt, Jessica Paxton, Christopher Rachor, Timothy Refi, Pamela Wentz, Joshua Wise, and Michael Zrncic. A list of errata and other information related to this book can be found at a web site I created:

10:56:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

xiv

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 0

An Undergraduate Introduction to Financial Mathematics

http://banach.millersville.edu/∼bob/book/ Please feel free to share your comments, criticism, and (I hope) praise for this work through the email address that can be found at that site. J. Robert Buchanan Lancaster, PA, USA October 31, 2005

10:56:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

About the Author

J. Robert Buchanan is a Professor in the Department of Mathematics at Millersville University of Pennsylvania. He received his Ph.D. in applied mathematics from North Carolina State University in 1993. Since his undergraduate days, he has had a keen interest in applications of mathematics in physics, biology, economics and finance. Teaching applications of mathematics and making advanced mathematics accessible to undergraduate students is his professional goal. In his spare time, he enjoys mathematical puzzles and challenges.

10:57:54.

465

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 0

Contents

Preface

vii

Preface to the Second Edition

ix

Preface to the First Edition

xi

1.

The Theory of Interest 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

2.

Simple Interest . . . . . . . . . . . Compound Interest . . . . . . . . . Continuously Compounded Interest Present Value . . . . . . . . . . . . Time-Varying Interest Rates . . . . Rate of Return . . . . . . . . . . . Continuous Income Streams . . . . Exercises . . . . . . . . . . . . . .

1 . . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

Discrete Probability 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Events and Probabilities . . . . . . . . . . . . . . Addition Rule . . . . . . . . . . . . . . . . . . . . Conditional Probability and Multiplication Rule Random Variables and Probability Distributions Binomial Random Variables . . . . . . . . . . . . Expected Value . . . . . . . . . . . . . . . . . . . Variance and Standard Deviation . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . .

10:56:13.

1 3 4 6 12 14 15 17 21

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

. . . . . . . .

22 23 24 27 28 30 38 42

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

Chapter 1

The Theory of Interest

One of the first types of investments that people learn about is some variation on the savings account. In exchange for the temporary use of an investor’s money, a bank or other financial institution agrees to pay interest, a percentage of the amount invested, to the investor. There are many different schemes for paying interest. In this chapter we will describe some of the most common types of interest and contrast their differences. Along the way the reader will have the opportunity to renew their acquaintanceship with exponential functions and the geometric series. Since an amount of capital can be invested and earn interest and thus numerically increase in value in the future, the concept of present value will be introduced. Present value provides a way of comparing values of investments made at different times in the past, present, and future. As an application of present value, several examples of saving for retirement and calculation of mortgages will be presented. Sometimes investments pay the investor varying amounts of money which change over time. The concept of rate of return can be used to convert these payments in effective interest rates, making comparison of investments easier. 1.1

Simple Interest

In exchange for the use of a depositor’s money, banks pay a fraction of the account balance back to the depositor. This fractional payment is known as interest. The money a bank uses to pay interest is generated by investments and loans that the bank makes with the depositor’s money. Interest is paid in many cases at specified times of the year, but nearly always the fraction of the deposited amount used to calculate the interest is called the interest rate and is expressed as a percentage paid per year.

10:56:51.

May 25, 2012

2

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

For example, a credit union may pay 6% annually on savings accounts. This means that if a savings account contains $100 now, then exactly one year from now the bank will pay the depositor $6 (which is 6% of $100) provided the depositor maintains an account balance of $100 for the entire year. In this chapter and those that follow, interest rates will be denoted symbolically by r. To simplify the formulas and mathematical calculations, when r is used it will be converted to decimal form even though it may still be referred to as a percentage. The 6% annual interest rate mentioned above would be treated mathematically as r = 0.06 per year. The initially deposited amount which earns the interest will be called the principal amount and will be denoted P . The sum of the principal amount and any earned interest will be called the capital or the amount due. The symbol A will be used to represent the amount due. The reader may even see the amount due referred to as the compound amount, though this use of the adjective “compound” is independent of its use in the term “compound interest” to be explored in Section 1.2. The relationship between P , r, and A for a single year period is A = P + P r = P (1 + r). In general, if the time period of the deposit is t years then the amount due is expressed in the formula A = P (1 + rt).

(1.1)

This implies that the average account balance for the period of the deposit is P and when the balance is withdrawn (or the account is closed), the principal amount P plus the interest earned P rt is returned to the investor. No interest is credited to the account until the instant it is closed. This is known as the simple interest formula. Some financial institutions credit interest earned by the account balance at fixed points in time. Banks and other financial institutions “pay” the depositor by adding the interest to the depositor’s account. The interest, once paid to the depositor, is the depositor’s to keep. Unless the depositor withdraws the interest or some part of the principal, the process begins again for another interest earning period. If P is initially deposited, then after one year, the amount due according to Eq. (1.1) with t = 1 would be P (1 + r). This amount can be thought of as the principal amount for the account at the beginning of the second year. Thus, two years after the

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Theory of Interest

BC8495/Chp. 1

3

initial deposit the amount due would be A = P (1 + r) + P (1 + r)r = P (1 + r)2 . Continuing in this way we can see that t years after the initial deposit of an amount P , the capital A will grow to A = P (1 + r)t .

(1.2)

A mathematical “purist” may wish to establish Eq. (1.2) using the principle of induction. Banks and other interest-paying financial institutions often pay interest more than a single time per year. The yearly interest formula given in Eq. (1.2) must be modified to track the compound amount for interest periods of other than one year.

1.2

Compound Interest

The typical interest bearing savings or checking account will be described by an investor as earning a nominal annual interest rate compounded some number of times per year. Investors will often find interest compounded semi-annually, quarterly, monthly, weekly, or daily. In this section we will compare and contrast compound interest to the simple interest case of the previous section. Whenever interest is allowed to earn interest itself, an investment is said to earn compound interest. In this situation, part of the interest is paid to the depositor once or more frequently per year. Once paid, the interest begins earning interest. We will let n denote the number of compounding periods per year. For example for interest “compounded monthly” n = 12. Only two small modifications to the interest formula in Eq. (1.2) are needed to calculate the compound interest. First, it is now necessary to think of the interest rate per compounding period. If the annual interest rate is r, then the interest rate per compounding period is r/n. Second, the elapsed time should be thought of as some number of compounding periods rather than years. Thus, with n compounding periods per year, the number of compounding periods in t years is nt. Therefore, the formula for compound interest is  r nt A=P 1+ . n 10:56:51.

(1.3)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

4

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

Eq. (1.3) simplifies to the formula for the amount due given in Eq. (1.2) when n = 1. Example 1.1 Suppose an account earns 5.75% annually compounded monthly. If the principal amount is $3104 then after three and one-half years the amount due will be  (12)(3.5) 0.0575 A = 3104 1 + = 3794.15. 12 The reader should verify using Eq. (1.1) that if the principal in the previous example earned only simple interest at an annual rate of 5.75% then the amount due after 3.5 years would be only $3728.68. Thus happily for the depositor, compound interest builds capital faster than simple interest. Frequently it is useful to compare an annual interest rate with compounding to an equivalent simple interest, i.e. to the simple annual interest rate which would generate the same amount of interest as the annual compound rate. This equivalent interest rate is called the effective interest rate. For the rate mentioned in the previous example we can find the effective interest rate by solving the equation 12  0.0575 = 1 + re 1+ 12 0.05904 = re . Thus, the nominal annual interest rate of 5.75% compounded monthly is equivalent to an effective annual rate of 5.90%. In general, if the nominal annual rate r is compounded n times per year the equivalent effective annual rate re is given by the formula:  r n re = 1 + − 1. (1.4) n Intuitively it seems that more compounding periods per year implies a higher effective annual interest rate. In the next section we will explore the limiting case of frequent compounding going beyond semiannually, quarterly, monthly, weekly, daily, hourly, etc. to continuously. 1.3

Continuously Compounded Interest

Mathematically, when considering the effect on the compound amount of more frequent compounding, we are contemplating a limiting process. In

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Theory of Interest

BC8495/Chp. 1

5

symbolic form we would like to find the compound amount A which satisfies the equation  r nt A = lim P 1 + . (1.5) n→∞ n Fortunately, there is a simple expression for the value of the limit on the right-hand side of Eq. (1.5). We will find it by working on the limit  r n lim 1 + . n→∞ n

This limit is indeterminate of the form 1∞ . We will evaluate it through a standard approach using the natural logarithm and l’Hˆopital’s Rule. The reader should consult an elementary calculus book such as [Smith and Minton (2002)] for more details. We see that if y = (1 + r/n)n , then  r n ln y = ln 1 + n = n ln(1 + r/n) ln(1 + r/n) = 1/n which is indeterminate of the form 0/0 as n → ∞. To apply l’Hˆopital’s Rule we take the limit of the derivative of the numerator over the derivative of the denominator. Thus lim ln y = lim

n→∞

d dn

(ln(1 + r/n)) d dn

(1/n) r = lim n→∞ 1 + r/n =r n→∞

Thus, limn→∞ y = er . Finally we arrive at the formula for continuously compounded interest, A = P ert .

(1.6)

This formula may seem familiar since it is often presented as the exponential growth formula in elementary algebra, precalculus, or calculus. The quantity A has the property that A changes with time t at a rate proportional to A itself. Example 1.2 Suppose $3585 is deposited in an account which pays interest at an annual rate of 6.15% compounded continuously. After two and

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

6

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

one half years the principal plus earned interest will have grown to A = 3585e(0.0615)(2.5) = 4180.82. The effective simple interest rate is the solution to the equation e0.0615 = 1 + re which implies re ≈ 6.34%. 1.4

Present Value

One of the themes we will see many times in the study of financial mathematics is the comparison of the value of a particular investment at the present time with the value of the investment at some point in the future. This is the comparison between the present value of an investment versus its future value. We will see in this section that present and future value play central roles in planning for retirement and determining loan payments. Later in this book present and future values will help us determine a fair price for stock market derivatives. The future value t years from now of an invested amount P subject to an annual interest rate r compounded continuously is A = P ert . Thus, by comparison with Eq. (1.6), the future value of P is just the compound amount of P monetary units invested in a savings account earning interest r compounded continuously for t years. By contrast the present value of A in an environment of interest rate r compounded continuously for t years is P = Ae−rt . In other words, if an investor wishes to have A monetary units in savings t years from now and they can place money in a savings account earning interest at an annual rate r compounded continuously, the investor should deposit P monetary units now. There are also formulas for future and present value when interest is compounded at discrete intervals, not continuously. If the interest rate is r annually with n compounding periods per year then the future value of P is  r nt A=P 1+ . n 10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

7

The Theory of Interest

Compare this equation with Eq. (1.3). Simple algebra shows then the present value of P earning interest at rate r compounded n times per year for t years is  r −nt P =A 1+ . n

Example 1.3 Suppose an investor will receive payments at the end of the next six years in the amounts shown in the table below. Year Payment

1 465

2 233

3 632

4 365

5 334

6 248

If the interest rate is 3.99% compounded monthly, what is the present value of the investments? Assuming the first payment will arrive one year from now, the present value is the sum  −12  −24  −36 0.0399 0.0399 0.0399 465 1 + + 233 1 + + 632 1 + 12 12 12  −48  −60  −72 0.0399 0.0399 0.0399 + 365 1 + + 334 1 + + 248 1 + 12 12 12 = 2003.01. Notice that the present value of the payments from the investment is different from the sum of the payments themselves (which is 2277). Unless the reader is among the very fortunate few who can always pay cash for all purchases, you may some day apply for a loan from a bank or other financial institution. Loans are always made under the assumptions of a prevailing interest rate (with compounding), an amount to be borrowed, and the lifespan of the loan, i.e. the time the borrower has to repay the loan. Usually portions of the loan must be repaid at regular intervals (for example, monthly). Now we turn our attention to the question of using the amount borrowed, the length of the loan, and the interest rate to calculate the loan payment. A very helpful mathematical tool for answering questions regarding present and future values is the geometric series. Suppose we wish to find the sum S = 1 + a + a2 + · · · + an

(1.7)

where n is a positive whole number. If both sides of Eq. (1.7) are multiplied

10:56:51.

May 25, 2012

8

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

by a and then subtracted from Eq. (1.7) we have S − aS = 1 + a + a2 + · · · + an − (a + a2 + a3 + · · · + an+1 )

S(1 − a) = 1 − an+1 1 − an+1 S= 1−a

(1.8)

provided a 6= 1. Now we will apply this tool to the task of finding out the monthly amount of a loan payment. Suppose someone borrows P to purchase a new car. The bank issuing the automobile loan charges interest at the annual rate of r compounded n times per year. The length of the loan will be t years. The monthly installment can be calculated if we apply the principle that the present value of all the payments made must equal the amount borrowed. Suppose the payment amount is the constant x. If the first payment must be made at the end of the first compounding period, then the present value of all the payments is r −1 r r ) + x(1 + )−2 + · · · + x(1 + )−nt n n n r −1 1 − (1 + nr )−nt = x(1 + ) n 1 − (1 + nr )−1

x(1 +

=x

1 − (1 + nr )−nt r n

.

Therefore, the relationship between the interest rate, the compounding frequency, the period of the loan, the principal amount borrowed, and the payment amount is expressed in the following equation.   h n r i−nt P =x 1− 1+ (1.9) r n Example 1.4 If a person borrows $25000 for five years at an interest rate of 4.99% compounded monthly and makes equal monthly payments, the payment amount will be  −1 −(12)(5) x = 25000(0.0499/12) 1 − [1 + (0.0499/12)] = 471.67.

Similar reasoning can be used when determining how much to save for retirement. Suppose a person is 25 years of age now and plans to retire at age 65. For the next 40 years they plan to invest a portion of their monthly income in securities which earn interest at the rate of 10% compounded

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

9

The Theory of Interest

monthly. After retirement the person plans on receiving a monthly payment (an annuity) in the absolute amount of $1500 for 30 years. The amount of money the person should invest monthly while working can be determined by equating the present value of all their deposits with the present value of all their withdrawals. The first deposit will be made one month from now and the first withdrawal will be made 481 months from now. The last withdrawal will be made 840 months from now. The monthly deposit amount will be be denoted by the symbol x. The present value of all the deposits made into the retirement fund is −480 −i  −1 480  X 1 − 1 + 0.10 0.10 0.10 12 1+ x = x 1+ −1 12 12 1 − 1 + 0.10 i=1 12

≈ 117.765x.

Meanwhile, the present value of all the annuity payments is 1500

840  X

i=481

1+

0.10 12

−i

−360  −481 1 − 1 + 0.10 0.10 12 = 1500 1 + −1 12 1 − 1 + 0.10 12 ≈ 3182.94.

Thus, x ≈ 27.03 dollars per month. This seems like a small amount to invest, but such is the power of compound interest and starting a savings plan for retirement early. If the person waits ten years (i.e., until age 35) to begin saving for retirement, but all other factors remain the same, then −i 360  X 0.10 x 1+ ≈ 113.951x 12 i=1 −i 720  X 0.10 1500 1+ ≈ 8616.36 12 i=361 which implies the person must invest x ≈ 75.61 monthly. Waiting ten years to begin saving for retirement nearly triples the amount which the future retiree must set aside for retirement. The initial amounts invested are of course invested for a longer period of time and thus contribute a proportionately greater amount to the future value of the retirement account. Example 1.5 Suppose two persons will retire in twenty years. One begins saving immediately for retirement but due to unforeseen circumstances must abandon their savings plan after four years. The amount they

10:56:51.

May 25, 2012

10

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

put aside during those first four years remains invested, but no additional amounts are invested during the last sixteen years of their working life. The other person waits four years before putting any money into a retirement savings account. They save for retirement only during the last sixteen years of their working life. Let us explore the difference in the final amount of retirement savings that each person will possess. For the purpose of this example we will assume that the interest rate is r = 0.05 compounded monthly and that both workers will invest the same amount x, monthly. The first worker has upon retirement an account whose present value is −i 48  X 0.05 x 1+ ≈ 43.423x. 12 i=1 The present value of the second worker’s total investment is −i 240  X 0.05 x 1+ ≈ 108.102x. 12 i=49 Thus, the second worker retires with a larger amount of retirement savings; however, the ratio of their retirement balances is only 43.423/108.102 ≈ 0.40. The first worker saves, in only one fifth of the time, approximately 40% of what the second worker saves. The discussion of retirement savings makes no provision for rising prices. The economic concept of inflation is the phenomenon of the decrease in the purchasing power of a unit of money relative to a unit amount of goods or services. The rate of inflation (usually expressed as an annual percentage rate, similar to an interest rate) varies with time and is a function of many factors including political, economic, and international factors. While the causes of inflation can be many and complex, inflation is generally described as a condition which results from an increase in the amount of money in circulation without a commensurate increase in the amount of available goods. Thus, relative to the supply of goods, the value of the currency is decreased. This can happen when wages are arbitrarily increased without an equal increase in worker productivity. We now focus on the effect that inflation may have on the worker planning to save for retirement. If the interest rate on savings is r and the inflation rate is i we can calculate the inflation-adjusted rate or as it is sometimes called, the real rate of interest. This derivation will test your understanding of the concepts of present and future value discussed earlier in this chapter. We will let the symbol ri denote the inflation-adjusted

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

11

The Theory of Interest

interest rate [Broverman (2004)]. Suppose at the current time one unit of currency will purchase one unit of goods. Invested in savings, that one unit of currency has a future value (in one year) of 1 + r. In one year the unit of goods will require 1 + i units of currency for purchase. The difference (1 + r) − (1 + i) = r − i will be the real rate of growth in the unit of currency invested now. However, this return on saving will not be earned until one year from now. Thus, we must adjust this rate of growth by finding its present value under the inflation rate. This leads us to the following formula for the inflationadjusted interest rate. ri =

r−i 1+i

(1.10)

Note that when inflation is low (i is small), ri ≈ r −i and this latter approximation is sometimes used in place of the more accurate value expressed in Eq. (1.10). Returning to the earlier example of the worker saving for retirement, consider the case in which r = 0.10, the worker will save for 40 years and live on a monthly annuity whose inflation adjusted value will be $1500 for 30 years, and the rate of inflation will be i = 0.03 for the entire lifespan of the worker/retiree. Thus ri ≈ 0.0680. Assuming the worker will make the first deposit in one month the present value of all deposits to be made is −480 −i  −1 480  X 1 − 1 + 0.068 0.068 0.068 12 x 1+ = x 1+ −1 12 12 1 − 1 + 0.068 i=1 12 ≈ 164.756x. The present value of all the annuity payments is given by −360 −i  −481 840  X 1 − 1 + 0.068 0.068 0.068 12 1500 1+ = 1500 1 + −1 12 12 1 − 1 + 0.068 i=481 12 ≈ 15273.80. Thus, the monthly deposit amount is approximately $92.71. This is roughly four times the monthly investment amount when inflation is ignored. However, since inflation does tend to take place over the long run, ignoring a 3% inflation rate over the lifetime of the individual would mean that the

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

12

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

present purchasing power of the last annuity payment would be  −840 0.03 1500 1 + ≈ 184.17. 12 This is not much money to live on for an entire month. Retirement planning should include provisions for inflation, varying interest rates, the period of retirement, the period of savings, and desired monthly annuity during retirement.

1.5

Time-Varying Interest Rates

All of the discussion so far has assumed that interest rates remain constant during the life of a loan or a deposit. However, interest rates change over time due to a variety of economic and political factors. In this section we will extend ideas of present and future value to handle the case of a time-varying interest rate. We will call the continuously compounded interest rate r(t) (where the dependence on time t is explicit) the spot rate. While the behavior of the spot rate can be quite complex, for the moment we will assume that it is a continuous function of time. Assuming the amount due on a deposit earning interest at the spot rate r(t) at time t is A(t) then on the interval from t to t+∆t we can assume the interest rate remains near r(t) and simple interest accrues. Thus we may approximate the amount due at t + ∆t as A(t + ∆t) ≈ A(t)(1 + r(t)∆t). Rearranging terms in this approximation produces A(t + ∆t) − A(t) ≈ r(t)A(t) ∆t which upon taking the limit of both sides as ∆t → 0 yields the equation A0 (t) = r(t)A(t).

(1.11)

This is an example of a first-order linear homogeneous differential equation. Many elementary calculus textbooks and most undergraduate-level texts on ordinary differential equations discuss solving this type of equation. For an extensive discussion the reader is referred to [Smith and Minton (2002)] or [Boyce and DiPrima (2001)]. The approach is to multiply both sides of

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

13

The Theory of Interest

Eq. (1.11) by an integrating factor and integrate with respect to t. Suppose we define the integrating factor as µ(t) = e−

Rt 0

r(s) ds

then, multiplying both sides of Eq. (1.11) by µ(t) allows us to write the following. e−

Rt 0

r(s) ds

Rt

µ(t)A0 (t) = r(t)µ(t)A(t)

A0 (t) − r(t)e− 0 r(s) ds A(t) = 0 i d h − R t r(s) ds e 0 A(t) = 0 dt

Integrating both sides from 0 to t produces the formula for the amount due at time t. e−

Rt 0

r(s) ds

A(t) − e−

R0 0

r(s) ds

A(0) = 0 A(t) = A(0)e

Rt 0

r(s) ds

(1.12)

The present value of amount A due at time t under the time-varying schedule of interest rate r(t) is P (t) = Ae−

Rt 0

r(s) ds

.

(1.13)

Closely associated with the definite integral of the spot rate is the average of the spot rate over the interval [0, t]. The average interest rate written as Z 1 t r(t) = r(s) ds (1.14) t 0 is referred to as the yield curve. Thus, the formulas for amount due and present value can be written as A(t) = A(0)er(t) t P (t) = Ae−r(t) t . If the spot rate is constant, these formulas revert to the earlier forms. Example 1.6

Suppose the spot rate is r(t) =

r1 r2 t + 1+t 1+t

and find a formula for the yield curve and the present value of $1 due at time t.

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

14

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

By Eq. (1.14)  r2 s r1 + ds 1+s 1+s 0 r1 − r2 = r2 + ln(1 + t). t

r(t) =

1 t

Z t

Thus the present value of $1 is P (t) = e−r(t) t = e−t(r2 +

r1 −r2 t

ln(1+t))

= (1 + t)r2 −r1 e−r2 t .

1.6

Rate of Return

The present value of an item is one way to determine the absolute worth of the item and to compare its worth to that of other items. Another way to judge the value of an item which an investor may own or consider purchasing is known as the rate of return. If a person invests an amount P now and receives an amount A one time unit from now, the rate of return can be thought of as the interest rate per time unit that the invested amount would have to earn so that the present value of the payoff amount is equal to the invested amount. Since the rate of return is going to be thought of as an equivalent interest rate, it will be denoted by the symbol r. Then, by definition P = A(1 + r)−1

or equivalently r =

A − 1. P

Example 1.7 If you loan a friend $100 today with the understanding that they will pay you back $110 in one year’s time, then the rate of return is r = 0.10 or 10%. In a more general setting, a person may invest an amount P now and receive a sequence of positive payoffs {A1 , A2 , . . . , An } at regular intervals. In this case the rate of return per period is the interest rate such that the present value of the sequence of payoffs is equal to the amount invested. In this case P =

n X i=1

10:56:51.

Ai (1 + r)−i .

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

15

The Theory of Interest

It is not clear from this definition that r has a unique value for all choices of P and payoff sequences. Defining the function f (r) to be f (r) = −P +

n X

Ai (1 + r)−i

(1.15)

i=1

we can see that f (r) is continuous on the open interval (−1, ∞). In the limit as r approaches −1 from the right, the function values approach positive infinity. On the other hand as r approaches positive infinity, the function values approach −P < 0 asymptotically. Thus by the Intermediate Value Theorem (p. 108 of [Smith and Minton (2002)]) there exists r∗ with −1 < r∗ < ∞ such that f (r∗ ) = 0. The reader is encouraged to show that r∗ is unique in the exercises. Rates of return can be either positive or negative. If f (0) > 0, i.e., the sum of the payoffs is greater than the amount invested then r∗ > 0 since f (r) changes sign on the interval [0, ∞). If the sum of the payoffs is less than the amount invested then f (0) < 0 and the rate of return is negative. In this case the function f (r) changes sign on the interval (−1, 0]. Example 1.8 Suppose you loan a friend $100 with the agreement that they will pay you at the end of each year for the next five years amounts {21, 22, 23, 24, 25}. The rate of return per year is the solution to the equation, −100 +

21 22 23 24 25 + + + + = 0. 2 3 4 1+r (1 + r) (1 + r) (1 + r) (1 + r)5

Newton’s Method (Sec. 3.2 of [Smith and Minton (2002)]) can be used to approximate the solution r∗ ≈ 0.047. 1.7

Continuous Income Streams

The treatment of interest, present value and future value has focused on discrete sums of money paid or received at distinct times spread throughout an interval. A large company may be receiving thousands or even hundreds of thousands of payments from customers each day. With income being received all the time, it is preferable to think of the payments as a continuous income stream rather than as a sequence of distinct payments. Other situations in which it is natural to think of a continuous income stream could be the owner of an oil well. The well produces oil continuously and thus income is generated continuously. In this section we

10:56:51.

May 25, 2012

16

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

will develop the means to determine the present value and future value of continuous income streams. Suppose the income received per unit time is the function S(t). Over the short time interval from t to t + ∆t we can assume that S(t) is nearly constant and thus the income earned is approximately S(t)∆t. If we wish to determine the total income generated during an interval [a, b] we may create a partition of the interval a = t0 ≤ t1 ≤ · · · ≤ tn−1 ≤ tn = b and approximate the total income as n X

k=1

S(tk )(tk − tk−1 ).

In elementary calculus this quantity is known as a Riemann sum. According to the definition of the definite integral, as n → ∞ the total income is Stot =

Z

b

S(t) dt.

a

A Riemann sum can be used to determine the present value of the income stream. Assuming that the continuously compounded interest rate is r, the present value at time t = 0 of the income S(t)∆t is e−rt S(t)∆t. Therefore, the present value of the income stream S(t) over the interval [0, T ] is P =

Z

T

e−rt S(t) dt.

(1.16)

0

Similarly, the future value at t = T of the income stream is A=e

rT

Z

T

e 0

−rt

S(t) dt =

Z

T

er(T −t) S(t) dt.

(1.17)

0

Example 1.9 Suppose the slot machine floor of a new casino is expected to bring in $30, 000 per day. What is the present value of the first year’s slot machine revenue assuming the continuously compounded annual interest rate is 3.55%?

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Theory of Interest

BC8495/Chp. 1

17

Using Eq. (1.16) we have Z 1 P = (30000)(365)e−0.0355t dt 0

1 (30000)(365) −0.0355t = e ≈ 10, 757, 917.19. −0.0355 0

The formulas for present value and future value in Eq. (1.16) and (1.17) can be generalized further by assuming the interest rate is time dependent, though in these cases the definite integral may have to be approximated by some numerical method. 1.8

Exercises

(1) Suppose that $3659 is deposited in a savings account which earns 6.5% simple interest. What is the amount due after five years? (2) Suppose that $3993 is deposited in an account which earns 4.3% interest. What is the compound amount after two years if the interest is compounded (a) (b) (c) (d)

monthly? weekly? daily? continuously?

(3) Suppose $3750 is invested today. Find the amount due in 8 years if the interest rate is (a) (b) (c) (d)

1.5% simple annual interest, 1.5% effective annual compound interest, 0.75% six-month interest compounded every six months, 0.375% three-month interest compounded every three months.

(4) Find the effective annual interest rate which is equivalent to 8% interest compounded quarterly. (5) You are preparing to open a bank which will accept deposits into savings accounts and which will pay interest compounded monthly. In order to be competitive you must meet or exceed the interest paid by another bank which pays 5.25% compounded daily. What is the minimum interest rate you can pay and remain competitive? (6) Suppose you have $1000 to deposit in one of two types of savings accounts. One account pays interest at an annual rate of 4.75% com-

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

18

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

pounded daily, while the other pays interest at an annual rate of 4.75% compounded continuously. How long would it take for the compound amounts to differ by $1? (7) Many textbooks determine the formula for continuously compounded interest through an argument which avoids the use of l’Hˆopital’s Rule (for example [Goldstein et al. (1999)]). Beginning with Eq. (1.5) let h = r/n. Then  r nt P 1+ = P (1 + h)(1/h)rt n

and we can focus on finding the limh→0 (1 + h)1/h . Show that (1 + h)1/h = e(1/h) ln(1+h) and take the limit of both sides as h → 0. Hint: you can use the definition of the derivative in the exponent on the right-hand side. (8) Which of the two investments described below is preferable? Assume the first payment will take place exactly one year from now and further payments are spaced one year apart. Assume the continually compounded annual interest rate is 2.75%. Year Investment A Investment B

1 200 198

2 211 205

3 198 211

4 205 200

(9) Suppose you wish to buy a house costing $200000. You will put a down payment of 20% of the purchase price and borrow the rest from a bank for 30 years at a fixed interest rate r compounded monthly. If you wish your monthly mortgage payment to be $1500 or less, what is the maximum annual interest rate for the mortgage loan? (10) If the effective annual interest rate is 5.05% and the rate of inflation is 2.02%, find the nominal annual real rate of interest compounded quarterly. (11) Confirm by differentiation that i Rt Rt d h − R t r(s) ds e 0 A(t) = e− 0 r(s) ds A0 (t) − r(t)e− 0 r(s) ds A(t). dt

(12) Use the Mean Value Theorem (p. 235 of [Stewart (1999)]) to show the rate of return defined by the root of the function in Eq. (1.15) is unique.

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

19

The Theory of Interest

(13) Suppose for an investment of $10000 you will receive payments at the end of each of the next four years in the amounts {2000, 3000, 4000, 3000}. What is the rate of return per year? (14) Suppose you have the choice of investing $1000 in just one of two ways. Each investment will pay you an amount listed in the table below at the end of each year for the next five years. Year Investment A Investment B

1 225 220

2 215 225

3 250 250

4 225 250

5 205 210

(a) Using the present value of the investment to make the decision, which investment would you choose? Assume the annual interest rate is 4.33%. (b) Using the rate of return per year of the investment to make the decision, which investment would you choose? (15) Over the next three years an oil well will produce income at a rate of 50, 000e−0.01t. If the continuous compounded interest rate is 4.25%, what is the present value of the income to be generated by the oil well? (16) In six years a company must pay a fine of $1, 000, 000. The continuously compounded interest rate is 2.49%. At what continuous and constant rate must the company invest money so that the fine can be paid? (17) Suppose Alice puts $12,000 into a savings account that pays an effective annual interest of 3% compounded annually for 15 years. The interest is credited to her account at the end of each year. If Alice withdraws any money from her account during the first 10 ten years there will be a penalty of 5% of the withdrawal amount. To help pay for the education of her son, Alice withdraws T from her account at the end of years 8, 9, 10, and 11. The balance of her account at the end of the 15th year is $12,000. Find the value of T . (18) A homeowner receives a property tax bill on July 1 in the amount of $4500. There are two schedules of payment described on the bill. The full amount minus 2% can be paid by August 31, or $1500 can be paid on each of August 31, October 31, and December 31. If the homeowner can invest $4500 in a savings account earning effective annual interest at rate r compounded monthly, what is the minimum value of r at which the homeowner would prefer the “three equal payments” plan?

10:56:51.

May 25, 2012

20

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 1

An Undergraduate Introduction to Financial Mathematics

(19) Gail has $1500 to invest on July 1. She decides to invest in a Treasury Bill. From her perspective a Treasury Bill is like a loan to the government that will be paid back in one lump sum (including principal and interest) at a specified time in the future. Gail has two options to consider: (a) She can buy a 6-month Treasury Bill which will pay her $1600 on December 31 and then she can invest that amount in a savings account earning simple interest at rate r until June 30 of the following year. (b) She can buy for $1450 a 1-year Treasury Bill which will pay her $1600 on June 30 and with the remaining $50 she can open a savings account which will earn interest at rate r compounded semiannually. If the two options have the same present values, find the interest rate r. (20) Helen thinks that interest rates will rise over the next five years according to the function r(t) = 0.04 + for 0 ≤ t ≤ 5.

0.005t t+1

(a) What is the average annual compound rate for 0 ≤ t ≤ 5? (b) What is the effective annual interest rate for the third year 2 ≤ t ≤ 3? (c) If the amount due at time t = 5 is $1750, what is its present value at time t = 1?

10:56:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

Chapter 2

Discrete Probability

Since the number and interactions of forces driving the values of investments are so large and complex, development of a deterministic mathematical model of a market is likely to be impossible. In this book a probabilistic or stochastic model of a market will be developed instead. This chapter presents some elementary concepts of probability and statistics. Here the reader will find explanations of discrete events and their outcomes. A discrete outcome can take on only one value from a list of a finite (or countable) number of values. For example the outcome of a roll of a fair die can be only one of the six values in the set {1, 2, 3, 4, 5, 6}. No one ever rolls a die and discovers the outcome to be π for example. Basic methods for determining the probabilities of outcomes will be presented. The concept of the random variable, a numerical quantity whose value is not known until an experiment is conducted, will be explained. There are many different kinds of discrete random variables, but one that frequently arises in financial mathematics is the binomial random variable. While statistics is a field of study unto itself, two important descriptive statistics will be introduced in this chapter, expected value (or mean) and standard deviation. The expected value provides a number which is representative of typical values of a random variable. The standard deviation is a number which provides a measure related to the width of an interval centered at the mean into which values of the random variable are likely to fall. As will be seen when discussing specific experiments, the standard deviation measures the degree to which values of the random variable are “spread out” around the mean.

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

22

2.1

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

Events and Probabilities

To the layman an event is something that happens. To the statistician, an event is an outcome or set of outcomes of an experiment. This brings up the question of what is an experiment? For our purposes an experiment will be any activity that generates an observable outcome. Some simple examples of experiments include flipping a coin, rolling a pair of dice, and drawing cards from a deck. An outcome of each of these experiments could be “heads”, 7, or the ace of hearts respectively. For the example of the coin flip or drawing a card from the deck, the example outcomes given can be thought of as “atomic” in the sense that they cannot be further broken down into simpler events. The outcome of achieving a 7 on a roll of a pair of dice could be thought of as consisting of a pair of outcomes, one for each die. For example the 7 could be the result of 2 on the first die and 5 on the second. Having a flipped coin land heads up cannot be similarly decomposed. An event can also be thought of as a collection of outcomes rather than just a single outcome. For example the experiment of drawing a card from a standard deck, the events could be segregated into hearts, diamonds, spades, or clubs depending on the suit of the card drawn. Then any of the atomic events 2, 3, . . . , 10, jack, queen, king, or ace of hearts would be a “heart” event for the experiment of drawing a card and observing its suit. In this chapter the outcomes of experiments will be thought of as discrete in the sense that the outcomes will be from a set whose members are isolated from each other by gaps. The discreteness of a coin flip, a roll of a pair of dice, and card draw are apparent due to the condition that there is no outcome between “heads” and “tails”, or between 6 and 7, or between the two of clubs and the three of clubs respectively. Also in this chapter the number of different outcomes of an experiment will be either finite or countable (meaning that the outcomes can be put into one-to-one correspondence with a subset of the natural numbers). The probability of an event is a real number measuring the likelihood of that event occurring as the outcome of an experiment. To begin the more formal study of events and probabilities, let the symbol A represent an event. The probability of event A will be denoted P (A). By convention, probabilities are always real numbers in the interval [0, 1], that is, 0 ≤ P (A) ≤ 1. If A is an event for which P (A) = 0, then A is said to be an impossible event. If P (A) = 1, then A is said to be a certain event. Impossible events never occur, while certain events always occur. Events

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Discrete Probability

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

23

with probabilities closer to 1 are more likely to occur than events whose probabilities are closer to 0. There are two approaches to assigning a probability to an event, the classical approach and the empirical approach. Adopting the empirical approach requires an investigator to conduct (or at least simulate) the experiment N times (where N is usually taken to be as large as practical). During the N repetitions of the experiment the investigator counts the number of times that event A occurred. Suppose this number is x. Then the probability of event A is estimated to be P (A) = x/N . The classical approach is a more theoretical exercise. The investigator must consider the experiment carefully and determine the total number of different outcomes of the experiment (call this number M ), assume that each outcome is equally likely, and then determine the number of outcomes among the total in which event A occurs (suppose this number is y). The probability of event A is then assigned the value P (A) = y/M . When the assumption that each outcome is equally likely is true, the two methods closely agree, especially when N is very large. Some experiments involve events which can be thought of as the result of two or more outcomes occurring simultaneously. For example, suppose a red coin and a green coin will be flipped. One compound outcome of the experiment is the red coin lands on “heads” and the green coin lands on “heads” also. The next section contains some simple rules for handling the probabilities of these compound events.

2.2

Addition Rule

Suppose A and B are two events which may occur as a result of conducting an experiment. An investigator may wish to know the probability that A or B occurs. Symbolically this would be represented as P (A ∨ B). If an investigator rolls a pair of fair dice they may want to know the probability that a total of 2 or 12 results. Let event A be the outcome of 2 and B be the outcome of 12. Since P (A) = 1/36 and P (B) = 1/36 and the two events are mutually exclusive, that is, they cannot both simultaneously occur, P (A ∨ B) = P (A) + P (B) = 1/18. Suppose instead that the investigator wants to know the probability that a total of less than 6 or an odd total results. We can let event A be the outcome of a total less than 6 (that is, a total of 2, 3, 4, or 5) and let event B be the outcome of an odd total (specifically 3, 5, 7, 9, or 11). We see this time that events A an B are not

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

24

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

mutually exclusive, there are outcomes which overlap both events, namely the odd numbers 3 and 5 are less than 6. To adjust the calculation the probabilities of the non-exclusive events should be counted only once. Thus P (A ∨ B) = P (A) + P (B) − P (A ∧ B) where P (A ∧ B) is the probability that one of the outcomes in the overlapping non-exclusive set of outcomes occurs. Hence       1 1 1 1 1 1 1 1 1 1 1 P (A ∨ B) = + + + + + + + + − + 36 18 12 9 18 9 6 9 18 18 9 11 = . 18 Thus the calculation of the probability of event A or event B occurring is different depending on whether A and B are mutually exclusive. The concept outlined above is known as the Addition Rule for Probabilities and can be stated in the form of a theorem. Theorem 2.1 (Addition Rule) For events A and B, the probability of A or B occurring is P (A ∨ B) = P (A) + P (B) − P (A ∧ B) .

(2.1)

If A and B are mutually exclusive events then P (A ∧ B) = 0 and the Addition Rule simplifies to P (A ∨ B) = P (A) + P (B) . Determining the probability of the occurrence of A or B rests on determining the probability that both A and B occur. This topic is explored in the next section. 2.3

Conditional Probability and Multiplication Rule

During the past decade a very famous puzzle involving probability has come to be known as the “Monty Hall Problem”. This paradox of probability was published in a different but equivalent form in Martin Gardner’s “Mathematical Games” feature of Scientific American [Gardner (1959)] in 1959 and in the American Statistician [Selvin (1975)] in 1975. In 1990 it appeared in its present form in the “Ask Marilyn” column of Parade Magazine [vos Savant (1990)]. A game show host hides a prize behind one of three doors. A contestant must guess which door hides the prize. First, the

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Discrete Probability

BC8495/Chp. 2

25

contestant announces the door they have chosen. The host will then open one of the two doors, not chosen, in order to reveal the prize is not behind it. The host then tells the contestant they may keep their original choice or switch to the other unopened door. Should the contestant switch doors? At first glance when faced with two identical unopened doors, it may seem that there is no advantage to switching doors; however, if the contestant switches they will win with probability 2/3. When the contestant makes the first choice they have a 1/3 chance of being correct and a 2/3 chance of being incorrect. When the host reveals the non-winning, unchosen door, the contestant’s first choice still has a 1/3 chance of being correct, but now the unchosen unopened door has a 2/3 probability of being correct, so the contestant should switch. A more detailed explanation of the reason for switching doors is given in [Barrow (2008), Chap. 30]. This example illustrates the concept known as conditional probability. Essentially the decision the contestant faces is “given that I have seen that one of the doors I did not choose is not the winning door, should I alter my choice?” The probability that one event occurs given that another event has occurred is called conditional probability. The probability that event A occurs given that event B has occurred is denoted P (A|B). One of the classical thought experiments of discrete probability involves selecting balls from an urn. Suppose an urn contains 20 balls, 6 of which are blue and the remaining 14 are green. Two balls will be drawn, the second will be drawn without replacing the first. The question “what is the probability that the second ball is green, given that the first ball was green?” could be asked. The answer to this question will motivate the statement of the multiplication rule of probability. One approach to the answer involves determining the probability that when two balls are drawn without replacement they are both green. The probability that both selections are green would be the number of two green ball outcomes divided by the total number of outcomes. There are 20 candidates for the first ball selected and there are 19 candidates for the second ball selected. Thus the total number of outcomes is 380. Of those outcomes (14)(13) = 182 are both green balls. Thus the probability that both balls are green is 182/380 = 91/190. The reader may be asking what this situation has to do with the question originally posed. The outcome in which both balls are green is a subset of all the outcomes in which the first ball is green. Consider the diagram in Fig. 2.1. Thus the probability that both balls are green is the product of

10:57:00.

May 25, 2012

26

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

1 Blue

1 Green

+

1 Green

1 Blue

2 Blue

2 Green

Fig. 2.1 The sets of outcomes of drawing one or two balls from an urn containing blue and green balls.

the probability that the first ball is green multiplied by the probability the second ball is green. Let event A be the set of outcomes in which the first ball is green and event B be the set of outcomes in which the second ball is green. Numerically P (A) = 7/10 and symbolically P (A ∧ B) = P (A) P (B|A) Thus P (B|A) = P (A ∧ B) /P (A) = (91/190)/(7/10) = 13/19. The concept illustrated above is known as the Multiplication Rule for Probabilities and can be stated in the form of a theorem. Theorem 2.2 (Multiplication Rule) For events A and B, the probability of A and B occurring is P (A ∧ B) = P (A) P (B|A) .

(2.2)

Equation (2.2) can be used to find P (B|A) directly P (B|A) =

P (A ∧ B) . P (A)

This expression is meaningful only when P (A) > 0. Example 2.1 One type of roulette wheel, known as the American type, has 38 potential outcomes represented by the integers 1 through 36 and two special outcomes 0 and 00. The positive integers are placed on alternating

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Discrete Probability

BC8495/Chp. 2

27

red and black backgrounds while 0 and 00 are on green backgrounds. What is the probability that the outcome is less than 10 and more than 3 given that the outcome is an even number? Let event A be the set of outcomes in which the number is even. P (A) = 10/19 if 0 and 00 are treated as even numbers. Let B be the set of outcomes in which the number is greater than 3 and less than 10. Then P (A ∧ B) = 3/38 and P (B|A) =

3/38 = 3/20. 10/19

To expand on the previous example, suppose the roulette wheel will be spun twice. One could ask what is the probability that both spins have a red outcome. If event A is the outcome of red on the first spin and event B is the outcome of red on the second spin, then we have as before P (A ∧ B) = P (A) P (B|A). However there is no reason to believe that the wheel somehow “remembers” the outcome of the first spin while it is being spun the second time. The first outcome has no effect on the second outcome. In any experiment, if event A has no effect on event B then A and B are said to be independent. In this situation P (B|A) = P (B). Thus for independent events the Multiplication Rule can be modified to P (A ∧ B) = P (A) P (B) . Therefore the probability that both spins will have red outcomes is P (A ∧ B) = (9/19)(9/19) = 81/361. 2.4

Random Variables and Probability Distributions

The outcome of an experiment is not known until after the experiment is performed. For example the number of people who vote in an election is not known until the election is concluded. In a more formal sense we can describe a random variable as a function which maps the set of outcomes of an experiment to some subset of the real numbers. In the election example (assuming the number of registered voters is N and that there will be no fraudulent voting) the sample space of outcomes of voter turnout is the set S = {0, 1, 2, . . . , N }. Symbolically a random variable for the voter turnout is the function X : S → R. Often X is thought of as the eventual numeric result of the experiment. A probability distribution (or probability function) is a function

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

28

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

which assigns a probability to each element in the sample space of outcomes of an experiment. If S is the set of outcomes of an experiment and f is the associated probability function, then f maps each element in S to a unique real number in the interval [0, 1]. If x is a potential outcome of an experiment with sample space S then f (x) = P (X = x), in other words f (x) is the probability that x occurs as the outcome of the experiment. Since a probability function maps an outcome to a probability then the following two characteristics are true of the function. (1) If xi is one of the N outcomes of an experiment then 0 ≤ f (xi ) ≤ 1. (2) The sum of the values of the probability function is unity, i.e. 1=

N X

f (xi ).

i=1

Example 2.2 Consider a family with four children. The random variable X will represent the number of children who are male. The sample space for this experiment (having four children and counting the number of boys) is the set S = {0, 1, 2, 3, 4}. The gender of each child is independent from the genders of their siblings, so assuming that the probability of a child being male is 1/2 then the 16 events shown in table 2.1 are equally likely. There is one outcome in which there are no male children, thus f (0) = P (X = 0) = 1/16. There are four cases in which there is a single male child and hence f (1) = 1/4. The reader can readily determine from the table that f (2) = 3/8, f (3) = 1/4, and f (4) = 1/16. Several common types of random variables and their associated probability distributions will be important to the study of financial mathematics. The binomial random variable will be discussed in the next section. It will be seen to be related to the Bernoulli random variable which takes on only one of two possible values, often thought of as true or false (or sometimes as success and failure). It is mathematically convenient to designate the outcomes as 0 and 1. The probability function of a Bernoulli random variable is particularly simple, f (1) = P (X = 1) = p where 0 ≤ p ≤ 1 and f (0) = 1 − p. 2.5

Binomial Random Variables

Returning to the last example of the previous section, we can think of the births of four children as four independent events. The gender of one child

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Discrete Probability

BC8495/Chp. 2

29

Table 2.1 The genders of four children born to the same set of parents. 1 B G B B B G G G B B B B G G G G

Child 2 3 B B B B G B B G B B G B B G B B G G G B B G G G B G G B G G G G

4 B B B B G B B G B G G G G G B G

in no way influences the genders of children born before or after. Since the gender of a child can take on only one of two values it is possible to think of the birth of each child as a Bernoulli event. The probability of having a male or female child does not change between births. Thus the experiment of producing four children in a family is the same as repeating the experiment of having a single child four times or, repeating a Bernoulli experiment four times. This is the idea of the binomial random variable defined next. A binomial random variable X is the number of successful outcomes out of n independent Bernoulli random trials. A binomial random variable is parametrized by the number of repetitions of the Bernoulli experiment (referred to as trials from here on) and by the probability of success on a single trial. If the number of trials is n and the probability of success on a single trial is p, then the set of possible outcomes of the binomial experiment is the set {0, 1, . . . , n}. The number of combinations of x successes out of n trials is   n n! = . x x!(n − x)! The probability of x successes out of n independent trials in a specified combination is, according to the Multiplication Rule, px (1 − p)n−x . Since 10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

30

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

the various combinations are mutually exclusive, by the Addition Rule the probability of x successes out of n trials is given by the function   n x n! P (X = x) = p (1 − p)n−x = px (1 − p)n−x (2.3) x x!(n − x)! Thus if the probability of an individual child being born male or female is 1/2, the probability that a family with four children will have two female children is  2  4−2 3 4! 1 1 P (X = 2) = = . 2! (4 − 2)! 2 2 8 This result agrees with the result of the more cumbersome method used to determine this probability in the previous section. Example 2.3 The probability that a computer memory chip is defective is 0.02. A SIMM (single in-line memory module) contains 16 chips for data storage and a 17th chip for error correction. The SIMM can operate correctly if one chip is defective, but not if two or more are defective. The probability that a SIMM will not function is P (X ≥ 2) = 2.6

17   X 17 (0.02)x (0.98)17−x ≈ 0.044578. x x=2

Expected Value

When faced with experimental data, summary statistics are often useful for making sense of the data. In this context “statistics” refers to numbers which can be calculated from the data rather than the means and algorithms by which these numbers are calculated. In financial mathematics the statistical needs are somewhat more specialized than in a general purpose course in statistics. Here we wish to answer the hypothetical question, “if an experiment was to be performed an infinite number of times, what would be the typical outcome?” Thus we will introduce only the statistical concepts to be used later in this text. A reader interested in a broader, deeper, and more rigorous background in statistics should consult one of the many textbooks devoted to the subject for example [Ross (2006)]. To take an example, if a fair die was rolled an infinite number of times, what would be the typical result? The notion to be explored in this section is that of expected value. In some ways expected value is synonymous

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Discrete Probability

BC8495/Chp. 2

31

with the mean or average of a list of numerical values; however, it can differ in at least two important ways. First, the expected value usually refers to the typical value of a random variable whose outcomes are not necessarily equally likely whereas the mean of a list of data treats each observation as equally likely. Second, the expected value of a random variable is the typical outcome of an experiment performed an infinite number of times whereas the statistical mean is calculated based on a finite collection of observations of the outcome of an experiment. If X is a discrete random variable with probability distribution P (X) then the expected value of X is denoted E [X] and defined as X E [X] = (X · P (X)). (2.4) X

It is understood that the summation is taken over all values that X may assume. In the case that X takes on only a finite number of values with nonzero probability, then this sum is well-defined. If X may assume an infinite number of values with probabilities greater than zero, we will assume that the sum converges. Since each value of X is multiplied by its corresponding probability, the expected value of X is a weighted average of the variable X. Returning to the question posed in the previous paragraph as to the typical outcome achieved when rolling a fair die an infinite number of times, we may determine this number from the formula for expected value. Since X ∈ {1, 2, 3, 4, 5, 6} and P (X) = 1/6 for all possible values of X, then the expected value of X is E [X] =

6 6 X X 1 X 1 (6)(7) 7 = X= · = . 6 6 6 2 2

X=1

X=1

Thus the average outcome of rolling a fair die is 3.5. Example 2.4 Let random variable X represent the number of female children in a family of four children. Assuming that births of males and females are equally likely and that all births are independent events, what is the E [X]? The sample space of X is the set {0, 1, 2, 3, 4}. Using the binomial probability formula P (0) = 1/16, P (1) = 1/4, P (2) = 3/8, P (3) = 1/4, and P (4) = 1/16, we have 1 1 3 1 1 +1· +2· +3· +4· = 2. 16 4 8 4 16 In families having four children, typically there are two female children and consequently two male children. E [X] = 0 ·

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

32

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

The notion of the expected value of a random variable X can be extended to the expected value of a function of X. Thus we say that if F is a function applied to X, then X E [F (X)] = F (X)P (X) . X

When the function F is merely multiplication by a constant then the expected value takes on a simple form. Theorem 2.3 If X is a random variable and a is a constant, then E [aX] = aE [X]. Proof.

By the definition of expected value X X E [aX] = ((aX) · P (X)) = a (X · P (X)) = aE [X] . X

X



Later in this work sums of random variables will become important. Thus some attention must be given to the expected value of the sum of random variables. However, this requires that the probability of two or more random variables be considered simultaneously. The reader should already be familiar with one example of this situation, namely the rolling of a pair of dice. Suppose that the two dice can be distinguished from one another (imagine that one of them is red while the other is green). Let X be the random variable denoting the outcome of the green die while Y is a random variable denoting the outcome of the red die. If the experiment to be performed is rolling the pair of dice and considering the total of the upward faces then the random variable denoting the outcome of this experiment is X + Y . This naturally leads us to the issue of describing the probabilities associated with various values of the random variable X + Y . The joint probability function is denoted P (X, Y ) and we will understand it to mean P (X, Y ) = P (X ∧ Y ). Thus P (1, 3) symbolizes the probability that the outcome of the red die is 1 while the outcome of the green die is 3. If the individual dice are independent then P (1, 3) = P (1) P (3) = 1/36 according to the multiplication rule. A couple of additional comments are in order. First, joint probabilities exist even for random events which are not independent. Second, realize that in general P (X + Y ) 6= P (X) + P (Y ). This is an abuse of notation, but is not likely to cause confusion in what follows. P (X + Y ) refers to the probability of the sum X + Y which depends on the joint probabilities of X and Y . P (X) and P (Y ) refer respectively to the individual probabilities of random variable X and Y . The following is true,

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Discrete Probability

BC8495/Chp. 2

33

if we wish to know the probability that the sum of the discrete random variables X and Y is m then by using the addition rule for probabilities X P (X + Y = m) = P (X, Y ) . X+Y =m

The summation is taken over all combinations of X and Y such that X + Y = m. Returning to the dice example introduced earlier in the paragraph we see that the probability that the sum of the dice is 4 is X P (X + Y = 4) = P (X, Y ) X+Y =4

= P (1, 3) + P (2, 2) + P (3, 1) 1 1 1 1 = + + = . 36 36 36 12

The joint probability distribution of a pair of random variables possesses many of the same properties that the probability distribution of a single random variable possesses. For example 0 ≤ P (X, Y ) ≤ 1 for all X and Y in the discrete sample space. It is also true that XX XX P (X, Y ) = P (X, Y ) = 1. X

Y

Y

X

An important property will be used in the proof of the next theorem. The sum of the joint probability of X and Y where Y is allowed to take on each of its possible values is called the marginal probability of X. Without confusion we will denote the marginal probability of X as P (X) and realize that X P (X) = P (X, Y ) . Y

Similarly the marginal probability of Y is denoted P (Y ) and defined as X P (Y ) = P (X, Y ) . X

Conveniently the expected value of a sum of random variables is the sum of the expected values of the random variables. This notion is made more precise in the following theorem. Theorem 2.4

If X1 , X2 , . . . , Xk are random variables then

E [X1 + X2 + · · · Xk ] = E [X1 ] + E [X2 ] + · · · + E [Xk ] . 10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

34

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

Proof.

If k = 1 then the proposition is certainly true. If k = 2 then

E [X1 + X2 ] =

X

((X1 + X2 )P (X1 , X2 ))

X1 ,X2

=

XX

((X1 + X2 )P (X1 , X2 ))

X1 X2

=

XX

X1 P (X1 , X2 ) +

X

X1

X1

=

X

X

P (X1 , X2 ) +

X2

X

X2

X2

X1 P (X1 ) +

X1

X2 P (X1 , X2 )

X2 X1

X1 X2

=

XX

X

X

P (X1 , X2 )

X1

X2 P (X2 )

X2

= E [X1 ] + E [X2 ] .

For a finite value of k > 2 the result is true by induction. Suppose the result is true for n < k where k > 2, then E [X1 + · · · + Xk−1 + Xk ] = E [X1 + · · · + Xk−1 ] + E [Xk ]

= E [X1 ] + · · · + E [Xk−1 ] + E [Xk ]

The last step is true by the induction hypothesis.



We can use Theorem 2.4 to determine the expected value of a binomial random variable. Along the way we will also find the expected value of a Bernoulli random variable. Suppose n trials of a Bernoulli experiment will be conducted for which the probability of success on a single trial is 0 ≤ p ≤ 1. Random variable X represents the number of successes out of n trials. By assumption the trials are independent of one another and the outcomes are mutually exclusive. The result of the binomial experiment can be thought of as the sum of the results of n Bernoulli experiments. Let the random variable Xi be the number of successes of the ith Bernoulli trial, then E [X] = E [X1 + · · · + Xn ] = E [X1 ] + · · · + E [Xn ] = p + · · · + p = np. If functions are applied to random variables a corollary to Theorem 2.4 can be stated. Corollary 2.1

Let X1 , X2 , . . . , Xk be random variables and let Fi be a

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

35

Discrete Probability

function defined on Xi for i = 1, 2, . . . , k then E [F1 (X1 ) + F2 (X2 ) + · · · + Fk (Xk )]

= E [F1 (X1 )] + E [F2 (X2 )] + · · · + E [Fk (Xk )] .

Proof. We will prove this result for the case when k = 2 and leave it to the reader to apply the principle of mathematical induction to extend the result to the case when k > 2. E [F1 (X1 ) + F2 (X2 )] =

X

((F1 (X1 ) + F2 (X2 ))P (X1 , X2 ))

X1 ,X2

=

X

F1 (X1 )

X1

+

X

X X1

P (X1 , X2 )

X2

X2

=

X

F2 (X2 )

X

P (X1 , X2 )

X1

F1 (X1 )P (X1 ) +

X

F2 (X2 )P (X2 )

X2

= E [F1 (X1 )] + E [F2 (X2 )]  Later we will have need to calculate the expected value of a product of random variables. This situation is not as straightforward as the case of a sum of random variables. Theorem 2.5 ables, then

Let X1 , X2 , . . . , Xk be pairwise independent random vari-

E [X1 X2 · · · Xk ] = E [X1 ] E [X2 ] · · · E [Xk ] . Proof. Naturally we see that when k = 1 the theorem is true. Next we will consider the case when k = 2. Let X1 and X2 be independent random variables with joint probability distribution P (X1 , X2 ). Since the random variables are assumed to be independent then P (X1 , X2 ) = P (X1 ) P (X2 ). Once again we are lax in our use of notation, since in the previous equation the symbol P is used in three senses ((1) the joint probability distribution of X1 and X2 , (2) the probability distribution of X1 , and (3) the probability distribution of X2 ); however, there is little chance of confusion in this

10:57:00.

May 25, 2012

36

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

elementary proof. E [X1 X2 ] =

X

X1 X2 P (X1 , X2 )

X1 ,X2

=

XX

X1 X2 P (X1 ) P (X2 )

X1 X2

=

X

X1 P (X1 )

X1

X

X2 P (X2 )

X2

= E [X1 ] E [X2 ]

For a finite value of k > 2 the result is true by induction. Suppose the result is true for n < k where k > 2, then E [X1 · · · Xk−1 Xk ] = E [X1 · · · Xk−1 ] E [Xk ]

= E [X1 ] · · · E [Xk−1 ] E [Xk ]

The last step is true by the induction hypothesis.



A corollary to Theorem 2.5 holds for functions of pairwise independent random variables as well. Corollary 2.2 Let X1 , X2 , . . . , Xk be pairwise independent random variables and let Fi be a function defined on Xi for i = 1, 2, . . . , k then E [F1 (X1 )F2 (X2 ) · · · Fk (Xk )] = E [F1 (X1 )] E [F2 (X2 )] · · · E [Fk (Xk )] . Proof. Once again we will prove this result for the case when k = 2 and leave it to the reader to extend the result to the case when k > 2. X E [F1 (X1 )F2 (X2 )] = F1 (X1 )F2 (X2 )P (X1 , X2 ) X1 ,X2

=

X X1

F1 (X1 )P (X1 )

X

F2 (X2 )P (X2 )

X2

= E [F1 (X1 )] E [F2 (X2 )]



If the reader is interested in more properties of the expected value of sum and products of random variables, consult a textbook on probability such as [Ross (2003)]. The notion of conditional expected value follows from an understanding of conditional probability. Since P (X|Y ) is understood to mean the probability of event X given event Y , then the notation E [X|Y = y] or the equivalent, but more compact notation E [X|y] will mean the expected

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

37

Discrete Probability

value of random variable X given that random variable Y has the value y. The condition expected value can be calculated according to the formula E [X|y] =

X X

(X · P (X|Y = y)) .

(2.5)

For example if a pair of fair dice consists of one red die and one green die the expected value of their sum given that the red die shows “3” can be calculated as follows.

E [G + R|R = 3] =

6 X

G=1

((G + R) · P (G|R = 3)) =

6 1X 13 (G + 3) = 6 2 G=1

The conditional expected value E [X|Y = y] can also be thought of as a random variable itself. E [X|Y = y] =

X X

(X · P (X|Y = y)) = f (y)

The expression f (Y ) is a new random variable. Example 2.5 Suppose integer random variables (X, Y ) satisfy the inequality −5 ≤ Y ≤ X ≤ 5 and each ordered pair is equally likely. To find E [X|Y ] first note that since there are 11 potential values for Y ∈ {−5, −4, . . . , 5} and when Y = y then there are 6 − y potential values for X. Thus the set of outcomes of a random selection of (X, Y ) contains 5 X

(6 − y) = 66 elements.

y=−5

Therefore P ((X, Y )) = 1/66. The conditional probability P (X|Y = y) can be found from the marginal probability for Y .

P (X|Y = y) =

P ((X, Y )) P ((X, Y )) 1 1 = P5 = P5 = . P (Y = y) 6−y X=y P ((X, y)) X=y 1 10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

38

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

Applying the definition of conditional expectation in Eq. (2.5) yields

E [X|Y = y] =

5  X X·

X=y

=

1 6−y



5 1 X X 6−y X=y

1 30 − y 2 + y · 6−y 2 y+5 = . 2

=

Conditional expected value also shares the linearity property with the original notion of expected value (see exercise (20)). The expected value of a random variable specifies the average outcome of an infinite number of repetitions of an experiment. In the next section the notions of variance and standard deviation are introduced. They specify measures of the spread of the outcomes from the expected value.

2.7

Variance and Standard Deviation

The variance of a random variable is a measure of the spread of values of the random variable about the expected value of the random variable. The variance is defined as   V (X) = E (X − E [X])2 .

(2.6)

As the reader can see from Eq. (2.6), the variance is always non-negative. The expression X − E [X] is the signed deviation of X from its expected value. The variance may be interpreted as the average of the squared deviation of a random variable from its expected value. An alternative formula for the variance is sometimes more convenient in calculations. Theorem 2.6 Let X be a random variable, then the variance of X is   2 E X 2 − E [X] . 10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Discrete Probability

Proof.

BC8495/Chp. 2

39

By definition,   V (X) = E (X − E [X])2 i h   = E X 2 − E [2XE [X]] + E E [X]2   2 = E X 2 − 2E [X] E [X] + E [X]   = E X 2 − E [X]2 .

The third and fourth steps of this derivation made use of theorems 2.4 and 2.3 respectively.  Returning to the previous example of the hypothetical family with four children, we can now investigate the variance in the number of female children. We already know that E [X] = 2. If we make use of the result of Theorem 2.6 then   V (X) = E X 2 − E [X]2

= (02 )(1/16) + (12 )(1/4) + (22 )(3/8) + (32 )(1/4) + (42 )(1/16) − 22 = 1.

Before investigating the variance of a binomial random variable, we should determine the variance of a Bernoulli random variable. If the probability of success is 0 ≤ p ≤ 1 then according to Eq. (2.6),   V (X) = E (X − E [X])2   = E (X − p)2

= (1 − p)2 p + (0 − p)2 (1 − p) = p(1 − p).

The following theorem provides an easy formula for calculating the variance of independent random variables. Theorem 2.7 ables, then

Let X1 , X2 , . . . , Xk be pairwise independent random vari-

V (X1 + X2 + · · · + Xk ) = V (X1 ) + V (X2 ) + · · · + V (Xk ) . 10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

40

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

Proof. If k = 1 then the result is trivially true. Take the case when k = 2. By the definition of variance,   V (X1 + X2 ) = E ((X1 + X2 ) − E [X1 + X2 ])2   = E ((X1 − E [X1 ]) + (X2 − E [X2 ]))2     = E (X1 − E [X1 ])2 + E (X2 − E [X2 ])2 + 2E [(X1 − E [X1 ])(X2 − E [X2 ])]

= V (X1 ) + V (X2 ) + 2E [(X1 − E [X1 ])(X2 − E [X2 ])] . Since we are assuming that random variables X1 and X2 are independent, then by Theorem 2.5 E [(X1 − E [X1 ])(X2 − E [X2 ])] = E [X1 − E [X1 ]] E [X2 − E [X2 ]]

= (E [X1 ] − E [X1 ])(E [X2 ] − E [X2 ])

= 0, and thus

V (X1 + X2 ) = V (X1 ) + V (X2 ) . The result can be extended to any finite value of k by induction. Suppose the result has been shown true for n < k with k > 2. Then V (X1 + · · · + Xk−1 + Xk ) = V (X1 + · · · + Xk−1 ) + V (Xk )

= V (X1 ) + · · · + V (Xk−1 ) + V (Xk )

where the last equality is justified by the induction hypothesis.



Readers should think carefully about the validity of the claim that X1 − E [X1 ] and X2 − E [X2 ] are independent in light of the assumption that X1 and X2 are independent. Example 2.6 Suppose a binomial experiment is characterized by n independent repetitions of a Bernoulli trial for which the probability of success on a single trial is 0 ≤ p ≤ 1. Random variable X denotes the total number 10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Discrete Probability

BC8495/Chp. 2

41

of successes accrued over the n trials.   n X V (X) = V  Xj  j=1

= =

n X j=1 n X j=1

V (Xj )

(since trials are independent)

p(1 − p)

= np(1 − p) So far we have made no mention of the other topic in the heading for this section, namely standard deviation. There is little more that must be said since by definition the standard deviation is the square root of the variance. Standard deviation of a random variable X is denoted by σ (X) and thus σ (X) =

p V (X).

The reader may also be left wondering about the possible existence of a result regarding the variance of a product of random variables. The general result for the variance of a product would take us too far afield, but we can state and prove a result for the product of pairwise independent random variables. Theorem 2.8 ables, then

Let X1 , X2 , . . . , Xk be pairwise independent random vari-

      2 V (X1 X2 · · · Xk ) = E X12 E X22 · · · E Xk2 − (E [X1 ] E [X2 ] · · · E [Xk ]) .

Proof. The case when k = 1 follows from Theorem 2.6. Take the case when k > 1.   V (X1 X2 · · · Xk ) = E (X1 X2 · · · Xk )2 − (E [X1 X2 · · · Xk ])2   = E X12 X22 · · · Xk2 − (E [X1 ] E [X2 ] · · · E [Xk ])2       = E X12 E X22 · · · E Xk2 − (E [X1 ] E [X2 ] · · · E [Xk ])2

The last equation holds as a result of Corollary 2.2.

10:57:00.



May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

42

2.8

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

Exercises

(1) Suppose the four sides of a regular tetrahedron are labeled 1 through 4. If the tetrahedron is rolled like a die, what is the probability of it landing on 3? (2) Use the classical approach and the assumption of a fair die to find the probabilities of the outcomes obtained by rolling a pair of dice and summing the dots shown on the the upward faces. (3) If the probability that a batter strikes out in the first inning of a baseball game is 1/3 and the probability that the batter strikes out in the fifth inning is 1/4, and the probability that the batter strikes out in both innings is 1/10, then what is the probability that the batter strikes out in either inning? (4) Part of a well-known puzzle involves three people entering a room. As each person enters, at random either a red or a blue hat is placed on the person’s head. The probability that an individual receives a red hat is 1/2. No person can see the color of their own hat, but they can see the color of the other two persons’ hats. The three will split a prize if at least one person guesses the color of their own hat correctly and no one guesses incorrectly. A person may decide to pass rather than to guess. The three people are not allowed to confer with one another once the hats have been placed on their heads, but they are allowed to agree on a strategy prior to entering the room. At the risk of spoiling the puzzle, one strategy the players may follow instructs a player to pass if they see the other two persons wearing mis-matched hats and to guess the opposite color if their friends are wearing matching hats. Why is this a good strategy and what is the probability of winning the game? (5) Suppose cards will be drawn without replacement from a standard 52-card deck. What is the probability that the first two cards will be aces? (6) Suppose cards will be drawn without replacement from a standard 52-card deck. What is the probability that the second card drawn will be an ace and that the first card was not an ace? (7) Suppose cards will be drawn without replacement from a standard 52-card deck. What is the probability that the fourth card drawn will be an ace given that the first three cards drawn were all aces? (8) Suppose cards will be drawn without replacement from a standard 52-card deck. On which draw is the card mostly likely to be the first

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Discrete Probability

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

43

ace drawn? (9) A biased coin has a probability p of landing on heads and probability of 1 − p of landing on tails. An experiment is conducted in which the coin is repeatedly flipped until it lands on heads. (a) What is the sample space of outcomes of this experiment? (b) What is the probability that the coin will be flipped exactly 5 times? (10) Suppose that four DVDs are removed from their cases and then placed back into the empty cases in random order. What is the probability that at least one of the DVDs is in the correct case? (11) On the last 100 spins of an American style roulette wheel, the outcome has been black. What is the probability of the outcome being black on the 101st spin? (12) On the last 5000 spins of an American style roulette wheel, the outcome has been 00. What is the probability of the outcome being 00 on the 5001st spin? (13) Suppose that a random variable X has a probability distribution function f so that f (x) = c/x for x = 1, 2, . . . , 10 and is zero otherwise. Find the appropriate value of the constant c. (14) Suppose that a box contains 15 black balls and 5 white balls. Three balls will be selected without replacement from the box. Determine the probability function for the number of black balls selected. (15) Quality control for a manufacturer of integrated circuits is done by randomly selecting 25 chips from the previous days manufacturing run. Each of the 25 chips is tested. If two or more chips are faulty, then the entire run is discarded. Previously gathered evidence indicates that the defect rate for chips is 0.0016. What is the probability that a manufacturing run of chips will be discarded? (16) The probabilities of a child being born male or female are not exactly equal to 1/2. Typically there are nearly 105 live male births per 100 live female births. Determine the expected number of female children in a family of 6 total children using these birth ratios and ignoring infant mortality. (17) One version of the table game called “craps” is played by having participants roll a pair of dice. • If the player rolls 7 or 11, the player wins. • If the player rolls 2, 3, or 12, the player loses. • If the player rolls 4, 5, 6, 8, 9, or 10 then the player must keep 10:57:00.

May 25, 2012

44

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

An Undergraduate Introduction to Financial Mathematics

rolling the dice until they roll their original number or 7. If the player rolls their original number before they roll 7, they win; otherwise they lose. (a) What is the probability the player will roll a 7 or 11 on the first roll of the dice? (b) What is the probability the player will roll a 2, 3, or 12 on the first roll of the dice? (c) Suppose the player rolls a 4 on the first roll of the dice, what is the probability the player will roll another 4 before rolling a 7? (d) What is the probability the player will roll a 4 on the first roll of the dice and then win the game? (e) What is the probability the player will roll a 5 (6, 8, 9, 10) on the first roll of the dice and then win the game? (f) What is the probability that a player will win at the game of craps? (18) Suppose a standard deck of 52 cards is well shuffled and one card at a time will be drawn without replacement from the deck. What is the expected value of the first ace drawn (in other words, of the first, second, third, etc cards drawn, on average which will be the first ace drawn)? (19) Show that for constants a and b and discrete random variable X that E [aX + b] = aE [X] + b. (20) Suppose that X, Y and Z are jointly distributed discrete random variables. (a) Show that E [X + Y |Z = z] = E [X|Z = z] + E [Y |Z = z]. (b) Suppose c is a constant and show that E [cX|Y = y] c E [X|Y = y].

=

(21) For the situation described in exercise (18) determine the variance in the occurrence of the first ace drawn. (22) Show that for constants a and b and discrete random variable X that V (aX + b) = a2 V (X). (23) Suppose that X is a binomial random variable representing the number of successful trials out of M trials where the probability of success on a single trial is 0 < p < 1. Likewise suppose that Y is also a binomial random variable for which the probability of success on a single trial is also p but the number of trials is N , where M and N can be

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 2

Discrete Probability

45

different. Assume X and Y are independent and find a formula for P (X = x|X + Y = n) . (24) Suppose n people are in a group. None of the people was born during a leap year. Show that the probability that at least two of the people share a birthday (so-called “birthday buddies”) is p=1−

365! . (365 − n)! 365n

Enrollment in mathematics courses is holding steady at 30 students per class section. What is the probability of birthday buddies in such a class? (25) Consider the following three random variables. • X is a Bernoulli random variable with probability of success p and probability of failure 1 − p. • Y =1−X • Z =XY (a) (b) (c) (d)

Find the sample space of outcomes of random variable Y . Find the sample space of outcomes of random variable Z. Find P ((X = x) ∧ (Y = y)) for all possible outcomes of X and Y . Find P ((X = x) ∧ (Z = z)) for all possible outcomes of X and Z.

10:57:00.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

This page intentionally left blank

10:57:00.

46

BC8495/Chp. 2

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

Chapter 3

Normal Random Variables and Probability

Whereas in Chapter 2 random variables could take on only a finite number of values taken from a set with gaps between the values, in the present chapter, continuous random variables will be described. A continuous random variable can take on an infinite number of different values from a range without gaps. Calculus-based methods for determining the expected value and standard deviation of a continuous random variable will be described. A very useful type of continuous random variable for the study of financial mathematics is the normal random variable. We will see that when many random factors and influences come together (as in the complex situation of a financial market), the sum of the influences can be modeled by a normal random variable. Finally we will examine some stock market data and detect the presence of normal randomness in the fluctuations of stock prices. 3.1

Continuous Random Variables

Our understanding of probability must change if we are to understand the differences between discrete and continuous random variables. Suppose we consider the interval [0, 1]. If we think of only the integers contained in this interval, then only the discrete set {0, 1} need be considered. If the probability of selecting either integer from this set is 1/2 then we remain in the realm of discrete probability, random variables, and distributions. If we think of selecting, with equal likelihood, a number from the interval [0, 1] of the form k/10 where k ∈ {0, 1, . . . , 10}, then once again our approach to 1 probability remains discrete. The P (X = k/10) = 11 . Continuing in this way we see that if we consider only the real numbers in [0, 1] of the form k/n 1 where k ∈ {0, 1, . . . , n}, then P (X = k/n) = n+1 . So long as n is finite we 10:57:10.

May 25, 2012

48

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

are dealing with the familiar concept of discrete probability. What happens as n → ∞? In one sense we can think of this limiting case as the case of continuous probability. We say continuous because in the limit, the gaps between the outcomes in the sample space disappear. However, if we retain the old notion of probability, the likelihood of choosing a particular number from the interval [0, 1] becomes lim

n→∞

1 = 0. n+1

This is correct but brings to mind a paradox. Does this imply that the probability of choosing any number from [0, 1] is 0? Surely we must choose some number. What is needed is a shift in our notion of probability when dealing with continuous random variables. Instead of determining the probability that a continuous random variable equals some real number, the meaningful determination is the likelihood that the random variable lies in a set, usually an interval or finite union of intervals on the real line. These notions are defined next. A random variable X has a continuous distribution if there exists a non-negative function f : R → R such that for an interval [a, b] the Z b P (a ≤ X ≤ b) = f (x) dx. (3.1) a

The function f which is known as the probability distribution function or probability density function must, in addition to satisfying f (x) ≥ 0 on R, have the following property, Z ∞ f (x) dx = 1. (3.2) −∞

These properties are the analogues of properties of discrete random variables and their distributions. The probabilities of individual values of the random variable are non-negative. In fact, as explained above P (X = x) = 0 for a continuous random variable provided f (t) is continuous at t = x. By contrast, P (X = x) can be greater than zero for a discrete random variable. The sum, expressed now as an integral over the real number line, of the values of the probability distribution function must be one. Interpreted graphically, if X represents a continuous random variable then P (a ≤ X ≤ b) represents the area of the region bounded by the graph of the probability distribution function, the x-axis, and the lines x = a and x = b. See Fig. 3.1. The total area under the graph of the probability

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

49

Normal Random Variables and Probability

fHxL

x a

b

Fig. 3.1 An example of a probability distribution function. The shaded area represents the P (a ≤ X ≤ b).

density function is unity. Perhaps the most elementary of the continuous random variables is the uniformly distributed continuous random variable. A continuous random variable X is uniformly distributed in the interval [a, b] (with b > a) if the probability that X belongs to any subinterval of [a, b] is equal to the length of the subinterval divided by b − a. The definition of the continuous uniform random variable allows us to determine the probability density function for X. Since the length of the subinterval is proportional to the probability that X lies in the subinterval then the probability density function f (x) must be constant. Suppose this constant is k. From the property expressed in Eq. (3.2) we know that Z ∞ Z b Z b 1 dx. 1= f (x) dx = k dx = b − a −∞ a a We are assuming here that the probability distribution function vanishes outside of the interval [a, b]. This simplifies the evaluation of the improper integral. Thus the probability density function for a continuously, uniformly distributed random variable on interval [a, b] is the piecewise-defined function  1 if a ≤ x ≤ b f (x) = b−a 0 otherwise. 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

50

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

Example 3.1 Random variable X is continuously uniformly randomly distributed in the interval [−1, 12]. Find the probability that 2 ≤ X ≤ 7. Since according to the previous discussion of uniformly distributed random variables, the probability density function of X is  1 if −1 ≤ x ≤ 12 f (x) = 13 0 otherwise, we have P (2 ≤ X ≤ 7) =

Z

7

f (x) dx =

2

Z

2

7

1 5 dx = . 13 13

Now that continuous random variables and their probability distributions have been introduced we can begin to discuss formulae for determining their means and variances. 3.2

Expected Value of Continuous Random Variables

By definition the expected value or mean of a continuous random variable X with probability density function f (x) is Z ∞ E [X] = xf (x) dx. (3.3) −∞

This equation is analogous to Eq. (2.4) for the case of a discrete random variable. In the continuous case the product of the value of the random variable and its probability density function is summed (this time through the use of a definite integral) over all values of the random variable. The expected value is only meaningful in cases in which the improper integral in (3.3) converges. Example 3.2 Find the expected value of X if X is a continuously uniformly distributed random variable on the interval [−50, 75]. Using Eq. (3.3) we have Z ∞ Z 75  25 x 1 1 E [X] = dx = x dx = 752 − (−50)2 = . 125 −50 250 2 −∞ 75 − (−50)

The results which follow in this section will be similar to the results that were presented for discrete random variables. In most cases the notion of a discrete sum need only be replaced by a definite integral to justify these new results. To keep the exposition as brief as possible, many of the proofs

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Normal Random Variables and Probability

BC8495/Chp. 3

51

of results in this section will be left to the reader. The expected value of a function g of a continuously distributed random variable X which has probability distribution function f is defined as Z ∞ E [g(X)] = g(x)f (x) dx, (3.4) −∞

provided the improper integral converges absolutely, i.e., E [g(X)] is defined if and only if Z ∞ |g(x)|f (x) dx < ∞. −∞

Joint probability distributions for continuous random variables are defined similarly to those for discrete random variables. A joint probability distribution for a pair of random variables, X and Y , is a non-negative function f (x, y) for which Z ∞ Z ∞ f (x, y) dx dy = 1. −∞

−∞

Consequently the expected value of a function g : R2 → R of the two random variables X and Y is defined as Z ∞Z ∞ E [g(X, Y )] = g(x, y)f (x, y) dx dy, −∞

−∞

again, provided the integral is absolutely convergent. Before turning our attention to sums and products of continuous random variables we will define the notion of a marginal distribution. If X and Y are continuous random variables with joint distribution f (x, y) then the marginal distribution for X is defined as the function Z ∞ fX (x) = f (x, y) dy. −∞

The marginal distribution fY (y) for the random variable Y is defined similarly. Recall that for jointly distributed discrete random variables A and B we said that A and B are independent if and only if P (A ∧ B) = P (A) P (B). For continuous random variables we will use a similar definition. Two continuous random variables are independent if and only if the joint probability distribution function factors into the product of the marginal distributions of X and Y . In other words X and Y are independent if and only

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

52

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

if f (x, y) = fX (x)fY (y) for all real numbers x and y. Example 3.3 Consider the jointly distributed random variables (X, Y ) ∈ [1, ∞) × [−1, 2] whose distribution is the function f (x, y) = 3x2 3 . We can find the mean of X + Y as follows. Z ∞Z ∞ 2 E [X + Y ] = (x + y) 3 dx dy 3x −∞ −∞ Z 2Z ∞ 2 = (x + y) 3 dx dy 3x −1 1 Z 2Z ∞ Z 2Z ∞ 2x 2y = dx dy + dx dy 3 3 3x 3x −1 1 −1 1 Z ∞Z 2 Z 2Z ∞ 2x 2y = dy dx + dx dy 3 3 3x 3x 1 −1 −1 1 Z ∞ Z 2 2 y =3 dx + dy 2 3x 1 −1 3 1 5 = 2+ = 2 2 If we examine the example above we recognize that E [X + Y ] = E [X]+ E [Y ]. This is true in general, not just for the previous example. The additivity property of the expected value of continuous random variables is stated in the following theorem. Theorem 3.1 If X1 , X2 , . . . , Xk are continuous random variables with joint probability distribution f (x1 , x2 , . . . , xk ) then E [X1 + X2 + · · · Xk ] = E [X1 ] + E [X2 ] + · · · + E [Xk ]. Proof.

See exercise (7).



The reader should again note that, just as was the case in the previous chapter, the previous theorem is true for random variables which are dependent or independent. Theorem 3.2 Let X1 , X2 , . . . , Xk be pairwise independent random variables with joint distribution f (x1 , x2 , . . . , xk ), then E [X1 X2 · · · Xk ] = E [X1 ] E [X2 ] · · · E [Xk ] . Proof.

See exercise (8).

10:57:10.



May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Normal Random Variables and Probability

BC8495/Chp. 3

53

If the continuous random variable X is conditional on another continuous random variable Y having value Y = y, the conditional probability density will be denoted f (x|y). The conditional density of X given Y = y can be calculated as f (x|y) =

f (x, y) . fY (y)

(3.5)

To put this notion to use, assume the ordered pairs (X, Y ) are uniformly distributed over the triangle T = {(x, y) | 0 ≤ y ≤ x ≤ 1}. Since the distribution is uniform and the area of T is 1/2, the joint probability distribution can be defined as f (x, y) = 2 on T and zero elsewhere. The conditional probability distribution is f (x|y) = R 1 y

f (x, y) f (x, y) dx

= R1 y

1 1 dx

=

1 1−y

provided 0 < y < 1. The conditional expected value for continuously distributed random variables is found by using the conditional probability density. Z ∞ E [X|Y = y] = xf (x|y) dx (3.6) −∞

The reader will be asked to show in the exercises that conditional expected value has the linearity property of the familiar expected value. Example 3.4 The joint probability distribution of ordered pairs (X, Y ) on the triangle T = {(x, y) | − 3 ≤ y ≤ x ≤ 3} is given by f (x, y) = 5xy 2 /162. The conditional expected value E [X|Y = y] can be found in two steps. First the conditional probability density function f (x|y) must be determined for −3 < y < 3. f (x|y) = R 3 y

=

y

10:57:10.

f (x, y) dx

5xy 2 162 R 3 5xy 2 y 162

= R3 =

f (x, y)

x

x dx

2x 9 − y2

dx

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

54

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

Second the formula in Eq. (3.6) for conditional expectation yields

E [X|Y = y] =

Z

y

3

x



2x 9 − y2



3 2(y 2 + 3y + 9) 2x3 = dx = , 3(9 − y 2 ) y 3(y + 3)

which holds for −3 < y < 3. Note once again that the conditional expected value can be thought of as another random variable depending on Y . Just as was the case for discrete random variables in the previous chapter, the expected value of a continuous random variable can be thought of as the average value of the outcome of an infinite number of experiments. The variance and standard deviation of a continuous random variable again mirror the concept earlier defined for discrete random variables.

3.3

Variance and Standard Deviation

The variance of a continuous random variable is a measure of the spread of values of the random variable about the expected value of the random variable. The variance is defined as   V (X) = E (X − µ)2 =

Z



−∞

(x − µ)2 f (x) dx,

(3.7)

where µ = E [X] and f (x) is the probability distribution function of X. The variance may be interpreted as the squared deviation of a random variable from its expected value. An alternative formula for the variance is given in the following theorem. Theorem 3.3 Let X be a random variable  with  probability distribution f and mean µ, then the variance of X is E X 2 − µ2 .

Example 3.5 Find the variance of the continuous random variable whose probability distribution is given by

f (x) =

10:57:10.



2x if 0 ≤ x ≤ 1, 0 otherwise.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Normal Random Variables and Probability

BC8495/Chp. 3

55

The reader may readily check that µ = E [X] = 2/3. Now by use of Theorem 3.3 we have  2 2 V (X) = x f (x) dx − 3 −∞ Z 1 4 = 2x3 dx − 9 0 1 4 1 = − = . 2 9 18 Z



2

In the exercises the reader will be asked to prove the following two theorems which extend to continuous random variables results we have already seen for discrete random variables. Theorem 3.4 Let X be a continuous random variable with probability distribution f (x) and let a, b ∈ R, then V (aX + b) = a2 V (X) . Theorem 3.5 Let X1 , X2 , . . . , Xk be pairwise independent continuous random variables with joint probability distribution f (x1 , x2 , . . . , xk ), then V (X1 + X2 + · · · + Xk ) = V (X1 ) + V (X2 ) + · · · + V (Xk ) . By definition the standard deviation of a continuous random variable is the square root of its variance. The standard deviation is sometimes denoted by σ. On occasion we will denote the variance of a random variable as σ 2 .

3.4

Normal Random Variables

For our purposes one of the most important and useful continuous random variables will be the normally distributed random variable. A continuous random variable obeying a normal distribution is frequently said to follow the “bell curve”. Many measurable quantities found in nature seem to have normal distributions, for example adult heights and weights. Statisticians, mathematicians, and physical scientists frequently assume any quantity subject to a large number of small independently acting forces (regardless of their distributions) is normally distributed. The normal dis-

10:57:10.

May 25, 2012

56

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

tribution even finds its way into the financial arena via the assumption that movements in the price of an asset are subject to a large number of incompletely understood political, economic, and social forces, thus justifying the assumption that these changes in value are related to a normal random variable (this assumption will be explored further later). The continuous normal random variable can also be thought of as the limiting behavior of the discrete binomial random variable introduced in Chapter 2. Recall the relatively simple probability function for a binomial random variable given in Eq. (2.3) and reproduced below. P (X = x) =

n! px (1 − p)n−x x!(n − x)!

With suitable assumptions we can develop the probability density function for a continuous normally distributed random variable from the probability function of a binomial random variable. The remainder of this section is dedicated to this derivation and to some elementary properties of normal random variables. Before delving into the details of the development we should outline the steps to be taken. We start with a binomial experiment of n trials with outcomes ±∆x where ∆x > 0. The probabilities of various outcomes of the experiment will be calculated using Eq. (2.3). To transition from the discrete case to the continuous realm we will take the limit of these probabilities as ∆x → 0, but this brings up the paradox mentioned at the beginning of Section 3.1. As ∆x → 0 the discrete probabilities will converge to zero unless the concept of a random variable equaling a specific value is replaced with the idea of a random variable lying in an interval around a specific value. We will also assume there exists a relationship between n and ∆x in order to ensure the mean and variance of the random variable remain constant as ∆x → 0. This derivation is outlined in [Bleecker and Csordas (1996), pg. 139]. We begin by supposing that a particle sits at the origin of the x-axis and may move to the left or to the right a distance ∆x in each of n independent identically distributed trials. We will assume that n(∆x)2 = 2kt, a constant. The particular form of constant is chosen to simplify the ultimate result. It is convenient to think of the time necessary to conduct the n trials as t and hence the time required by one trial is ∆t = t/n. The (discrete) probability of moving left is p = 1/2 which also happens to be the probability of moving to the right. We will show that if n ∈ N and m ∈ Z with −n ≤ m ≤ n then the 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

57

Normal Random Variables and Probability

probability that the particle is at location X = m∆x at time t = n∆t is n n! 21  1 . 1 2 (n + m) ! 2 (n − m) !

(3.8)

We will assume that all the steps taken by the particle are independent and identically distributed. Let the total number of steps taken be n and the number of steps taken to the right be r. Consequently the number of steps taken to the left will be n − r. The position of the particle after n steps will be (r − (n − r))∆x = (2r − n)∆x = m∆x, where we have assigned m = 2r − n. Consequently n + m is even (exercise (16) asks the reader to confirm that likewise n − m is even). Using the binomial probability distribution given in Eq. (2.3) we can state that the probability that the particle is at position m∆x after n steps is P (X = m∆x) = P (X = (2r − n)∆x)    r  n−r 1 n 1 = 2 2 r  n n! 1 = r!(n − r)! 2 n n! 12   = 1 1 2 (n + m) ! 2 (n − m) !

since n + m = 2r. This is the same probability as in Eq. (3.8). This claim can also be proved by induction on n. If n = 1 then m = ±1 (the particle is not allowed to remain in place). 1 1! 12   1 1 2 (1 + (−1)) ! 2 (1 − (−1)) ! 1 1! 12 1  1  P (X = ∆x) = = 1 2 2 (1 + 1) ! 2 (1 − 1) !

1 P (X = −∆x) = = 2

Thus the claim is true for n = 1. Now suppose the claim is true for k ≤ n−1. If the particle will move to m∆x at time n∆t, then at time (n − 1)∆t the particle must be at either (m − 1)∆x or (m + 1)∆x. Therefore at time 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

58

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

t = n∆t, 1 1 P (X = (m − 1)∆x) + P (X = (m + 1)∆x) 2 2 ! n−1 (n − 1)! 12 1  1  = 1 2 2 (n − 1 + m − 1) ! 2 (n − 1 − (m − 1)) ! ! n−1 (n − 1)! 21 1  1  + 1 2 2 (n − 1 + m + 1) ! 2 (n − 1 − (m + 1)) ! n (n − 1)! 12  1  = 1 2 (n − m − 2) ! 2 (n + m) ! n (n − 1)! 12   + 1 1 2 (n − m) ! 2 (n + m − 2) ! n n! 12  1 , = 1 2 (n + m) ! 2 (n − m) !

P (X = m∆x) =

which is the probability given in expression (3.8). We can treat each step taken by the particle as a Bernoulli experiment with outcomes ∆x and −∆x with equal probabilities. Thus the mean of this Bernoulli “step” is 0. The variance in the outcome of this Bernoulli experiment is 1 1 (−∆x)2 + (∆x)2 = (∆x)2 . 2 2 Since the steps are independent and identically distributed then the mean position of the particle after n steps is the sum of the means of each of the steps, or again 0. Thus on average the particle will be at the origin at the end of the experiment. Calculating the variance directly using Theorem 2.6 is tedious, but if we think of the motion of the particle as the sum of the outcomes of n independent Bernoulli experiments, then the variance of the location of the particle is σ 2 = n(∆x)2 . Hence the constant 2kt can be interpreted as the variance in the final position of the particle. Next we will make use of Stirling’s Formula which approximates n! when n is large. n! ≈



2πe−n nn+1/2

(3.9)

If we apply this formula to every factorial present in expression (3.8) we

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

59

Normal Random Variables and Probability

obtain the following sequence of equivalent expressions. √ n 2πe−n nn+1/2 12 √ (n+m+1)/2 √ (n−m+1)/2 2πe−(n+m)/2 12 (n + m) 2πe−(n−m)/2 12 (n − m) n nn+1/2 21 1 = √   2π 1 (n + m) (n+m+1)/2 1 (n − m) (n−m+1)/2 2 2 nn+1/2 2 = √ (n+m+1)/2 (n − m)(n−m+1)/2 2π (n + m) 2 n(n+m+1)/2 n(n−m+1)/2 = √ 2nπ (n + m)(n+m+1)/2 (n − m)(n−m+1)/2 (n+m+1)/2  (n−m+1)/2  2 n n = √ n−m 2nπ n + m    −(n+m+1)/2 2 m m −(n−m+1)/2 = √ 1+ 1− n n 2nπ  −(n+1)/2     2 m m/2 m2 m −m/2 = √ 1− 1− 2 1+ n n n 2nπ

Now we will make use of the fact that m = x/∆x and n = t/∆t. The last expression above is then equal to √  x − 2∆x  x  2 ∆t x∆t x∆t 2∆x √ 1+ 1− t∆x t∆x 2πt



x∆t 1− t∆x

2 !− 1+t/∆t 2

.

Earlier we assumed that n and ∆x were related by the equation n(∆x)2 = 2kt = σ 2 , the constant variance in the final position of the particle. Thus we see that n and ∆x are inversely related, as one becomes large (typically n), the other must become small i.e. ∆x. We can also rewrite n(∆x)2 = 2kt as (∆x)2 = 2k∆t. Using this relationship to replace the ∆t’s in the expression above yields kt 1  − x   x " 2 #− (∆x)  2 −2 x ∆x 2∆x x ∆x 2∆x x ∆x ∆x √ 1+ 1− 1− . 2kt 2kt 2kt kπt

So far what we have derived is an approximation (which we will treat as an equality for large n or, which is equivalent, for small ∆x) for the probability that the particle is located at position x = m∆x at time t = n∆t. The reader should realize that these are still discrete probabilities. Next we would like to examine the case in which ∆x → 0. However, if we take the 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

60

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

limit of the last expression we will obtain a probability of 0 just as in the elementary example discussed at the beginning of this chapter. In the limit the likelihood of the particle being at precisely any specified location is 0. What we must consider is the probability that the particle is in an interval, but what interval? Notice that for the discrete random random variable X P (X = m∆x) = P ((m − 1)∆x < X < (m + 1)∆x) . According to the Mean Value Theorem [Smith and Minton (2002)], the right-hand side of this equation is 2∆x times the probability density function of a continuous random variable. To derive a probability distribution function for this continuously distributed random variable we must divide the expression giving us P (X = m∆x) above by 2∆x and take the limit as ∆x → 0. Readers accustomed to numerical approximations for derivatives will recognize that we have used the centered difference formula for the first derivative (see exercise (17)) in order to develop a difference quotient whose limit will be the probability distribution. Background information on finite difference approximations to derivatives can be found in [Burden and Faires (2005)]. We will make use of the result that lim∆x→0 (1 + a∆x)1/∆x = ea . Let us define f (x, t) as f (x, t) = = = =

1 kt   −x   x " 2 #− (∆x)  2 −2 1 x ∆x 2∆x x ∆x x ∆x 2∆x √ 1− 1− lim 1 + 2kt 2kt 2kt 2 kπt ∆x→0   −kt x2 1 x − x x x √ e 2kt 2 e− 2kt 2 e− 4k2 t2 2 kπt x2 x2 x2 1 √ e− 4kt e− 4kt e 4kt 2 kπt x2 1 √ e− 4kt . 2 kπt

We must verify that this expression satisfies the properties of a probability distribution function. For k, t > 0 the expression is non-negative. Since for 2 |x| ≥ 1 it is true that 0 < e−x ≤ e−|x| , then the integral of f (x, t) over the entire real number line converges, suppose we write: 0
0 the function f (x, t) = resembling the often discussed “bell curve.”

x √1 e− 4kt 2 kπt

will have a graph

no preference for movement to the left or right. In a similar manner we may determine the variance of the position of the particle. Z ∞ x2 x2 2 √ V (X) = e− 4kt dx − (E [X]) −∞ 2 kπt Z M x2 1 lim x2 e− 4kt dx = √ kπt M→∞ 0 ! Z M 1 −M 2 /4kt −x2 /4kt = √ − e dx lim (−2kt) M e kπt M→∞ 0 Z ∞ 2 2kt e−x /4kt dx = √ kπt 0 Z ∞ 2 1 e−x /4kt dx = 2kt · √ 2 kπt −∞ = 2kt We were able to avoid evaluating the last improper integral by making use of the unit area property of the probability distribution. From the fact that σ 2 = 2kt we know that the “spread” in the location of the particle increases with time. This can be readily seen in the surface plot shown in Fig. 3.3. If the particle were initially placed at location µ but all other assumptions remained the same then the probability distribution would simply be

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

63

Normal Random Variables and Probability

z

t

x

x2

Fig. 3.3 For t > 0 the function f (x, t) = √1 e− 4kt retains the bell-shaped cross 2 kπt section, but the profile flattens and broadens as t increases.

shifted in the x direction by µ. Thus we may define the normal probability distribution with mean µ and variance σ 2 to be the function (x−µ)2 1 φ (x) = √ e− 2σ2 . σ 2π

(3.10)

When µ = 0 and σ = 1, this is referred to as the standard normal probability distribution. In the sequel the cumulative distribution function Φ (x) where Z x t2 1 √ e− 2 dt Φ (x) = P (X < x) = (3.11) 2π −∞ will be frequently used in discussions involving probabilities of normal random variables. For example, if X is a standard normal random variable then P (X < 0) = Φ (0) = 1/2. It is frequently helpful when dealing with normally distributed random variables to perform a mathematical change of variable which produces a standard normal random variable. Theorem 3.6 If X is a normally distributed random variable with expected value µ and variance σ 2 , then Z = (X − µ)/σ is normally distributed with an expected value of zero and a variance of one.

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

64

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

The proof of this result is left to the reader in exercise (24). In the sequel the symbol Z will be used to denote a normally distributed random variable with standard normal distribution. 3.5

Central Limit Theorem

The name Central Limit Theorem is given to several results concerned with the distribution of sample means or the distribution of the sum of random variables. The proofs of these theorems are beyond the scope of this work. The interested reader should consult a book on the mathematical underpinnings of statistics for proofs of the various versions of the Central Limit Theorem (for example [DeGroot (1975)]). Nevertheless we can observe the consequences of the Central Limit Theorem on data from a random number simulation. The reader should keep the following points in mind. Given a random variable X which can be either discrete or continuous and which may have any probability distribution, we may collect a sample of size n and denote the mean of that sample X n . If the process of collecting multiple samples and calculating their means is repeated then we can treat the sample means as random variables in their own right. One version of the Central Limit Theorem due to Lindeberg and L´evy implies that the sample means become normally distributed as the sample size becomes large. Theorem 3.7 If random variables X1 , X2 , . . . , Xn form a random sample of size n from a probability distribution with mean µ and standard deviation σ then for all x √  n(X n − µ) lim P ≤ x = Φ (x) , n→∞ σ where X n is the mean of a random sample of size n. Example 3.6 The reader can replicate this example on most programmable computing devices. Collect 5000 samples of size n (where n ∈ {2, 5, 10, 20, 40}) of a uniformly distributed continuous random variable on the interval [0, 1]. Compute the means of each sample of size n and plot the frequency histogram of the means. As can be seen in Fig. 3.4 as n increases the histograms take on the appearance of a normal probability distribution function. Another version of the Central Limit Theorem is due to Liapounov and concerns the asymptotic distribution of a sum of random variables.

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

65

Normal Random Variables and Probability

n=2

BC8495/Chp. 3

n=5

0.02 0.03 0.015

0.025

0.01

0.015

0.02

0.01 0.005 0.005 0

0 0

12 n=10

1

0

12 n=20

1

0

12

1

0.07 0.06

0.04

0.05 0.03 0.04 0.03

0.02

0.02 0.01 0.01 0

0 0

12

1 n=40

0.08 0.06 0.04 0.02 0 0

12

1

Fig. 3.4 An illustration of the Central Limit Theorem due to Lindeberg and L´ evy. As the sample size increases the distribution of the sample means becomes more normal in appearance.

Suppose the random variables X1 , X2 , . . . , Xn are pairwise independent but not necessarily identically distributed. We will assume that for each i ∈ {1, 2, . . . , n}, E [Xi ] = µi and that V (Xi ) = σi2 . Now define a new random variable Yn as Pn i=1 (Xi − µi ) Yn = p . Pn 2 i=1 σi

Using the assumption that the random variables are pairwise independent and Theorems 3.1, 3.4, and 3.5 we can determine that E [Yn ] = 0 and V (Yn ) = 1. The following theorem establishes that as n becomes large Yn is approximately normally distributed. Theorem 3.8 Suppose that the infinite collection {Xi }∞ i=1 of random variables are pairwise independent and that for each i ∈ N we have 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

66

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

  E |Xi − µi |3 < ∞. If in addition,   Pn 3 i=1 E |Xi − µi | lim =0 Pn 3/2 n→∞ ( i=1 σi2 )

then for any x ∈ R

lim P (Yn ≤ x) = Φ (x)

n→∞

where the random variable Yn is defined as above. Example 3.7 Once again, in place of a proof of Theorem 3.8 we will present a numerical simulation illustrating this result. The reader is encouraged to pursue their own similar numerical exploration. The data used in the following plots were generated by defining n continuous uniform distributions on [αi , βi ] for i = 1, 2, . . . , n where each of αi and βi were randomly, uniformly distributed in [−100, 100] and independently selected. The mean and variance of each uniform distribution were calculated. Then a random variable Xi was randomly chosen from each uniform distribution on [αi , βi ] and Yn was computed as above. A sample for Yn of size 5000 was collected and frequency histograms were created. These are shown in Fig. 3.5. It is readily seen that as n increases the distribution of Yn becomes more normal in appearance.

3.6

Lognormal Random Variables

While normal random variables are central to our discussion of probability and the upcoming Black-Scholes option pricing formula, of equal importance will be continuous random variables which are distributed in a lognormal fashion. A random variable X is a lognormal random variable with parameters µ and σ if ln X is a normally distributed random variable with mean µ and variance σ 2 . When referring to a lognormal random variable, the parameters µ and σ are often called the drift and volatility respectively. A lognormal random variable is a continuous random variable which takes on values in the interval (0, ∞). From Eq. (3.10) we see that for the lognormal random variable X, the probability that ln X < ln x is expressed as the integral Z ln x 2 2 1 P (ln X < ln x) = √ e−(t−µ) /2σ dt. σ 2π −∞ 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

67

Normal Random Variables and Probability

n=2

n=5

0.025

0.025

0.02

0.02

0.015

0.015

0.01

0.01

0.005

0.005

0 -3

BC8495/Chp. 3

-2

-1

0 n=10

1

2

0 -3

3

-2

-1

0 n=20

1

2

3

-2

-1

0

1

2

3

0.025 0.025 0.02

0.02

0.015

0.015 0.01

0.01

0.005

0.005

0 -3

-2

-1

0

1

2

0 -3

3 n=50

0.025 0.02 0.015 0.01 0.005 0 -3

-2

-1

0

1

2

3

Fig. 3.5 An illustration of the Central Limit Theorem due to Liapounov. As the sample size increases the distribution of the sum of the random variables becomes more normal in appearance.

Making the change of variable t = ln u we see then that the cumulative distribution function (CDF) for a lognormal random variable is P (X < x) = P (ln X < ln x) Z x 1 1 −(ln u−µ)2 /2σ2 √ = e du. σ 2π 0 u

(3.12)

Therefore a lognormally distributed random variable with parameters µ and σ 2 has a probability distribution function (PDF) of the form f (x) =

2 2 1 √ e−(ln x−µ) /2σ (σ 2π)x

(3.13)

for 0 < x < ∞. See Fig. 3.6 for the graph of the lognormal probability density function. The introduction of lognormal random variables does not complicate matters much since the probability that a lognor-

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

68

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

y

x

Fig. 3.6

The graph of the lognormal probability distribution function.

mally distributed random variable X is less than some x > 0 is then P (X < x) = Φ (ln x) where Φ (z) is the cumulative probability distribution function for a normal random variable with mean zero and variance one. Similarly the P (X > x) = 1 − Φ (ln x) for a lognormally distributed random variable. Using the definitions of expected value and variance we can prove the following lemma. Lemma 3.1 σ then

If X is a lognormal random variable with parameters µ and E [X] = eµ+σ

2

V (X) = e2µ+σ

/2 2

 2  eσ − 1 .

(3.14) (3.15)

Proof. According to the definition of the expected value of a continuous random variable and using the probability density function found in Eq. (3.13)  Z ∞  1 1 −(ln x−µ)2 /2σ2 E [X] = √ x e dx x σ 2π 0 Z ∞ 2 2 1 = √ et e−(t−µ) /2σ dt σ 2π −∞ Z ∞ 2 2 2 2 1 = eµ+σ /2 √ e−(t−(µ+σ )) /2σ dt σ 2π −∞ = eµ+σ

10:57:10.

2

/2

.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Normal Random Variables and Probability

BC8495/Chp. 3

69

In the second line of the equation we have made the substitution t = ln x. The last equality is true since the integral represents the area under the probability distribution curve for a normal random variable with mean µ + σ 2 and variance σ 2 . Likewise by Eq. (3.7)   V (X) = E X 2 − (E [X])2  Z ∞   2 2 1 1 −(ln x−µ)2 /2σ2 = √ x2 e dx − eµ+σ /2 x σ 2π 0 Z ∞ 2 2 1 = √ e2t e−(t−µ) /2 dt − e2µ+σ σ 2π −∞ Z ∞ 2 2 2 1 = e2(µ+σ ) √ e−(t−(µ+2σ)) /2 dt − e2µ+σ σ 2π −∞  2  2µ+σ2 =e eσ − 1 .

Between the second and third lines of the equation we used the substitution t = ln x.  Now we will apply the concept of the lognormally distributed random variable to the situation of the price of a stock or other security. Suppose that the selling price of a security will be measured daily and that the starting measurement will be denoted S(0). For n ≥ 1, we will let S(n) denote the selling price on day n. To a good approximation the ratios of consecutive days’ selling prices are lognormal random variables, i.e. the expressions X(n) = S(n)/S(n − 1) for n ≥ 1 are lognormally distributed. It is also generally assumed that the ratios X(n) and X(m) are identically distributed and independent when n 6= m. If a sufficient number of measurements have been made that a financial analyst estimates the parameters of the random variable X are µ = 0.0155 and σ = 0.0750, then questions regarding the likelihood of future prices of the security can be answered. Example 3.8 First, what is the probability that the selling price of the stock on the next day will be higher than the price on the present day? P (X(n) > 1) = P (ln X(n) > 0)   0−µ =P z> σ = P (z > −0.206667) = 1 − Φ (−0.206667) ≈ 0.582 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

70

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

Thus for a security with the parameters measured and reported above, there is a better than one half probability of the selling price increasing on any given day. Second, we may ask what is the probability that the selling price two days hence will be higher than the present selling price? At first glance the reader may be tempted to use the fact that each observation of the random variable is independent and hence declare that the sought after probability is 0.5822 ≈ 0.316. This implies the mistaken assumption that the price of the security increased on each of the two days. This ignores the possibility that the price of the security could decrease on either (but not both) of the two days and still make a net gain over the two-day period. To correctly approach the problem we must make use of the Central Limit Theorem.   S(n + 2) S(n + 1) P (S(n + 2)/S(n) > 1) = P · >1 S(n + 1) S(n) = P (X(n + 1)X(n) > 1) = P (ln X(n + 1) + ln X(n) > 0)   0 − 2(0.0155) √ =P z> 0.0750 2 = P (z > −0.292271) = 1 − Φ (−0.292271) ≈ 0.615

Thus the probability of a gain over a two-day period is nearly twice as large as may have been first suspected. The assumption that ratios of selling prices sampled at regular intervals for securities are lognormally distributed will be explored further in Chapter 5. Appendix A contains a sample of closing prices for Sony Corporation stock. Readers can collect their own data set from many corporations’ websites or from finance sites such as finance.yahoo.com. The eager reader may want to collect data for a particular stock to further test the lognormal hypothesis. 3.7

Properties of Expected Value

In later chapters the reader will encounter frequent references to the positive part of the difference of two quantities. For instance, if an investor may purchase a security worth X for price K, then the excess profit (if any)

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

71

Normal Random Variables and Probability

of the transaction is the positive part of the difference X − K. This is typically denoted (X−K)+ = max{0, X−K}. In this section a theorem and several corollaries are presented which will enable the reader to calculate the expected value of (X − K)+ when X is a continuous random variable and K is a constant. Theorem 3.9 Let X be a continuous random variable with probability distribution function f (x) and finite variance. If K is a constant then  Z ∞ Z ∞   + E (X − K) = f (t) dt dx. (3.16) K

x

Proof. Starting with the definition of the expected value of a function of the continuous random variable X with probability density function f (x), we have Z ∞   (x − K)+ f (x) dx E (X − K)+ = −∞ Z ∞ = (x − K)f (x) dx K

= lim

M→∞

Z

M

(x − K)f (x) dx.

K

The last definite integral can be evaluated using integration by parts with Rx u = x − K v = −∞ f (t) dt du = dx dv = f (x) dx This yields   E (X − K)+ M Z  Z x = lim f (t) dt (x − K) − M→∞

"Z

= lim

M→∞



= lim

M→∞

= lim

M→∞

=

Z



K

−∞

M

K

Z



x

f (t) dt (M − K) −

−∞

1−

Z

#

Z



Z





K

Z

f (t) dt dx − (M − K)

f (t) dt



dx.

10:57:10.



M

f (t) dt (M − K) −

M

x

Z

!

x

f (t) dt dx

−∞

K

K

M

Z

M

1−

M K

Z





M

Z





f (t) dt dx

x

1−

!

Z



x

f (t) dt

!



!

f (t) dt dx

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

72

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

The reader is asked in exercise (18) to provide the details used in evaluating the final limit.  Corollary 3.1 is a special case of Theorem 3.9. Corollary 3.1 If X is normal random variable with mean µ and variance σ 2 and K is a constant, then     σ −(µ−K)2 /2σ2 µ−K + E (X − K) = √ e + (µ − K)Φ . (3.17) σ 2π Proof.

The expected value of (X − K)+ is Z ∞   2 2 1 (x − K)+ e−(x−µ) /2σ dx E (X − K)+ = √ 2πσ −∞ Z ∞ 2 2 1 (x − K)e−(x−µ) /2σ dx = √ 2πσ K

If we make the substitution t = (x − µ)/σ then we have Z ∞   2 1 E (X − K)+ = √ (µ − K + tσ)e−t /2 dt 2π (K−µ)/σ Z Z ∞ 2 2 σ µ−K ∞ = √ e−t /2 dt + √ te−t /2 dt 2π (K−µ)/σ 2π (K−µ)/σ   2 2 µ−K σ = (µ − K)Φ + √ e(K−µ) /2σ . σ 2π The first integral on the right-hand side follows from the definition of the cumulative distribution function for the standard normal random variable. The reader is asked to evaluate the second improper integral in exercise (19).  For the purpose of completing the mean-variance analysis encountered later in Chapter 12 we must also be able to evaluate the expected value of the positive part of (X −K) when X is a lognormally distributed continuous random variable. Corollary 3.2 If X is a lognormally distributed random variable with parameters µ and σ 2 and K > 0 is a constant then       µ − ln K µ − ln K + µ+σ2 /2 E (X − K) = e Φ + σ − KΦ (3.18) σ σ 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Normal Random Variables and Probability

Proof.

BC8495/Chp. 3

73

By definition

  Z ∞   1 1 −(ln x−µ)2 /2σ2 E (X − K)+ = √ (x − K)+ e dx x 2πσ 0   Z ∞ 1 1 −(ln x−µ)2 /2σ2 (x − K) e dx. = √ x 2πσ K Making the substitution t = (ln x − µ)/σ allows us to write   E (X − K)+ Z ∞ 2 1 = √ (eσt+µ − K)e−t /2 dt 2π (ln K−µ)/σ ! Z ∞ Z ∞ 1 σt −t2 /2 −t2 /2 µ e e dt − K e dt = √ e 2π (ln K−µ)/σ (ln K−µ)/σ   Z 1 µ+σ2 /2 ∞ µ − ln K −(t−σ)2 /2 e dt − KΦ = √ e . σ 2π (ln K−µ)/σ The remaining integral can be evaluated using another substitution, z = t − σ. Therefore we have   Z ∞   µ − ln K + µ+σ2 /2 1 −z 2 /2 √ E (X − K) = e e dz − KΦ σ 2π ln K−µ −σ σ     2 µ − ln K µ − ln K + σ − KΦ . = eµ+σ /2 Φ σ σ 

3.8

Properties of Variance

In this section we will derive properties of the variance for the positive part of the difference of a continuous random variable and a constant. These results will be of use in later sections of this text. Lemma 3.2 Let X be a normally distributed random variable with mean µ and variance σ 2 . If K is a constant then   E ((X − K)+ )2 2

= (µ − 2K) + σ

2

10:57:10.



Φ



µ − 2K σ



+

2 (µ − 2K)σ − (µ−2K) 2σ2 √ e . (3.19) 2π

May 25, 2012

74

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

Proof. Since X is normally distributed with mean µ and variance σ 2 , then X − K is normally distributed with mean µ − K and variance σ 2 .   E ((X − K)+ )2 Z ∞ 2 2 1 = √ ((x − K)+ )2 e−(x−(µ−K)) /2σ dx 2πσ −∞ Z ∞ 2 2 1 = √ (x − K)2 e−(x−(µ−K)) /2σ dx 2πσ K Z ∞ 2 1 (σt + µ − 2K)2 e−t /2 dt = √ 2π (2K−µ)/σ upon making the substitution t = (x − (µ − K))/σ. Expanding the square in the last integrand and integrating produces   E ((X − K)+ )2   Z 2 µ − 2K 2(µ − 2K)σ ∞ √ = (µ − 2K)2 Φ te−t /2 dt + σ 2π (2K−µ)/σ Z ∞ 2 σ2 t2 e−t /2 dt +√ 2π (2K−µ)/σ   µ − 2K 2(µ − 2K)σ −(µ−2K)2 /2σ2 √ = (µ − 2K)2 Φ e + σ 2π   µ − 2K (µ − 2K)σ −(µ−2K)2 /2σ2 e + σ2 Φ − √ σ 2π    2 2 µ − 2K (µ − 2K)σ √ = (µ − 2K)2 + σ 2 Φ e−(µ−2K) /2σ . + σ 2π  Now the following result is immediately established. Corollary 3.3 Let X be a normally distributed random variable with mean µ and variance σ 2 . If K is a constant then  V (X − K)+    µ − 2K (µ − 2K)σ −(µ−2K)2 /2σ2 2 2 √ = (µ − 2K) + σ Φ + e σ 2π   2 2 2 σ µ−K − √ e−(µ−K) /2σ + (µ − K)Φ . (3.20) σ 2π Now we must establish a similar result for lognormally distributed random variables.

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Normal Random Variables and Probability

BC8495/Chp. 3

75

Lemma 3.3 Let X be a lognormally distributed random variable with parameters µ and σ 2 . If K > 0 is a constant then   E ((X − K)+ )2 2

= e2(µ+σ ) Φ (w + 2σ) − 2Keµ+σ

2

/2

Φ (w + σ) + K 2 Φ (w)

(3.21)

where w = (µ − ln K)/σ. Proof. Z ∞   2 2 1 1 E ((X − K)+ )2 = √ ((x − K)+ )2 e−(ln x−µ) /2σ dx x 2πσ 0 Z ∞ 2 2 1 1 = √ (x − K)2 e−(ln x−µ) /2σ dx x 2πσ K The expression σz = ln x − µ will be substituted into the last integral to yield: Z ∞   2 1 (eσz+µ − K)2 e−z /2 dz E ((X − K)+ )2 = √ 2π (ln K−µ)/σ 2 Z 2 e2(µ+σ ) ∞ = √ e−(z−2σ) /2 dz 2π (ln K−µ)/σ Z 2 2 2Keµ+σ /2 ∞ √ − e−(z−σ) /2 dz 2π (ln K−µ)/σ 2 Z ∞ 2 K +√ e−z /2 dz 2π (ln K−µ)/σ     µ − ln K µ − ln K 2 2(µ+σ2 ) + 2σ + K Φ =e Φ σ σ   2 µ − ln K − 2Keµ+σ /2 Φ +σ . σ  At last we have an expression for the variance of the positive part of the difference of a lognormal random variable and a constant. Corollary 3.4 Let X be a lognormally distributed random variable with parameters µ and σ 2 . If K > 0 is a constant then  2 2 V (X − K)+ = e2(µ+σ ) Φ (w + 2σ) − 2Keµ+σ /2 Φ (w + σ) + K 2 Φ (w)  2 2 − eµ+σ /2 Φ (w + σ) − KΦ (w) (3.22) where w = (µ − ln K)/σ.

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

76

3.9

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

Exercises

(1) Random variable X is continuously uniformly distributed in the interval [−4, 1]. Find P (X ≥ 0). (2) Suppose the probability distribution function of a continuous random variable X is f (x) =

C . 1 + x2

(a) Find the value of the constant C. (b) Find P (X > 1). (3) Random variable X is continuously distributed on the interval (1, ∞), with probability distribution function  c if 1 ≤ x f (x) = x3 0 otherwise. Determine the value of c. (4) Show using properties of the definite integral that for a continuous random variable X with probability distribution function f (x), P (X ≥ a) = 1 − P (X < a) . (5) A random variable X has a continuous Cauchy distribution with probability density function f (x) =

1 . π(1 + x2 )

Show that the mean of this random variable does not exist. (6) If X is a continuous random variable with probability density function f (x), show that E [aX + b] = aE [X] + b where a, b ∈ R. (7) Prove Theorem 3.1. (8) Prove Theorem 3.2. (9) Referring to the joint probability distribution for (X, Y ) given in Example 3.4 find E [Y |X = x]. (10) Suppose that X, Y and Z are jointly distributed continuous random variables. (a) Show that E [X + Y |Z = z] = E [X|Z = z] + E [Y |Z = z]. (b) Suppose c is a constant and show that E [c X|Y = y] c E [X|Y = y].

10:57:10.

=

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Normal Random Variables and Probability

BC8495/Chp. 3

77

(11) Suppose X and Y are continuously distributed random variables whose joint probability distribution function is 3 3 xy if 0 ≤ y ≤ x ≤ 2, f (x, y) = 8 0 otherwise.

(12) (13) (14) (15)

(a) Find E [XY ].  (b) Find E hX 2 Y .i √ (c) Find E X Y .

Prove Theorem 3.3. Prove Theorem 3.4. Prove Theorem 3.5. Find the expected value and variance of the continuous random variable X whose probability distribution function is given by 2 |x| if −1 ≤ x ≤ 2, f (x) = 5 0 otherwise.

(16) Show that if m and n are integers then n − m is even if and only if n + m is even. (17) Suppose f (x) is three times continuously differentiable at x = x0 . Use Taylor’s Theorem [Stewart (1999)] to expand f (x) about x = x0 . Then by using f (x0 + h) and f (x0 − h) show that f 0 (x0 ) ≈

f (x0 + h) − f (x0 − h) . 2h

(18) Show that lim (M − K)

M→∞

Z



f (t) dt = 0

M

where f is a probability density function for a continuous random variable with finite variance. (19) Show that Z ∞ 2 2 2 σ σ √ te−t /2 dt = √ e−(K−µ) /2σ . 2π (K−µ)/σ 2π (20) Evaluate the following indefinite integral. Z x2 x √ e− 4kt dx 2 kπt 10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

78

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 3

An Undergraduate Introduction to Financial Mathematics

(21) Use the technique of integration by parts to evaluate Z x2 x2 √ e− 4kt dx. 2 kπt (22) Evaluate the following limit with k > 0 and t > 0. lim 2ktM e−M

2

/4kt

M→∞

(23) Using a computer algebra system, graphing calculator, or some numerical method evaluate the following probabilities for a standard normal random variable: (a) (b) (c) (d)

P (−1 < X < 1) P (−2 < X < 2) P (−3 < X < 3) P (1 < X < 3)

(24) Prove Theorem 3.6. (25) The annual rainfall amount in a geographical area is normally distributed with a mean of 14 inches and a standard deviation of 3.2 inches. What is the probability that the sum of the annual rainfalls in two consecutive years will exceed 30 inches? (26) Suppose that Alice and Bob agree to meet in the library between 10:00AM and 11:00AM to study calculus. Because neither of them is very reliable, either may be late to the meeting or forget to come at all. They have an understanding that neither will wait more than 15 minutes for the other to arrive. If they arrive at independent times between 10:00AM and 11:00AM, what is the probability they will meet? (27) The ratio of selling prices of a security on consecutive days is a lognormally distributed random variable with parameters µ = 0.01 and σ = 0.05. What is the probability of a one-day increase in the selling price? What is the probability of a one-day decrease in the selling price? What is the probability of a four-day decrease in the selling price of the security? (28) Let X be a uniformly distributed continuous random variable on the interval [a, b]. If K is a constant, find an expression for E [(X − K)+ ]. (29) Show that for the standard normal random variable 1 − Φ (x) = Φ (−x). (30) Fill in the details of the proof of Corollary 3.3. (31) Fill in the details of the proof of Corollary 3.4.

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Normal Random Variables and Probability

BC8495/Chp. 3

79

(32) Let X be a continuously distributed random variable with E [X] = µ and V (X) = σ 2 . (a) Find the E [X(X + 1)].   (b) Find the E (X − C)2 where C is a constant.

10:57:10.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

This page intentionally left blank

10:57:10.

80

BC8495/Chp. 3

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

Chapter 4

The Arbitrage Theorem

The concept known as arbitrage is subtle and can seem counter-intuitive. Formally there are two types of arbitrage (creatively named type A and type B). Type A arbitrage is a trading strategy that results in an initial positive cash flow (the investor receives a positive amount of income initially) and there is no risk of loss in the future. Type B arbitrage is a trading strategy in which the investor needs no initial cash investment, has no risk of future loss, and has a positive probability of profit in the future [Cornuejols and T¨ ut¨ unc¨ u (2007)]. Informally arbitrage exists whenever two financial instruments are mis-priced relative to one another. Due to the mis-pricing, it becomes possible to make a financial gain. For example, suppose bank A issues loans at a 5% interest rate and bank B offers a savings account which pays 6% interest. A person could take out a loan from bank A and place the loan into savings with bank B. When it becomes time to repay bank A for the loan, the person closes the savings account with bank B, repays the principal and interest and still has 1% of the loaned amount as profit. The assumption that financial markets are efficient prevents such obvious arbitrage opportunities from being commonplace. When arbitrage opportunities arise, investors wanting to make a profit flock to the mis-priced instruments and the financial market reacts by correcting the pricing of the instruments. Readers often wonder if any true arbitrage opportunities exist. The recent (circa 2008-2010) heightened volatility in the financial markets has provided observers with interesting examples of mis-priced financial products. One such example can be drawn from the behavior of the stock of Accenture, PLC (a business consulting corporation, ticker symbol ACN) on May 6, 2010. During the two-minute-long period from 2:47PM to 2:49PM, Accenture stock traded on the New York Stock Exchange (NYSE) fell from

10:57:19.

May 25, 2012

82

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

nearly $40.00 per share to $0.01 per share and then returned to approximately $40.00. It is unlikely that a fundamental change in the value of the corporation was responsible for the 99.975% drop in the value of its stock in one minute, thus almost certainly for nearly a minute the stock was mis-priced. The reason for the temporary mis-pricing has been attributed to a combination of human error and the un-intended effects of computerized trading. This put downward pressure on the price of Proctor and Gamble stock (ticker symbol PG), one of the 30 stocks making up the Dow Jones Industrial (DJI) average, and thus this average fell as well. Computers monitoring the fall of the DJI average, executed stop-loss sell orders to prevent further losses in case the DJI fell further. Some firms using computer-driven trading suspected that something was amiss in the prices they were monitoring and slowed or stopped electronic trading. This had the effect of removing buyers and sellers from the stock market which led to further disruptions [Mehta, et al. (2010), ]. Many algorithmic trading firms serve as market makers, keeping track of the order book of bid and ask prices for stocks. A common practice of market makers is to enter “stub quotes” as place holders in the order book for the stock for which they create a market. Stub quotes can serve as bid prices, usually set very low, for example at $0.01, so that during the normal course of price fluctuations, there is always an order to buy the stock no matter how low the price may fall. This prevents the occurrence of a situation where no investor is willing to buy the stock at any price. Likewise stub quotes can serve as ask prices, set as high as $100, 000 per share, so that there are always parties willing to sell in the market. Stocks should never be traded at the stub quotes; however, when algorithmic trading exhausted the legitimate buyers of Accenture stock on May 6, 2010 the highest remaining bids in the market were the stub quotes at $0.01 and in the absence of human intervention trades were executed at this price. Accenture and its stockholders were not ruined on that day since the exchange overseeing trading canceled these orders. In the end an arbitrage opportunity did not exist for investors in Accenture stock, since buy orders at the temporary low price were not allowed to be executed. We will see that financial instruments such as options, bonds, and stocks must be priced so as to be “arbitrage free.” It is the absence of arbitrage which forms the basis of the derivation of the Black-Scholes equation found in Chapter 7.

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

4.1

BC8495/Chp. 4

83

The Concept of Arbitrage

As mentioned previously one way to think about arbitrage is as the situation which arises when two financial instruments are mis-priced relative to one another. In this section we will refine and make precise our definition. Suppose there is a set of possible outcomes for some experiment and that wagers can be placed on those outcomes. The Arbitrage Theorem states that either the probabilities of the outcomes are such that all bets are fair, or there is a betting scheme which produces a positive gain independent of the outcome of the experiment. Since we are adopting the language of wagering, a simple example will illustrate the Arbitrage Theorem. To simplify the process of placing and paying off bets, odds are used rather than probabilities, though the two concepts are related by a simple formula. Suppose the “odds” of a particular sporting events outcome are quoted as “2 : 1 against.” We can think of this as implying that there are three outcomes to the experiment and in two of them the desired outcome does not occur and in one it does. Thus the probability of the desired outcome arising is 1/3. In general then if the “odds against” a particular outcome are n : 1 then the probability of the outcome is 1/(n + 1). Odds simplify the paying off of bets in the following way. If the odds against an outcome are n : 1 then a unit bet will pay us n units if the outcome occurs. If the outcome does not occur, the unit bet is lost. The payoffs scale multiplicatively for non-unit bets. Odds against the occurrence of an event can also be expressed in the form n : m where n and m are natural numbers. In this case the probability of the event occurring is m/(n + m). n Odds of the form n : m are equivalent to the odds m : 1. Now consider this example. Suppose the odds against player A defeating player B in a tennis match are 3 : 1 and the odds against player B defeating player A are 1 : 1. Converting these to probabilities we see that player A defeats player B with probability 1/4 while player B defeats player A with probability 1/2. There is obviously something wrong with these probabilities since they should add to one, but do not. We will see that the Arbitrage Theorem implies there is a betting strategy which generates a positive gain regardless of the outcome of the tennis match. Now suppose we wager 1 on player A and 2 on player B. If player A wins we will win 3 on the first bet and lose 2 on the second, producing a net gain of 1. If player B wins we will lose 1 on the first bet and win 2 on the second, again yielding a positive payoff of 1. Notice, no matter which player wins the tennis match, we have a positive gain. If this scenario was real, we would borrow as much

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

84

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

money as possible and place these wagers. Once we collect our winnings we could pay back the loan and retire. This example is hardly transparent, so study of the Arbitrage Theorem would be beneficial in avoiding arbitrage opportunities in more complex financial situations. Before a statement and proof of the Arbitrage Theorem is presented a preliminary result from linear programming is needed.

4.2

An Introduction to Linear Programming

In this section we will introduce the common forms of linear programs and show their equivalencies. Linear programming is the name given to a branch of mathematics often applied in business and economics in which a linear function of some (usually large) number of variables must be optimized (either maximized or minimized) subject to a set of linear equations or inequalities. To sample the types of linear programming problems solved by businesses consider this simple example. A bank may invest its deposits in loans which earn 6% interest per year and in the purchase of stocks which increase in value by 13% per year. The bank wishes to maximize the total return on its investments. Assume the bank can invest a proportion x in loans and proportion y in stocks and must retain 10% in the form of cash (in other words 10% of the portfolio has a rate of return of 0). Any remaining proportion is simply held by the bank. The total return is therefore 0.06x + 0.13y. Suppose that government regulations require that the bank invest no more than 60% its deposits in stocks. As a good business practice the bank wishes to devote at least 25% of its deposits to loans. These constraints impose some inequalities on the bank’s investment strategy. The inequalities are x ≥ 0, y ≥ 0, x + y ≤ 0.90 (non-negative proportions of the deposits are invested and the total amount invested is no greater than 90% of the total amount on deposit), y ≤ 0.6 (government regulation), and x ≥ 0.25 (the bank’s business practice). An investment strategy can be represented by a vector hx, yi. The bank’s linear programming problem is in picking the vector which satisfies the constraints and maximizes the total return. Since this problem is two dimensional a plot can reveal the solution. As seen in Fig. 4.1 the optimal solution occurs when x = 0.3 and y = 0.6. In the remainder of this section we will give a more formal, but still brief, introduction to linear programming and the Duality Theorem. Other

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

85

The Arbitrage Theorem

1.0

0.8

0.06 x + 0.13 y = k

0.6 y

0.4

Feasible Region

0.2

0.0 0.0

0.2

0.4

0.6

0.8

1.0

x

Fig. 4.1 The shaded region represents the set of possible solutions to the bank’s investment decision. The total return is maximized at h0.3, 0.6i.

accessible introductions to linear programming and the Duality Theorem can be found in [Franklin (1980)], [Noble and Daniel (1988)], and [Strang (1986)]. Readers wishing for a more detailed introduction not only to linear programming, but also to linear algebra, should consult [Gale (1960)]. A generic linear programming problem, or linear program, consists of a set of linear equations or inequalities (constraints), possibly sign conditions on the solution (more constraints), and a linear expression (a weighted sum) which must be optimized (either maximized or minimized). We will use bold letters to represent vectors and the dot product (Euclidean inner product) to express linear expressions. Vectors will be thought of as matrices with a single column.   x1  x2    x= .   ..  xn 10:57:19.

May 25, 2012

86

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

If c and x are vectors with n components each, the notation cT x = c1 x1 + c2 x2 + · · · + cn xn represents a linear expression, a weighted sum of the components of vector x where the weights are given in vector c. The notation cT stands for the usual matrix transpose operation. The components of vector x are called the decision variables. The linear expression cT x is called the objective function. The standard form of a linear program involves finding a vector x with non-negative components such that for a given vector c the linear expression cT x is optimized (maximized or minimized) subject to equality constraints of the form aT x = b where a is a given vector and b is a constant [Winston (1994)]. The standard form of a linear program is often preferred when solving the problem numerically (such as by the simplex method). Another format for stating a linear program is known as the canonical form. In canonical form the decision variables (components of x) are non-negative, the constraints are in equation form, the constants on the right-hand sides of the constraints are non-negative, and one decision variable in each constraint has a “+1” coefficient, has a zero coefficient in any other constraint, and a zero coefficient in the objective function [Bradley et al. (1977)]. Yet a third common format for stating a linear program is known as the normal form or symmetric form [Williams (1970)]. In the symmetric form the decision variables are non-negative and the constraints are inequalities (all purely ≤ or all purely ≥). The symmetric form of a linear program is often convenient when proving a property or theoretical result about a linear program. In practice we will not obsess over the format of a linear program since there are algorithmic ways to convert one format into another. For denoting the comparison of vectors we will adopt a convenient extension of the equation and inequality symbols. We will say that vector u is less than (less than or equal to) vector v if the vectors have the same number of elements and ui < vi (ui ≤ vi ) for all i. Similarly we will say that vector u is greater than (greater than or equal to) vector v if ui > vi (ui ≥ vi ) for i = 1, 2, . . . , k. These inequalities will be denoted as appropriate u < v, u ≤ v, u > v, or u ≥ v. The positive part of a real number x will be denoted x+ and is defined as  x if x ≥ 0 x+ = 0 if x < 0. 10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

BC8495/Chp. 4

87

Similarly the negative part of x is denoted x− and defined as  −x if x ≤ 0 x− = 0 if x > 0. The positive part, as well as the negative part, of x is always a non-negative real number. Thus any real number x can be expressed as x = x+ − x− . This notation is extended to vectors by applying it to each component of the vector. This new notation will simplify the statements of the theorems in this chapter. For instance the requirement that x be a vector with nonnegative components can be succinctly stated as x ≥ 0. If there are several inequality constraints placed on a solution to a linear programming problem, we may compactly describe them using matrix notation. Suppose there are m inequality constraints on the solution x, a vector with n components. aT1 x ≤ b1 aT2 x ≤ b2 .. . aTm x ≤ bm This will be expressed as the vector inequality Ax ≤ b where      b1 x1 a11 a12 · · · a1n  a21 a22 · · · a2n   x2   b2       Ax =  . ..   ..  ≤  ..  = b. .. .  . .  .   .  . bm xn am1 am2 · · · amn

For our purposes a convenient starting point for describing a linear program will be to state it as “maximize cT x subject to the constraints Ax ≤ b and x ≥ 0”. The process of maximizing can be replaced that of minimizing by multiplying the objective function by −1 since   − max(cT x) = min(−cT x). x

x

The vector x is a feasible vector or feasible solution to the linear program if x ≥ 0 and x satisfies the set of constraint inequalities Ax ≤ b. The feasible solutions to a standard linear program form a convex set. A set is convex if for every pair of points P and Q contained in the set, the line segment connecting them also lies completely in the set. The linear program is called feasible if there exists a feasible solution to it. If the

10:57:19.

May 25, 2012

88

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

vector x is feasible and maximizes the objective function, then x is also called an optimal solution. As we begin a mathematical study of linear programs we first establish the equivalence of linear programs in symmetric and standard form. Starting with a linear program in symmetric form maximize cT x subject to Ax ≤ b and x ≥ 0 we introduce slack variables. To illustrate, the following inequality constraint of a linear program x1 + x2 + x3 ≤ 1 becomes

x1 + x2 + x3 + x ˆ4 = 1,

where slack variable, xˆ4 ≥ 0 “takes up the slack” to produce the equality constraint of a standard linear program. No modification of the objective function is necessary. The weight assigned to a slack variable in the objective function is zero. If there are multiple linear inequality constraints in the symmetric linear program, several slack variables may be introduced. Suppose A is an m × n matrix, x is a vector of n components, and b is a vector of m components, then by augmenting x with m slack variables and A with the m × m identity matrix the inequality constraint Ax ≤ b is equivalent to   x1  x2      .    ..  a11 a12 · · · a1n 1 0 · · · 0  b1    a21 a22 · · · a2n 0 1 · · · 0   x   b2    n     . = .  ..   .. .. .. ..   ..  xˆ  .. .  . . . .  n+1   x ˆ am1 am2 · · · amn 0 0 · · · 1  n+2  bm   ..   .  xˆn+m     x = Ax = b. A Im x ˆ

The symmetric form of a linear program has been shown to be equivalent to the standard linear program “maximize cT x subject to Ax = b and x ≥ 0”, where it is understood that A is an m × (n + m) matrix consisting of the original constraint matrix from the standard linear program augmented with the identity matrix and x is the solution vector to the standard linear program augmented with the slack variables.

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

BC8495/Chp. 4

89

Now suppose we start with a linear program in standard or canonical form maximize cT x subject to Ax = b and x ≥ 0. The equality constraint Ax = b is equivalent to the pair of inequalities Ax ≤ b and Ax ≥ b. The latter inequality involving the greater than or equal comparison is equivalent to the inequality −Ax ≤ −b which involves the less than or equal comparison. Therefore the original linear program is equivalent to the linear program in symmetric form T

maximize c x subject to



   A b x≤ and x ≥ 0. −A −b

Any vector x which is a feasible solution to a standard or canonical program is also a feasible solution to the equivalent symmetric program. Likewise if z is a feasible vector for a symmetric program it is also a feasible vector for the related standard or canonical program. Some linear programs may be stated in forms that contain inequality constraints (mixing greater than or equal and less than or equal) and equality constraints and some may omit the non-negativity requirement of the solution vector. Variables whose signs are not specified for determining feasible solutions are said to be unrestricted in sign or free variables. These linear programs can be referred to as general linear programs and may be stated in the format of maximizing cT x subject to the constraints Ax ≤ b ˆ ˆ ≥b Ax ˜ ˜ =b Ax

(4.1) (4.2) (4.3)

where A is a r × n matrix, b is a vector with r components, Aˆ is a s × n ˆ is a vector with s components, A˜ is a t × n matrix, and b ˜ is matrix, b a vector with t components. Every standard, canonical, and symmetric linear program is a general linear program. In fact the four types of linear program are equivalent. We need only show that a general linear program can be recast as a linear program in symmetric form. If x is unrestricted in sign then x = x+ − x− where x+ ≥ 0 and x− ≥ 0. 10:57:19.

May 25, 2012

90

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

The constraints in Eqs. (4.1)–(4.3) can now be written as A(x+ − x− ) ≤ b ˆ ˆ + − x− ) ≤ −b −A(x

˜ ˜ + − x− ) ≤ b A(x + − ˜ ˜ −A(x − x ) ≤ −b.

In matrix/vector form this system of inequalities can be expressed as 

   b A −A  + ˆ  −Aˆ Aˆ  x  −b      A˜ −A˜  x− ≤  b ˜ . ˜ −A˜ A˜ −b

(4.4)

The linear expression for the objective function of the general linear program can be written as (c, −c)T (x+ , x− ) = cT x+ − cT x− = cT (x+ − x− ) = cT x. Thus, we have recast the general linear program in the form of a symmetric linear program. Vector (x+ , x− ) is an optimal solution to the symmetric linear program if and only if x = x+ − x− is an optimal solution to the general linear program. Thus the two linear programming problems, the one with the non-negativity constraint and the one without it, are equivalent [Gonz´ alez-D´ıaz, et al. (2010)]. When necessary we will re-formulate linear programming problems in the form most convenient for establishing the result of interest, though the symmetric form is generally the most convenient for our purposes. Lastly the choice of optimizing the objective function by maximization is arbitrary. All the results to follow hold if minimization is used in place of maximization. If the dimensions m and n of the problem are not too large this optimization task is easily accomplished. For example suppose we try to maximize 5x1 + 4x2 + 8x3 subject to the constraints x1 + x2 + x3 ≤ 1 and x ≥ 0. The constraints require the solution to the optimization problem to lie in a subset of the positive orthant of R3 . This subset is a tetrahedron. See Fig. 4.2. If the maximum of the objective function is the, as yet unknown, value k, then 5x1 + 4x2 + 8x3 = k defines a plane. The largest value of k for which the level set of the objective function intersects the constrained set of points will be the maximum of the objective function. Thus the maximum

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

BC8495/Chp. 4

91

2.0 1.5

x2 1.0 0.5 0.0 1.0

x3

0.5

0.0 0.0 0.5 1.0 x1

1.5 2.0

Fig. 4.2

The set of points in R3 where the objective function’s maximum may occur.

of the objective function is 8 as can be seen in Fig. 4.3. The objective function is maximized at the point (x1 , x2 , x3 ) = (0, 0, 1). So far we have solved linear programming problems by graphical techniques. When n and/or m are large this is impractical, so now we turn our attention to developing necessary and sufficient conditions for determining if a linear programming problem has an optimal solution.

4.3

Dual Problems

Here we derive and discuss a result from which the Arbitrage Theorem easily follows. The Duality Theorem is a familiar result to readers having studied linear programming and operations research. In mathematics we frequently benefit from the ability to solve one problem by means of finding

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

92

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

2.0 1.5

x2 1.0 0.5 0.0 1.0

x3

0.5

0.0 0.0 0.5 1.0 x1

1.5 2.0

Fig. 4.3 The maximum of the objective function is the last level set of the objective function to intersect the tetrahedron of constrained solution points.

the solution to a related, but simpler, problem. This is certainly true of linear programming problems. For every linear programming problem there is an associated linear programming problem known as its dual. Henceforth the original problem will be known as the primal. For linear programs stated in symmetric form, these paired optimization problems are related in the following ways. Primal: maximize cT x subject to Ax ≤ b and x ≥ 0. Dual: minimize bT y subject to AT y ≥ c and y ≥ 0. We should note that: (1) the process of maximization in the primal is replaced with the process of minimization in the dual,

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

93

The Arbitrage Theorem Table 4.1 The symmetric relationship between the primal and dual linear programs. max cT x x

T

min b y

(x1 ≥ 0)

(x2 ≥ 0)

···

(xn ≥ 0)

x1 a11 a21 .. . am1 ≥ c1

x2 a12 a22 .. . am2 ≥ c2

··· ··· ···

xn a1n a2n .. . amn ≥ cn

y

(y1 ≥ 0) (y2 ≥ 0) .. . (ym ≥ 0)

y1 y2 .. . ym

··· ···

≤ b1 ≤ b2 .. . ≤ bm

(2) the unknown of the dual is a vector y with m components where m is the number of inequality constraints of the primal, (3) the vector b moves from the constraint of the primal to the objective function of the dual, (4) the vector c moves from the objective of the primal to the constraint of the dual, (5) the constraints of the dual are greater than or equal inequalities and there are n of them where n is the dimension of the decision vector x. The relationship between the primal and the dual for symmetric linear programs is summarized in Table 4.1 (adapted from [Winston (1994)]). When the primal problem is in general form, the dual problem can also be formulated (after the primal is first converted to its equivalent symmetric form). Consider the general linear program of maximizing cT x subject to the constraints in Eqs. (4.1)–(4.3) and for which the decision variables are unrestricted in sign. The general linear problem can be recast as a problem in symmetric form with the constraints in Eq. (4.4). The dual of this linear program is ˆ b, ˜ −b) ˜ T (y, y ˆ, y ˜+, y ˜−) minimize (b, −b, subject to the inequality constraints "

AT −AˆT A˜T −A˜T −AT AˆT −A˜T A˜T

#

 y     y  ˆ ≥ c , y ˜+  −c ˜− y 

ˆ ≥ 0 (with s components), y ˜+ ≥ 0 where y ≥ 0 (with r components), y − ˜ ≥ 0 (with t components also). If we define (with t components), and y 10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

94

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

˜=y ˜+ − y ˜ − then y ˜ is unrestricted in sign and the dual constraints can be y expressed as " #y   AT −AˆT A˜T   c ˆ ≥ y . −c −AT AˆT −A˜T ˜ y Note that this set of inequalities implies   h i y ˆ  = c. AT −AˆT A˜T  y ˜ y

Two observations from this exercise will be important later when proving the Arbitrage Theorem. First, decision variables unrestricted in sign present in the primal (dual) problem induce equality constraints in the dual (primal) problem. Second, equality constraints in the primal (dual) problem induce decision variables which are unrestricted in sign in the dual (primal) problem. Given a primal or a dual problem it is a routine matter to construct its partner. The primal and the dual form a set of “fraternal twin” problems. The following theorem will shed some light on their relationship. Theorem 4.1 Proof.

The dual of the dual is the primal.

Starting with the dual problem in symmetric form, minimize bT y subject to AT y ≥ c and y ≥ 0.

We can re-write the dual as a maximization problem with less than or equal constraints. Maximize (−b)T y subject to (−A)T y ≤ −c and y ≥ 0. Now the dual of this problem (i.e., the dual of the dual) is minimize (−c)T x subject to ((−A)T )T x ≥ −b and x ≥ 0. This problem is logically equivalent to the problem maximize cT x subject to Ax ≤ b and x ≥ 0, which is the primal problem.

10:57:19.



May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

BC8495/Chp. 4

95

Perhaps it is no surprise then that the solutions to the primal and dual are related. Theorem 4.2 (Weak Duality Theorem) If x and y are the feasible solutions of the primal and dual problems respectively, then cT x ≤ bT y. If cT x = bT y then these solutions are optimal for their respective problems. Proof. Feasible solutions to the primal and the dual problems must satisfy the constraints Ax ≤ b with x ≥ 0 (for the primal problem) and AT y ≥ c with y ≥ 0 (for the dual). We multiply the constraint in the dual by xT to obtain xT AT y ≥ xT c

⇐⇒

cT x ≤ yT Ax.

We multiply the constraint in the primal by yT to find yT Ax ≤ yT b = bT y. Note that the directions of the inequalities are preserved because each component of x and y is non-negative. Combining these last two inequalities produces cT x ≤ yT Ax ≤ bT y.

(4.5)

Therefore we have cT x ≤ bT y. Now suppose there exist solutions x∗ and y∗ to the primal and dual problems respectively for which cT x∗ = bT y∗ . Let x be any feasible solution to the primal problem, then cT x ≤ bT y∗ = cT x∗ by Eq. (4.5) and implying that x∗ generated the maximum of the objective function for the primal problem. Likewise if y is any feasible solution to the dual problem, then by Eq. (4.5) bT y ≥ cT x∗ = bT y∗ meaning y∗ generated the minimum of the objective function for the dual problem.  One consequence of the Weak Duality Theorem is that if x is a feasible solution for the primal and y is a feasible solution for the dual, then the linear expression for the primal cT x is bounded above and the linear expression for the dual bT y is bounded below. Since we are trying to maximize (minimize) a linear expression for the primal (dual) problem, it is helpful

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

96

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

to know that there exists an upper (a lower) bound on this expression. Another consequence is that for optimal solutions x and y to the primal and dual problems respectively, the following equation holds (from Eq. (4.5)). cT x = yT Ax = bT y Example 4.1

(4.6)

Consider the paired primal and dual problems:

Primal: maximize 4x1 + 3x2 subject to x1 + x2 ≤ 2 and x1 , x2 ≥ 0. Dual: minimize 2y1 subject to y1 ≥ 3 and y1 ≥ 4 and y1 ≥ 0. While both problems may appear simple, the dual is trivial. The minimum value of y1 subject to the constraints must be y1 = 4. According to the Weak Duality Theorem, the minimum of the objective function of the primal must be at least 8. Applying the level set argument as before, the largest value of k for which the level set 4x1 + 3x2 = k intersects the set of feasible points for the primal is k = 8. See Fig. 4.4. 2.0

4 x1 +3 x2 =8

1.5

x2

Feasible Region

1.0

0.5

0.0 0.0

0.5

1.0

1.5

2.0

x1

Fig. 4.4 The maximum of the objective function 4x1 + 3x2 subject to the constraints x1 + x2 ≤ 2 and x1 , x2 ≥ 0, occurs at the point with coordinates (x1 , x2 ) = (2, 0).

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

BC8495/Chp. 4

97

As a consequence of the Weak Duality Theorem we can determine that when x and y are optimal for their respective problems then bT y = yT Ax = cT x (yT A − cT )x = 0

(AT y − c)T x = 0.

(4.7)

Since x is feasible then each component of that vector is non-negative. Likewise the assumption in the dual, AT y ≥ c, implies that each component of the vector AT y − c is non-negative. Using Eq. (4.7) we can conclude that vector x must be zero in every component for which vector AT y − c is positive and vice versa. Thus we have proved the following theorem. Theorem 4.3 (Complementary Slackness) Optimality in the primal and dual problems requires either xj = 0 or (AT y − c)j = 0 for each j = 1, 2, . . . , n. Due to the symmetry between the primal and dual problems it can also be stated that either yi = 0 or (Ax − b)i = 0 for each i = 1, 2, . . . , m. The variable xj is the complementary variable to the variable (AT y − c)j . Likewise the variable yi is the complementary variable to the variable (Ax− b)i . Example 4.2 We will use Theorem 4.3 to find the optimal value of the following linear programming problem. Primal: maximize cT x = −3x1 + 2x2 − x3 + 3x4 subject to x ≥ 0 and 

 x1    1 1 −1 0   x2  ≤ 5 . 3 −2 0 1 1  x3  x4 



(4.8)

Readers unfamiliar with matrix notation may wish to re-write Eq. (4.8) as a system of two linear equations with four unknowns. Since the set of points described by the constraints exists in fourdimensional space, it will be more difficult to use geometrical and graphical thinking to analyze this problem. Fortunately we can also analyze this problem’s associated dual problem.

10:57:19.

May 25, 2012

98

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

Dual: minimize bT y = 5y1 + 3y2 subject to     1 −2 −3    1 0  y1  2       −1 1  y2 ≥  −1  . 0

1

(4.9)

3

The space of points on which the dual is to be optimized exists in twodimensional space and thus is easily pictured. The constraints of the dual can be thought of as a system of inequalities. y1 − 2y2 ≥ −3 y1 ≥ 2

−y1 + y2 ≥ −1 y2 ≥ 3

The solution to this set of inequalities can be pictured as the shaded region shown in Fig. 4.5. The minimum of the objective function for the dual occurs at the point with coordinates (y1 , y2 ) = (3, 3). Thus the minimum value is 24 which will also be the maximum value of the primal problem’s objective function. At the optimal point for the dual problem, strict inequality is present in the second and third constraints since y1 = 3 > 2 −y1 + y2 = 0 > −1. By the previous theorem then the second and third components of x in the primal problem must be zero. Therefore the primal can be recast as Primal: maximize −3x1 + 3x4 subject to x1 ≥ 0, x4 ≥ 0 and   x1        x1 5 1 1 −1 0   0 = ≤ (4.10) −2 0 1 1  0  −2x1 + x4 3 x4

By inspection we must have x1 = 5 and x4 = 13. Consequently the maximum of the objective function for the primal is seen to be 24 and it occurs at the point with coordinates (x1 , x2 , x3 , x4 ) = (5, 0, 0, 13).

A consequence of the Weak Duality Theorem is that the equality bT y = cT x is sufficient for optimality of the solutions y and x to the dual and primal problems respectively. This is also a necessary condition

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

99

The Arbitrage Theorem

5.0

4.5

4.0

y2

3.5

3.0

2.5

2.0 1

2

3

4

5

6

y1

Fig. 4.5 The shaded region in the plot denotes the constrained set of points from Eq. (4.9) for the dual problem.

for optimality as will be seen in the Duality Theorem, which is stated below. This theorem was originally proved in [Gale et al. (1951)]. Theorem 4.4 (Duality Theorem) One and only one of the following four cases can be true. (i) There exist optimal solutions for both the primal and dual problems with the maximum of cT x equal to the minimum of bT y. (ii) There exists no feasible solution to the primal problem and the dual problem has feasible solutions for which the minimum of bT y approaches −∞. (iii) There exists no feasible solution to the dual problem and the primal problem has feasible solutions for which the maximum of cT x approaches ∞. (iv) Neither the primal nor the dual problem has a feasible solution.

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

100

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

The reader should observe that the four conditions are mutually exclusive, and thus only one can be true for a given pair of primal and dual problems. Before delving into the details of the proof of the Duality Theorem, we will give the reader an overview of the structure of the proof. The proof intends to show that unless (i) is true, one of the remaining three conditions is true. In order to prove the Duality Theorem we will make use of the Farkas Alternative Lemma found in [Franklin (1980)]. Lemma 4.1 (Farkas Alternative) Exactly one of the following two statements is true. Either (1) Ax ≤ b has a solution x ≥ 0, or (2) AT y ≥ 0 with bT y < 0 has a solution y ≥ 0. The proof of Farkas Alternative Lemma is not difficult, but is rather technical and thus is omitted. We proceed to use it to prove the strong version of the Duality Theorem. Proof.

First consider the general form of the primal and dual problems.

Primal: maximize cT x subject to Ax ≤ b and x ≥ 0. Dual: minimize bT y subject to AT y ≥ c and y ≥ 0. Assuming there are feasible solutions to the primal and dual problems, we can re-write the constraint of the dual as (−A)T y ≤ −c with y ≥ 0. Thus according to the constraint on the primal and the re-written constraint on the dual, the following inequalities hold for x ≥ 0 and y ≥ 0. Ax ≤ b T

(−A) y ≤ −c

(4.11) (4.12)

Furthermore if we assume that cT x = bT y then by the Weak Duality Theorem, x and y are optimal solutions for the primal and dual problems respectively. If cT x = bT y then it is also true that cT x − bT y ≤ 0.

(4.13)

Inequalities (4.11), (4.12), and (4.13) can be combined into one inequality written in the block matrix form     A 0   b x  0 −AT  ≤  −c  . (4.14) y T T c −b 0 10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

101

The Arbitrage Theorem

To recap the argument so far, we have seen that inequality (4.14) is equivalent to condition (i) of the Duality Theorem. Either condition (i) holds, or it does not. According to the Farkas Alternative (Lemma 4.1) either inequality (4.14) has a solution hx, yi ≥ 0 or the alternative      T  u u  T  A 0 c   v ≥ 0 and b −cT 0  v  < 0 0 −A −b λ λ has a solution hu, v, λi ≥ 0. The expressions u and v are vectors while λ is a scalar. We may decompose this block matrix to obtain the following system of inequalities: AT u + λc ≥ 0,

−Av − λb ≥ 0,

bT u − cT v < 0

with u ≥ 0, v ≥ 0, and λ ≥ 0. If λ > 0 then this system of inequalities is equivalent to the following system.   1 T A u ≥ −c (4.15) λ   1 A v ≤ −b (4.16) λ     1 1 −bT u > −cT v (4.17) λ λ Since u ≥ 0 and v ≥ 0 the vectors λ1 u ≥ 0 and λ1 v ≥ 0 as well. Inequality (4.16) and inequality (4.15) form a primal problem and its dual respectively. Applying the Weak Duality Theorem (Theorem 4.2), it must be the case that −bT λ1 u ≤ −cT λ1 v , contradicting inequality (4.17). Therefore we know that λ = 0. Thus the Farkas Alternative simplifies to the following system: AT u ≥ 0,

Av ≤ 0,

and bT u < cT v

where u ≥ 0 and v ≥ 0. The last (strict) inequality implies that bT u < 0 or cT v > 0 (possibly both are true, see exercise (15)). Suppose that the primal problem possesses a feasible solution x ≥ 0 and suppose that bT u < 0, then the following sequence of inequalities holds.

T

Ax ≤ b T

x A ≤b

T

T

(constraint of the primal problem) T T

(apply transpose)

x A u≤b u 0 and assuming the dual has a feasible solution, then AT y ≥ c T

T

(constraint on the dual problem)

T

y A≥c

(apply transpose) T

y (−A)v ≤ −c v < 0. However, y ≥ 0 and −Av ≥ 0 and thus yT (−Av) ≥ 0, again contradicting the last inequality above. Therefore the dual problem has no feasible solution. As before, if the primal problem has no feasible solution then we are in case (iv). If the primal problem has a feasible solution x then Ax + Aλv = A(x + λv) ≤ b + 0 = b where x + λv ≥ 0 for all λ ≥ 0. Thus x + λv is a feasible solution to the primal problem for all λ ≥ 0. However, lim cT (x + λv) = cT x + lim (λcT v) = ∞

λ→∞

λ→∞

which implies the primal problem has no optimal solution (this is situation (iii)). Thus the strong version of the Duality Theorem is established.  Thus far in this chapter we have developed a great deal of background knowledge in linear programming and duality. Linear programming is a vast field of study in its own right. The previous material is merely an introduction intended to enable us to prove the Fundamental Theorem of Finance in the next section.

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

103

The Arbitrage Theorem

4.4

BC8495/Chp. 4

The Fundamental Theorem of Finance

Consider an experiment with m possible outcomes numbered 1 through m. Suppose we can place n wagers (numbered 1 through n) on the outcomes of the experiment. Let rji be the return for a unit bet on wager i when the outcome of the experiment is j. The vector x = hx1 , x2 , . . . , xn i is called a betting strategy. Component xi is the amount placed on wager i. In the context of finance, the components of vector x may be thought of as the long and short positions taken on various potential investments. Consistent with the theory of linear programming which has been developed in the chapter, we will assume x ≥ 0, but the non-negativity assumption does not hamper the analysis, since a primal linear program with the non-negativity constraint is equivalent to another primal linear program without it. The composite return from this betting strategy when outcome j occurs is then Pn i=1 xi rji . The Arbitrage Theorem states that the probabilities of the m outcomes of the experiment are such that for each bet the expected value of the payoff is zero, or there exists a betting strategy for which the payoff is positive regardless of the outcome of the experiment. Lemma 4.2 false.

If one of the following statements is true, the other must be

(1) There is a vector of probabilities p = hp1 , p2 , . . . , pm i for which m X

pj rji = 0,

for each i = 1, 2, . . . , n.

j=1

(2) There is a betting strategy x = (x1 , x2 , . . . , xn ) for which n X

xi rji > 0,

for each j = 1, 2, . . . , m.

i=1

Proof. Suppose (1) is true and let x = (x1 , x2 , . . . , xn ) be any betting strategy. m X

pj

j=1

n X

xi rji =

i=1

=

m X n X j=1 i=1 n X m X i=1 j=1

10:57:19.

xi pj rji xi pj rji

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

104

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

= =

n X

i=1 n X

xi

m X

pj rji

j=1

xi (0) = 0.

i=1

Pm Since each pj ≥ 0 and j=1 pj = 1 then for some j ∈ {1, 2, . . . , m} it must Pn be true that i=1 xi rji ≤ 0 which implies statement (2) is false. Now suppose (2) is true. If statement (1) is also true then (2) is false, a contradiction. Thus at most one of the two statements is true.  The situation in which statement (1) is true may be interpreted as stating that all betting strategies have an expected return of zero since " n # m n X X X pj xi rji = E xi rji = 0. j=1

i=1

i=1

Consider a collection of n stocks (or other investment instruments) whose prices are denoted S i for i = 1, 2, . . . , n. In addition to these stocks suppose an investor can deposit cash in an amount denoted S 0 in a riskfree savings account earning simple interest at rate r ≥ 0. After one unit of time passes the stock prices will be in one of m possible states denoted ω1 , ω2 , . . . , ωm . One of the states will be achieved at random after one unit of time passes. Let S i (0) represent the price of stock i at time t = 0 and let S i (ωj ) be the price of stock i at time t = 1 in state ωj . Since S 0 is the initial amount deposited in a risk-free savings account then S 0 (ωj ) = (1 + r)S 0 for all j = 1, 2, . . . , m. In other words, the amount due in the savings account is independent of the future state. Let pj is the probability of future state ωj for j = 1, 2, . . . , m, then if S i (0) =

1 X pj S i (ωj ) 1 + r j=1

then vector p = hp1 , p2 , . . . , pm i is called a risk-neutral probability measure on the set Ω = {ω1 , ω2 , . . . , ωm }. Naturally it is still assumed that P pj ≥ 0 for j = 1, 2, . . . , m and m j=1 pj = 1. Now we may state and prove the Arbitrage Theorem which is also called by some authors the First Fundamental Theorem of Asset Pricing. Theorem 4.5 (Arbitrage Theorem) A risk-neutral probability measure exists if and only if there is no arbitrage.

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

BC8495/Chp. 4

105

Proof. Without loss of generality assume that S 0 = 1 (or choose a unit of currency for which S 0 = 1). We may also safely ignore any state ωj which has a zero probability of occurrence, in other words we will assume pj > 0 for j = 1, 2, . . . , m. Let yi be the number of shares of stock i bought or sold at time t = 0 for i = 1, 2, . . . , n and let y0 be the units of cash initially deposited in the risk-free savings account. Let vector y = hy0 , y1 , . . . , yn i (the vector of unknowns) and let vector S(·) = hS 0 (·), S 1 (·), . . . , S n (·)i. The problem is then to minimize (S(0))T y subject to the m constraints (S(ωj ))T y ≥ 0 for j = 1, 2, . . . , m. This linear program is a dual problem. The constraints can be expressed in matrix form as 

    S 0 (ω1 ) S 1 (ω1 ) · · · S n (ω1 ) y0 0  S 0 (ω2 ) S 1 (ω2 ) · · · S n (ω2 )   y1   0       AT y =   .  ≥  . . .. .. ..    ..   ..  . . . 0 1 n S (ωm ) S (ωm ) · · · S (ωm ) yn 0 The program is feasible since yi = 0 for i = 1, 2, . . . , n satisfies all the constraints. The solution y = 0 also implies that the minimum value of the objective function for the dual is less than or equal to 0. Suppose there is a feasible solution y∗ for which (S(0))T y∗ = c < 0. This is the case of Type A arbitrage, since the initial cost of the portfolio is negative, meaning there is a net cash flow to the investor. Multiplying y∗ by a scalar M > 1 we can obtain another feasible solution for which the objective function of the dual has value cM < c < 0. Consequently the dual is feasible but unbounded. Therefore there is no Type A arbitrage if and only if the minimum of the dual problem is 0. If Type B arbitrage exists then for all j the expressions (S(ωj ))T y are non-negative and for some j ∈ {1, 2, . . . , m} the inequality (S(ωj ))T y > 0 holds, and the minimum of the objective function for the dual problem is 0. Thus there is no Type B arbitrage if and only if the minimum of (S(0))T y = 0 and (S(ωj ))T y = 0 for j = 1, 2, . . . , m. In this case the constraint of the dual problem is AT y = 0. The corresponding primal linear programming problem with vector of unknowns p = hp1 , p2 , . . . , pm i is that of maximizing 0T p = 0 subject to the constraints     0  0 S (0) p1 S (ω1 ) S 0 (ω2 ) · · · S 0 (ωm )  S 1 (ω1 ) S 1 (ω2 ) · · · S 1 (ωm )   p2   S 1 (0)       Ap =   .  =  .  .. .. .. . .      .  . . . . n n n n pm S (0) S (ω1 ) S (ω2 ) · · · S (ωm ) 10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

106

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

where p ≥ 0. The constraints must be of equality type since the decision variables of the dual problem are unrestricted in sign. The primal problem’s objective function is trivially maximized at value 0 and thus if there exists a feasible solution to the primal problem, it is an optimal solution to the primal. If there is no Type A arbitrage then the minimum of the dual is 0. By Theorem 4.4 there is an optimal, feasible solution to the primal problem for which the maximum is 0. If there is no Type B arbitrage then likewise the minimum of the dual is 0 and again Theorem 4.4 implies there is an optimal, feasible solution p∗ to the primal problem for which the maximum is 0. Consider the constraint of the primal problem corresponding to the cash deposit. hS 0 (ω1 ), S 0 (ω2 ), . . . , S 0 (ωm )iT hp∗1 , p∗2 , . . . , p∗m i = S 0 (0) (1 + r)h1, 1, . . . , 1iT hp∗1 , p∗2 , . . . , p∗m i = 1 m X (1 + r) p∗j = 1 j=1

Thus (1 + r)p∗ is a risk-neutral probability measure. To establish the converse, assume a risk-neutral probability measure p > 0 exists. In this case the primal linear program is feasible with a maximum of 0. According to Theorem 4.4 there exists an optimal solution to the dual problem whose minimum is 0. This implies there is no Type A arbitrage. Since p > 0 then by Theorem 4.3 we have ST (ωj )y = 0 for j = 1, 2, . . . , m which indicates there is no Type B arbitrage. 

4.5

Exercises

(1) Suppose the odds against the three possible outcomes of an experiment are as given in the table below. Outcome A B C

Odds 2:1 3:1 1:1

Find a betting strategy which produces a positive net profit regardless of the outcome of the experiment.

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

BC8495/Chp. 4

107

(2) Sketch the region in the plane which satisfies the following inequalities: x1 + 3x2 ≥ 6 3x1 + x2 ≥ 6 x1 ≥ 0 x2 ≥ 0

(3) (4) (5)

(6) (7) (8)

Is the region convex? What are the coordinates of the points at the corners of the region? Minimize the objective function x1 + x2 on the region described in exercise (2). Find inequality constraints which will describe the rectangular region with corners at (0, 0), (2, 0), (2, 3), and (0, 3). Introduce slack variables in the inequality constraints found in exercise (4) to produce equality constraints Ax = b. What are A and b? Minimize the objective function x1 + x2 + 2x3 on the set where x1 + 2x2 + 3x3 ≥ 15 and xi ≥ 0 for i = 1, 2, 3. Minimize the objective function 7x2 + 9x3 subject to the constraints x1 + x2 + x3 ≥ 5 and x2 + x3 + 2x4 ≥ 1 with xi ≥ 0 for i = 1, 2, 3. Consider the standard linear program of maximizing 2x1 −3x2 subject to the constraints: 3x1 + 5x2 ≤ 4

−2x1 + 4x2 ≤ −3. Show that this linear program is not feasible. (9) Consider the standard linear program of maximizing 2x1 −3x2 subject to the constraints: −2x1 + x2 ≤ −3 x1 − 2x2 ≤ 3.

Show that this linear program is feasible, but has no optimal solution. (10) State the dual of the primal standard linear program in Exercise (9). Since the primal problem has no optimal solution, what does the Duality Theorem (Th. 4.4) imply about the dual problem? Verify this condition directly from the dual problem. (11) Write down the dual problem of the following linear programming problem: minimize x1 + x2 + x3 subject to 2x1 + x2 = 4 and x3 ≤ 6 10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

108

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 4

An Undergraduate Introduction to Financial Mathematics

with xi ≥ 0 for i = 1, 2, 3. (12) Find the solutions to the primal and dual problems of exercise (11). (13) Write down the dual problem to the following linear programming problem: maximize 2y1 + 4y3 subject to y1 + y2 ≤ 1 and y2 + 2y3 ≤ 1. (14) Find the solutions to the primal and dual problems of exercise (13). (15) Show that if a < b then either a < 0 or b > 0. (16) Let x be a non-negative vector with n components, y be a non-negative vector of m components, and A be a fixed m × n matrix. Define the function φ(x, y) = cT x + bT y − xT AT y where c is a fixed vector of n components and b is a fixed vector of m ˆ ) is a saddle point of φ(x, y) if components. We shall say that (ˆ x, y ˆ ) ≤ φ(ˆ ˆ ) ≤ φ(ˆ φ(x, y x, y x, y) for all non-negative vectors x and y. Consider the primal/dual pair of linear programs, Primal: maximize cT x subject to Ax ≤ b and x ≥ 0. Dual: minimize bT y subject to AT y ≥ c and y ≥ 0.

ˆ and y ˆ are optimal solutions to the primal and dual Show that if x ˆ ) is a saddle point of φ(x, y). problems respectively, then (ˆ x, y ˆ ) is a saddle point of φ(x, y) defined in exercise (16) (17) Show that if (ˆ x, y ˆ and y ˆ are optimal solutions of their corresponding primal and then x dual problems. (18) A metal worker can make knives or scissors. A knife takes 15 minutes to make and a pair of scissors takes 25 minutes to make. Each knife requires 6 ounces of steel to make, while a pair of scissors requires 8 ounces of steel. The metal worker can sell each knife for a profit of $2.25 and each pair of scissors for a profit of $3.50. If the worker has 40 hours to make knives and scissors and 52 pounds of steel, how many knives and scissors should be made to maximize the worker’s profit? (19) A farmer has 440 acres of land on which two crops may be planted, corn and saw grass. Each acre of corn costs $75 to plant and each acre of saw grass costs $50 to plant. Each acre of corn planted requires 110 bushels of storage while each acres of saw grass planted requires 30 bushels of storage. The farmer has access to 30,000 bushels of storage space. Each acre of corn will generate a profit of $100 and each acre

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

The Arbitrage Theorem

BC8495/Chp. 4

109

of saw grass will generate a profit of $80. The farmer has $50,000 with which to plant. How much of each crop should be planted to maximize the profit of the farmer? (20) A company sells two sizes of tablet personal computer, a 7-inch model and a 10-inch model. The tablets can be manufactured in either China or Japan. The manufacturing facility in China has an operating budget of $50,000 per day and can produce a total of 250 tablet PCs per day and the facility in Japan has an operating budget of $60,000 per day and can make 290 tablet PCs per day. In China it costs $175 to make a 7-inch tablet and $250 to make a 10-inch tablet. In Japan it costs $205 to make a 7-inch tablet and $255 to make a 10-inch tablet. Each 7-inch tablet sold generates a profit of $65 and each 10-inch tablet sold brings in a profit of $85. The company needs no more than 250 of the 7-inch tablet and 290 of the 10-inch tablet each day. Describe the linear program of finding the maximum profit in the form of a primal problem.

10:57:19.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

This page intentionally left blank

10:57:19.

110

BC8495/Chp. 4

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

Chapter 5

Random Walks and Brownian Motion

In this chapter we will introduce and explain some of the concepts surrounding the probabilistic models used to capture the behavior of stock, security, option, and index prices. Random walks are related to properties of discrete random variables studied in Chap. 2. Brownian motion can be thought of as a continuous random process which is the limiting case of a random walk as the time and space steps become infinitesimally small. Topics covered here could be expanded into an entire book of their own. For our purposes we will explore just enough of the stochastic calculus to provide some justification for Itˆo’s Lemma, the main result of this chapter. 5.1

Intuitive Idea of a Random Walk

Earlier in Chapter 3 we analyzed the motion of a particle taking a discrete step along the x-axis during every “tick” of the clock. At first it was revealed that the location of the particle relative to the origin followed a binomial distribution. As the step size was decreased, the probability of the particle lying in a particular interval was seen to obey the normal distribution. In this chapter we will extend this discussion to a more general setting. Imagine a person standing at the origin of the real number line. They will flip a coin. For every time the coin lands on heads, they will take a unit step to the right (the positive direction). For every time the coin lands on tails, they will move a unit step to the left (the negative direction). The evolution of this stochastic discrete dynamical system is called a random walk. A plot of the person’s location versus the number of times the coin has been flipped might resemble the graph in Fig. 5.1. Random walks do not require that the probabilities of moving left or right be equal (i.e., the coin does not have to be fair). The magnitude of the movement at each stage does

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

112

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

10

20

30

40

50

-2

-4

-6

-8

-10 Fig. 5.1

A realization of a random walk on the real number line.

not have to be a single unit. The person could move left or right depending on the value of a random number chosen from a normal distribution for example. The bias in choosing to move left or right and the possibility of moving different distances during different steps are generalizations not present in the discussion presented in Chapter 3. In the remainder of this chapter we will apply the concept of the random walk to modeling the movement of the value of a stock or other security.

5.2

Discrete Random Walks

A simple first application of a random walk is the situation of a person owning a share of stock whose current value is denoted by S. At discrete intervals (for example, once per day) the stock’s value can increase or decrease by one unit, in other words if its current value on day n is S(n), then tomorrow the value will either be S(n+1) = S(n)+1 or S(n+1) = S(n)−1. To keep the following discussion simple, the probability of increase in value will be p = 1/2 and consequently the probability of decrease is the same. This assumption implies the random walk is unbiased. Other authors use the term symmetric to describe this behavior (for example [Grimmett and Stirzaker (1982)]). If we think of Xn as a random variable which takes on value 1 with probability p = 1/2 and value −1 with the same probability, 10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

113

Random Walks and Brownian Motion

we can think of the N th state of the random walk as a partial sum where for N > 0, S(N ) = S(0) + X1 + X2 + · · · + XN .

(5.1)

The random variable Xn is a functions of the Bernoulli random variable discussed earlier (recall that we had defined a Bernoulli random variable to have outcomes 0 and 1). S(N ) can be defined inductively by the formula S(N ) = S(N − 1) + XN for N > 0 where S(0) is a specified initial state of the walk. The mapping S : N ∪ {0} → Z is an example of a discrete random process or discrete stochastic process. If we assume that the random variables Xi and Xj have the same distribution and are independent for i 6= j, then we see that the transitions between states in the random walk, S(N )−S(N −1) = XN are independent random variables. In fact given any integers 0 ≤ k1 < k2 ≤ k3 < k4 the expressions Sk2 ,k1 = S(k2 ) − S(k1 ) = Sk4 ,k3 = S(k4 ) − S(k3 ) =

k2 X

Xi

i=k1 +1 k4 X

Xi

i=k3 +1

are themselves random variables called increments of the random walk S(k). The increments are independent random variables since they depend on different sets of independent Bernoulli random variables. If the initial state of the random walk is S(0) and n ≥ 0 steps are taken, where 0 ≤ k ≤ n of the steps are in the positive direction and n − k ≥ 0 are in the negative direction, then P (S(n) = S(0) + k − (n − k)) = P (S(n) − S(0) = 2k − n)    n n 1 = . k 2 The last expression is the familiar probability formula for a binomial random variable with p = 1/2. If S(0) 6= 0 then we can perform a change of variable via T (n) = S(n) − S(0) for n = 0, 1, . . . and define a related random walk where T (0) = 0. The transitions in the states of the new random walk are independent, identically distributed random variables just as before. Likewise we see

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

114

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

that    n n 1 P (T (n) = 2k − n) = . k 2

(5.2)

Thus without loss of generality we will assume that S(0) = 0. This is known as the spatial homogeneity property of the random walk. Suppose two random walks are described as follows. S(0) = 0 and S(k) evolves for n additional steps. T (m) = 0 and T (k) also evolves for n additional steps. The first m steps of the process T are unimportant so long as T (m) = 0. Making use of the notion of increments of the random walks Sn,0 =

n X

Xi

and Tm+n,m =

i=1

m+n X

Xi

i=m+1

and thus P (S(n) = N | S(0) = 0) = P (T (m + n) = N | T (m) = 0) . This is called the temporal homogeneity property of the symmetric random walk. Now that we have settled on a starting state for the random walk, what states can be visited in n steps? The next lemma provides the answer and the respective probabilities of reaching these states. Lemma 5.1 S(0) = 0,

For the random walk defined in Eq. (5.1) with initial state

(1) P (S(n) = m) = 0 if |m| > n, (2) P (S(n) = m) = 0 if n + m  is  odd, n 1 n (3) P (S(n) = m) = (n+m)/2 2 , otherwise.

Proof.

(1) According to Eq. (5.1), −n ≤ S(n) ≤ n, thus if |m| > n the partial sum cannot attain this value. (2) If we let m = 2k − n in Eq. (5.2) then n + m = 2k which implies n + m is even, contradicting the assumption that n + m is odd. (3) We may assume that n + m is even and thus the result is shown by Eq. (5.2).  The state of the random walk after n steps can be summarized in the following theorem.

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

Theorem 5.1 S(0) = 0,

BC8495/Chp. 5

115

For the random walk defined in Eq. (5.1) with initial state E [S(n)] = 0

and

V (S(n)) = n.

Proof. E [S(n)] = E [S(0)] + E [X1 ] + E [X2 ] + · · · + E [Xn ] =0

by Theorem 2.4 since E [Xi ] = 0 for i = 1, 2, . . . , n. According to the assumption that Xi and Xj are independent when i 6= j we have V (S(n)) = V (S(0)) +

n X

V (Xi ) = n

i=1

by Theorem 2.7.



A closely related result holds for the increments of a discrete symmetric random walk. Theorem 5.2 Suppose k1 < k2 are nonnegative integers and Sk2 ,k1 is an increment of the random walk defined in Eq. (5.1) then E [Sk2 ,k1 ] = 0 Proof.

and

V (Sk2 ,k1 ) = k2 − k1 .

See exercise (1).



Coupling the idea of conditional expected value and the discrete symmetric random walk, we can consider the following expectation. Suppose m and n are integers with 0 ≤ m < n then E [S(n) | X1 X2 · · · Xm ] denotes the expected value of the discrete symmetric random walk S(i) given the outcomes of the first m Bernoulli experiments driving the random walk. As before we will assume S(0) = 0. E [S(n) | X1 X2 · · · Xm ] = E [S(n) − S(m) + S(m) | X1 X2 · · · Xm ] = E [S(n) − S(m) | X1 X2 · · · Xm ] + E [S(m) | X1 X2 · · · Xm ]

= E [Sn,m | X1 X2 · · · Xm ] + S(m)

= S(m)

Several key thoughts have been used to establish this equality. On the second line of the equation, the linearity property of conditional expectation

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

116

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

was used. On the third line, we used the fact that knowledge of the outcomes of the first m Bernoulli experiments determines S(m). The increment Sn,m is independent of the first m outcomes of the Bernoulli experiment and thus E [Sn,m | X1 X2 · · · Xm ] = E [Sn,m ] = 0. If m = n the same result holds if we generalize the notion of the increment of a random walk to include Sm,m = 0. The property of random process that E [S(n) | X1 X2 · · · Xm ] = S(m) for all integers 0 ≤ m ≤ n is called the martingale property. Stochastic processes which have the martingale property, or for short are simply called martingales, have no tendency to rise or fall as they evolve. Though the intuitive definition of a martingale given above is sufficient for the purposes of this introductory text, more precise and technical definitions can be found in sources such as [Shreve (2004b)]. This section concludes with the discussion of one more property of stochastic processes. The quadratic variation of a discrete random process M (n) is denoted [M, M ] (n) and is defined as [M, M ] (n) =

n X i=1

2

(M (i) − M (i − 1)) .

(5.3)

For the discrete symmetric random walk S(n) the quadratic variation is [S, S] (n) = n for n ∈ N. This is true because each simple increment S(i) − S(i − 1) = ±1. The reader will note that the variance of the discrete symmetric random walk and its quadratic variation agree, this is not true for all stochastic processes. One reason to expect a difference in general is that the variance of a random process involves the expected value of the squared deviation of the process from its mean, while the quadratic variation does not involve expected value.

5.3

First Step Analysis

Consider an un-restricted random walk {S(j)}nj=0 which possesses the property that S(k) = 0 for some k between 0 and n. An example is graphed in Fig. 5.2. If we reflect the path of the random walk across the j-axis for n ˆ k < j ≤ n, we obtain another random walk {S(j)} j=0 with the properties 10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

BC8495/Chp. 5

117

S H jL

S H 0L

A j -A

Fig. 5.2 The probabilities that an unbiased random walk will follow either the solid or dashed (reflected) path are equal.

that ˆ S(j) =



S(j) for j = 0, 1, . . . , k −S(j) for j = k + 1, k + 2, . . . , n.

The probability that the original random walk ended up in state A equals the probability the reflected random walk ended up in state −A. To see this suppose the original random walk starts at 0 (and hence the reflected random walk also starts at 0). Since downward steps occur with equal probability to upward steps, then   ˆ P (S(n) = A) = P S(n) = −A . Random walks possess the Markov property, meaning that the history of movements of the random walk is irrelevant to the next random step. In other words the random walk cannot “remember” how it arrived at a particular state. The current state and only the current state influences the next state of the random walk. Therefore we can think of partitioning a random walk which crosses the j-axis into two segments, the initial segment from j = 0 to j = k at which time the random walk is in state 0, and the

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

118

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

final segment from j = k + 1 to j = n. Hence by using the Markov property   ˆ P (S(n) = A) = P S(n) = −A     ˆ P (S(k) = 0) P (T (n − k) = A) = P S(k) = 0 P Tˆ(n − k) = −A   = P (S(k) = 0) P Tˆ(n − k) = −A   P (T (n − k) = A) = P Tˆ(n − k) = −A . The last equation is true by the special case discussed above. Hence we have established the following theorem.

Theorem 5.3 If {S(j)}nj=0 is an unbiased random walk with initial state S(0) = i, if S(k) = 0 for some 0 ≤ k ≤ n, and if |A − i| ≤ n and |A + i| ≤ n then P (S(n) = A | S(0) = i) = P (S(n) = −A | S(0) = i) .

(5.4)

Note that according to Lemma 5.1 these probabilities could be 0 if n + A − i (and consequently n − A − i) are odd. In the previous discussion the random walk was free to move in either the positive or negative direction any integer amount as the number of steps increased. Suppose that bounds are placed in the path the particle may follow. Once again, for the sake of simplicity, we will assume the initial state of the random walk is positive, i.e., S(0) > 0 and we will assume the random walk has a lower boundary of 0. If the state of the random walk reaches the lower boundary in a finite number of steps then the state remains at that boundary value. In the study of random walks, this is known as an absorbing boundary condition. For a gambler, “going broke” can be thought of as an absorbing boundary condition. In a financial setting we may think of the value of a security as following a random walk (though in reality a much more complicated one than the type we are exploring here). If the value of the security drops to 0, the security becomes worthless to the owner. We must further generalize our discussion of random walks to include the absorbing boundary. There are two questions related to this situation that we wish to explore. (1) What is the probability that the state of the random walk crosses a threshold value of A > 0 before it hits the boundary at 0 (and hence remains there)?

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

119

Random Walks and Brownian Motion

(2) What is the expected value of the number of steps which will elapse before the state of the random variable first crosses the A > 0 threshold? Figure 5.3 shows a random walk of a stock whose value was initially 10. This situation has been so simplified and abstracted from the reality 25

20

15

10

5

10

20

30

40

50

Fig. 5.3 A realization of a simple random walk attempting to capture the changes in the price of a stock.

of the stock market that it is of little use in making investment decisions, but serves as an instructive example of the use of the concept of a random walk. The answers to the two primary questions will be found following the lines of reasoning laid out in [Kijima (2003)] and [Steele (2001)]. Since the boundary at 0 is absorbing we must keep track of the smallest value which S(n) takes on. Thus we define Smin (n) = min{S(k) : 0 ≤ k ≤ n}. If i is the smallest non-negative integer such that mi = 0 (and hence that S(i) = 0) then by the absorbing boundary condition S(k) = 0 for all k ≥ i. Our attention is now focused on developing an understanding of the following conditional probability: P (S(n) = A ∧ Smin (n) > 0 | S(0) = i) . We will assume that A > 0 and i > 0. This probability should depend on the three parameters A, i, and n. A formula for this conditional probability is derived in the proof of the following lemma.

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

120

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

Lemma 5.2 Suppose a random walk has the form described in Eq. (5.1) in which the Xi for i = 1, 2, . . . are independent, identically distributed random variables taking on the values ±1, each with probability p = 1/2. Suppose further that the boundary at 0 is absorbing, then if A, i > 0, P (S(n) = A ∧ Smin (n) > 0 | S(0) = i)      n n n 1 = − , (n + A − i)/2 (n − A − i)/2 2

(5.5)

provided |A − i| ≤ n, |A + i| ≤ n, and n + A − i is even. Proof. In order to prove the lemma we will start by considering a random walk with no boundary, that is, the random variable S(n) has an initial state of S(0) = i > 0 and S(k) is allowed to wander into negative territory (and back) arbitrarily. In this situation P (S(n) = A | S(0) = i)

= P (S(n) = A ∧ Smin (n) > 0 | S(0) = i)

+ P (S(n) = A ∧ Smin (n) ≤ 0 | S(0) = i)

by the Addition Rule (Theorem 2.1). Now let us consider the probability on the left-hand side of the equation. It possesses no boundary condition and by the spatial homogeneity of the random walk P (S(n) = A | S(0) = i) = P (T (n) = A − i) where {T (j)}nj=0 is an unbiased random walk with initial state T (0) = 0. Hence by Lemma 5.1, P (T (n) = A − i) = 0 unless n + A − i is even and |A − i| ≤ n, in which case P (S(n) = A | S(0) = i) =



  n n 1 . (n + A − i)/2 2

On the other hand if the random walk starts at a positive state i and finishes at −A < 0 then it is certain that Smin (n) ≤ 0. Consequently P (S(n) = A ∧ Smin (n) ≤ 0 | S(0) = i) = P (S(n) = −A | S(0) = i)    n n 1 = (n − A − i)/2 2 10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

BC8495/Chp. 5

121

provided |A + i| ≤ n and n − A − i is even. Finally we are able to determine that P (S(n) = A ∧ Smin (n) > 0 | S(0) = i)    n    n n 1 n 1 = − , (n + A − i)/2 2 (n − A − i)/2 2 which is equivalent to the expression in Eq. (5.5).



Lemma 5.2 provides an answer to the first of the two questions posed. Example 5.1 For an unbiased random walk with initial state S(0) = 10, what is the probability that S(50) = 16 and S(n) > 0 for n = 0, 1, . . . , 50? Making use of the formula found in Eq. (5.5) we have     50 50 P (S(50) = 16 ∧ m50 > 0 | S(0) = 10) = − 2−50 28 12 ≈ 0.0787178. Since the current notation is so bulky, we will define the function      n n n 1 fA,i (n) = − (n + A − i)/2 (n − A − i)/2 2 and use it from now on. We will call the first time that the random walk S(n) equals A the stopping time, denoted by TA . Mathematically this can be expressed as TA =

min

n∈{0,1,...}

{n | S(n) = A}.

The stopping time is a random variable. A stopping time for the discrete symmetric random walk can depend on the evolution of the walk S(0), S(1), . . . , S(n) and cannot depend on future values of the walk. One must be able to decide whether the stopping time of a random process has arrived based on the known history of the process and not on the future of the process. In the remainder of this section we will explore the probability that TA takes on the various nonnegative integer values given the initial state of the random walk is S(0) = i > 0, parameter A > 0, and the boundary at 0 is absorbing. Suppose the initial state of the random walk S(0) = i > A > 0, then due to the spatial homogeneity of the random walk P (S(n) = A ∧ mn−1 > A | S(0) = i)

= P (S(n) = 0 ∧ mn−1 > 0 | S(0) = i − A) . 10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

122

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

T0 = n if and only if S(n − 1) = 1, mn−1 > 0 and Xn = −1. Therefore P (T0 = n | S(0) = i − A)

= P (Xn = −1 ∧ S(n − 1) = 1 ∧ mn−1 > 0 | S(0) = i − A) 1 = P (S(n − 1) = 1 ∧ mn−1 > 0 | S(0) = i − A) 2 1 = f1,(i−A) (n − 1). 2

Consequently P (TA = n | S(0) = i) = 12 f1,(i−A) (n − 1) as well. Next we take up the case in which the initial state of the random walk 0 < S(0) = i < A. Since the stopping time is the first time that S(n) = A then we can think of the spatial domain of the random walk as having two absorbing boundaries: the usual one at S = 0 and a new one at S = A. The analysis of this situation is modeled after the discussion in the first two chapters of [Redner (2001)]. The symbol pi→A will denote any random walk {S(j)} in the discrete interval [0, A] starting at i > 0, terminating at A, and which avoids 0. The symbol Ppi→A will denote the probability that the random walk starting at S(0) = i follows pi→A . Finally the symbol PA (i) will denote the probability that a random walk which starts at S(0) = i will achieve state S = A while avoiding the state S = 0. By the Addition Rule for probability X PA (i) = Ppi→A . pi→A

Since at each step of the random walk, the increment to the left or right is independent of the previous increments, PA (i) = P (S(1) = i − 1 | S(0) = i) PA (i − 1)

+ P (S(1) = i + 1 | S(0) = i) PA (i + 1) 1 1 = PA (i − 1) + PA (i + 1) 2 2

and thus we have PA (i − 1) − 2PA (i) + PA (i + 1) = 0.

(5.6)

Equation (5.6) is the discrete difference equation approximation to the second derivative [Burden and Faires (2005)]. It is also sometimes referred to as the discrete Laplacian equation. Since we know a priori that PA (0) = 0 and PA (A) = 1 then we may derive a system of linear equations 10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

123

Random Walks and Brownian Motion

from Eq. (5.6) whose solution will give us PA (i) for i = 1, 2, . . . , A − 1. In matrix form the system of equations resembles:      PA (1) 0 1 −2 1 0 · · · 0 0 0  0 1 −2 1 · · · 0 0 0   PA (2)   0       .  =  . . .. ..    ..     ..  . . 0 0 0 0 · · · 1 −2 1 PA (A − 1) 0

We can use a bit of intuition about Eq. (5.6) to hypothesize and then confirm a form of the solution to this matrix equation. The left-hand side of Eq. (5.6) represents an approximation to the second derivative. The right-hand side of the equation is zero, thus the solution is likely to be a linear function of i. If PA (i) = ai + b then upon substituting this linear function in Eq. (5.6) we obtain a(i − 1) + b − 2(ai + b) + a(i + 1) + b = 0

confirming that a linear function will express PA (i). We can use the two boundary conditions to determine the proper values for the coefficients a and b. PA (0) = b = 0 PA (A) = aA = 1

=⇒

a=

1 A

Consequently PA (i) = i/A. By the same type of argument we can also determine that the probability of a random walk finishing at state 0 while avoiding state A > 0 and starting at state 0 ≤ i ≤ A is P0 (i) = 1 − i/A. We can summarize these results in the following theorem. Theorem 5.4 Suppose a random walk has the form described in Eq. (5.1) in which the Xi for i = 1, 2, . . . are independent, identically distributed random variables taking on the values ±1, each with probability p = 1/2. Suppose further that the boundaries at 0 and A are absorbing, then if 0 ≤ S(0) = i ≤ A, (1) the probability that the random walk achieves state A without achieving state 0 is PA (i) = i/A, (2) the probability that the random walk achieves state 0 without achieving state A is P0 (i) = 1 − i/A. Before answering the question as to the expected stopping time for the random walk attaining state A, we will explore the simpler issue of the

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

124

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

stopping time of reaching either of the two absorbing boundaries. Let B = {0, A} represent the discrete boundary set of the one-dimensional, discrete random walk. To help in the derivation we will define the following quantities: ωpi→B : the exit time of the random walk which starts at S(0) = i, where 0 ≤ i ≤ A and which follows path pi→B . ΩB (i): the expected value of the exit time for a random walk which starts at S(0) = i, where 0 ≤ i ≤ A. To be clear ΩB (i) = E [TB | 0 ≤ S(0) = i ≤ A] . By the definition of expected value, X ΩB (i) = Ppi→B ωpi→B . pi→B

Certainly it is the case that if the random walk starts on the boundary the expected stopping time is 0. Thus it is true that ΩB (0) = ΩB (A) = 0. There is also a recursive relationship between the stopping times at neighboring starting points in space. The path from i → B can be decomposed into paths from (i − 1) → B and (i + 1) → B with the addition of a single step. The expected value of the exit time of a random walk starting at i is one more than the expected value of a random walk starting at i ± 1. Therefore ΩB (i) =

1 1 (1 + ΩB (i − 1)) + (1 + ΩB (i + 1)) 2 2

which is equivalent to the equation below. ΩB (i − 1) − 2ΩB (i) + ΩB (i + 1) = −2

(5.7)

Equation (5.7) is often called the Poisson equation for the stopping time. From the Poisson equation we can derive a system of linear equations for the stopping times which are parametrized by the various initial conditions of the random walk. In matrix form the system of equations has the form: 

    1 −2 1 0 · · · 0 0 0 ΩB (1) −2  0 1 −2 1 · · · 0 0 0   ΩB (2)   −2       .  =  . . ..   .. .  ..     . . .  0 0 0 0 · · · 1 −2 1 ΩB (A − 1) −2 10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

BC8495/Chp. 5

125

The matrix generated by the Poisson equation (as well as the earlier matrix generated by the Laplacian equation) are known as tridiagonal matrices due to the fact that all entries in the matrices are zero except for possibly the entries on the diagonal, the first sub-diagonal, and the first super-diagonal. There are many methods for solving tridiagonal linear systems. Some of the simplest methods involve Gaussian elimination followed by backwards substitution. This technique as well as many others is explained in detail in [Golub and Van Loan (1989)]. An alternative to the matrix solution technique is to observe that the left-hand side of Eq. (5.7) is an approximation to the second derivative of ΩB with respect to i. A reasonable hypothesis is that ΩB (i) is a quadratic function of i, since the second derivative is the constant −2. Suppose ΩB (i) = ai2 + bi + c where a, b, and c are constants. Substituting this quadratic expression into Eq. (5.7) produces −2 = a(i − 1)2 + b(i − 1) + c − 2(ai2 + bi + c) + a(i + 1)2 + b(i + 1) + c = 2a

which implies a = −1. Recall that on the boundary ΩB (0) = 0 = ΩB (A) and therefore 0=c 0 = −A2 + bA

(the case i = 0) (the case i = A) ,

therefore b = A. Hence the solution to the Poisson equation for the expected value of the exit time is ΩB (i) = i(A − i) for i = 0, 1, . . . , A. The results are summarized in the following theorem. Theorem 5.5 Suppose a random walk has the form described in Eq. (5.1) in which the Xi for i = 1, 2, . . . are independent, identically distributed random variables taking on the values ±1, each with probability p = 1/2. Suppose further that the boundaries at 0 and A are absorbing, then if 0 ≤ S(0) = i ≤ A the random walk intersects the boundary (S = 0 or S = A) after a mean number of steps given by the formula ΩB (i) = i(A − i).

(5.8)

Example 5.2 Suppose an unbiased random walk takes place on the discrete interval {0, 1, 2, 3, 4} for which the boundaries at 0 and 4 are absorbing. If S(0) = i then the expected value of the stopping time of the random walk is

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

126

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

i ΩB (i)

0 0

1 3

2 4

3 3

4 0

Theorem 5.5 yields a formula for the mean number of steps required for an unbiased random walk to reach either boundary point of the interval [0, A] based on the starting point. Now we want to determine a related quantity, the conditional exit time. This is the average number of steps required for an unbiased random walk to reach the boundary at A while avoiding the other boundary at 0. The symbol ΩA (i) will denote the expected value of the conditional exit time of a random walk in the initial state S(0) = i which exits at boundary A while avoiding the boundary at 0. By definition this conditional exit time is X X Ppi→A ωpi→A Ppi→A ωpi→A ΩA (i) =

pi→A

X

Ppi→A

pi→A

=

pi→A

PA (i)

.

Note that the sums above are taken over all random walks which start at S(0) = i and exit at A and avoid 0. This implies the conditional exit times and path probabilities are related by the equation: X Ppi→A ωpi→A . (5.9) ΩA (i)PA (i) = pi→A

Once again the idea of decomposing a random walk into a first step and the remainder of the steps will be used on the right-hand side of Eq. (5.9). The conditional exit time of a random walk starting in state i will be one more than the conditional exit times of random walks starting in states i ± 1. Therefore ΩA (i) = 1 +

1 2 ΩA (i

− 1)PA (i − 1) + 21 ΩA (i + 1)PA (i + 1) PA (i) 1 1 ΩA (i)PA (i) = PA (i) + ΩA (i − 1)PA (i − 1) + ΩA (i + 1)PA (i + 1) 2 2 i i i−1 i+1 ΩA (i) = + ΩA (i − 1) + ΩA (i + 1) A A 2A 2A = 1+

1 2 ΩA (i

− 1)PA (i − 1) + 21 ΩA (i + 1)PA (i + 1) 1 1 2 PA (i − 1) + 2 PA (i + 1)

by Theorem 5.4. The last equation is equivalent to (i − 1)ΩA (i − 1) − 2iΩA (i) + (i + 1)ΩA (i + 1) = −2i. 10:57:29.

(5.10)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

BC8495/Chp. 5

127

The hypothesis that the solution is a quadratic function of i was profitable once before, so we will try it again. If we let ΩA (i) = ai2 + bi + c and substitute this into Eq. (5.10) we obtain     −2i = (i − 1) a(i − 1)2 + b(i − 1) + c − 2i ai2 + bi + c   + (i + 1) a(i + 1)2 + b(i + 1) + c 0 = (3a + 1)i + b.

Therefore a = − 31 and b = 0. To evaluate the remaining coefficient c, we will use the boundary condition that ΩA (A) = 0 which implies. 1 0 = − A2 + c 3

=⇒

c=

1 2 A . 3

Finally we have established the following theorem. Theorem 5.6 Suppose a random walk has the form described in Eq. (5.1) in which the Xi for i = 1, 2, . . . are independent, identically distributed random variables taking on the values ±1, each with probability p = 1/2. Suppose further that the boundary at 0 is absorbing. The random walk that avoids state 0 will stop the first time that S(n) = A. The expected value of the stopping time is ΩA (i) =

 1 2 A − i2 , 3

for i = 1, 2, . . . , A.

(5.11)

If the random walk starts in state 0, an absorbing boundary, the expected value of the exit time is defined to be infinity. Example 5.3 Suppose an unbiased random walk takes place on the integer-valued number line. Suppose the boundary at 0 is absorbing. If 0 < S(0) = i ≤ 5 then the expected values of the first passage, or stopping, times at which the random walk attains a value of 5 are given in the table below. i Ω5 (i) 5.4

1 8

2 7

3 16 3

4 3

5 0

Continuous Random Walks

In this section we will bridge the gap between the discrete random walk and the continuous random walk. A fully rigorous treatment of this topic is beyond the undergraduate level, so this section will depend somewhat

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

128

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

on intuition. Readers interested in more of the details should consult any of the references on stochastic calculus such as [Lawler (2006)], [Shreve (2004b)], or [Wilmott (2006)]. Consider once again the unbiased or symmetric random walk defined in Eq. (5.1). Suppose the n selections of the Bernoulli random variables Xj will take place equally spaced in time in the interval [0, T ]. To keep the discussion simple, suppose as well that S(0) = 0. Our goal in this section is to derive a continuous random walk from the discrete random walk {S(jT /n)}nj=0 as n → ∞. At each tick of the clock the random variable takes a step either to the left or to the right, each with probability 1/2. All steps are identically distributed and pairwise independent. The sizepof the p step is constrained to be T /n. In other words for each j, Xj = ± T /n, each with probability 1/2. Therefore the final statep S(T ) is a binomial random variable with E [S(T )] = 0 and V (S(T )) = n( T /n)2 = T . To be precise, the random process just described is still a discrete random process, but one with finer (think smaller) steps in time and space. The process can be made continuous by interpolating between the already described discrete values. If 0 < t < T and nt/T ∈ / Z then there exists a largest integer v ∈ {1, 2, . . . , n} depending on t such that v − 1 < nt/T < v. This implies (v − 1)T vT t1 ≥ 0 E [W (t2 ) − W (t1 )] = E [W (t2 )] − E [W (t1 )] = 0 − 0 = 0.

(5.12)

The variance of the change in the Wiener process can also be found by making use of the fact that t = V (W (t))

Hence

= V (W (t) − W (0)) (by property (2))   = E (W (t) − W (0))2 (by Eq. (5.12))   2 = E (W (t)) .

  2 V (W (t2 ) − W (t1 )) = E (W (t2 ) − W (t1 ))2 − E [W (t2 ) − W (t1 )]     = E (W (t2 ))2 + E (W (t1 ))2 − 2E [W (t1 )W (t2 )] = t2 + t1 − 2E [W (t1 )(W (t2 ) − W (t1 ) + W (t1 ))] = t2 + t1 − 2E [W (t1 )(W (t2 ) − W (t1 ))]   − 2E (W (t1 ))2 = t2 + t1 − 2t1 = t2 − t1 .

(5.13)

The reader should examine each step in the derivation of the variance of the difference W (t2 )−W (t1 ) in Eq. (5.13) making note of where the various properties of the Wiener process were used. If we let ∆t = t2 − t1 where t2 > t1 ≥ 0 and let ∆W =  W (t2 ) − W (t1 ) then we may note that by Eq. (5.13) we have E (∆W )2 = ∆t. So far we have established these results for changes in a Wiener process over the interval [t1 , t2 ] in the discrete sense. We made use only of quantities evaluated at times t1 and t2 . (7) Quadratic variation: for a partition 0 = t0 < t1 < · · · < tn = t of the 10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

131

Random Walks and Brownian Motion

interval [0, t] into n non-overlapping subintervals, [W, W ] (t) = lim

n→∞

n X

k=1

2

[W (tk ) − W (tk−1 )] = t

(5.14)

where the limit is taken in the mean square sense. The quadratic variation property is important for the development of a stochastic integral and thus will be explored further later in this section. Keep in mind that W (s) is a random variable, so it is only meaningful to refer to its distribution, expected value, and variance. (8) The continuous random walk W (t) is a martingale. More precisely we can state that if W (t) is defined for t ≥ 0 and if 0 ≤ s < t then E [W (t) | W (τ ), 0 ≤ τ ≤ s] = W (s). The expression W (τ ), 0 ≤ τ ≤ s should be interpreted as the evolution of the continuous symmetric walk from τ = 0 until τ = s. Before proceeding to the main result of this chapter we should explore where a continuous version of Eq. (5.13) holds. The following lemma (adapted from [Seydel (2002), Chap. 1]), while technical in nature, will establish the necessary result. Lemma 5.3 Let {P (n) } for n ∈ N be a sequence of partitions of the interval [0, t] such that (n)

0 = t0 (n)

For each i let ∆ti (n)

(n)

= ti

δn = maxi {∆ti }. Then 

Proof.

E

n  X i=1

< · · · < t(n) n = t.

(n)

(n)

− ti−1 and ∆Wi

(n) 2

(∆Wi

(n)

< t1

(n)

) − ∆ti



(n)

!2  →0

as

δn → 0.

The proof will make use of the calculation that h i (n) (n) E (∆Wi )4 = 3(∆ti )2 10:57:29.

(n)

= W (ti ) − W (ti−1 ) and let

(5.15)

(5.16)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

132

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

(see exercise (16)). To simplify the notation we can drop the superscripts of the form ·(n) . The reader may verify by expanding the product that n X i=1

(∆Wi )2 − ∆ti

!2 

=

n X

(∆Wi )4 +

i=1

n X i=1

+2

n−1 X

n X

(∆ti )2 − 2

n X n X

(∆Wi )2 ∆tj

i=1 j=i

(∆Wi )2 (∆Wj )2 .

i=1 j=i+1

We can evaluate the following expected values: E

" n X

E

4

(∆Wi )

i=1 " n X

(∆ti )

i=1





E 2

E 2

n−1 X

n X n X i=1 j=i

n X

i=1 j=i+1

2

#

=3

#

=



n X

(∆ti )2

(by Eq. (5.16))

i=1 n X

(∆ti )2

i=1

(∆Wi )2 ∆tj  = 2 

(∆Wi )2 (∆Wj )2  = 2

n n X X

(∆ti )(∆tj )

i=1 j=i

n−1 X

n X

(∆ti )(∆tj )

i=1 j=i+1

The last equation is true since the ∆Wi and ∆Wj are independent for i 6= j. Therefore  !2  n X   E (∆Wi )2 − ∆ti i=1

=3

n X

(∆ti )2 +

i=1

=4 =4

n X

i=1 n X i=1

=2

n X

n X i=1

(∆ti )2 − 2 (∆ti )2 − 2

(∆ti )2 − 2

n X n X

n X n X

(∆ti )(∆tj ) + 2

i=1 j=i

(∆ti )(∆tj ) + 2

i=1 j=i n X

n−1 X

n X

(∆ti )(∆tj )

i=1 j=i+1

n−1 X

n X

(∆ti )(∆tj )

i=1 j=i+1

(∆ti )2

i=1

(∆ti )2 .

(5.17)

i=1

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

133

Random Walks and Brownian Motion

Now we can see that 2

n X i=1

(∆ti )2 = 2 (∆t1 )2 + (∆t2 )2 + · · · + (∆tn )2 ≤ 2δn (∆t1 + ∆t2 + · · · + ∆tn )



= 2δn t. Finally since 

0 ≤ E

n X i=1

!2    ≤ 2δn t, (∆Wi )2 − ∆ti

the limit in Eq. (5.15) follows as a result of the Squeeze Theorem as n → ∞ or equivalently as δn → 0.  Hence we may now pass to the limit as ∆t becomes infinitesimally small and write (dW (t))2 = dt.

(5.18)

This result will be used to prove Itˆo’s Lemma in the next section. (9) The derivative dW/dt does not exist. The final property addresses the “roughness” of the path followed by the Wiener process as seen in Fig. 5.4. Technically we should say that the probability that dW/dt exists is zero. Recall the limit definition of the derivative from calculus, df f (s + h) − f (s) = lim . h→0 dt h Suppose f (t) is a Wiener process W (t). Since     E (W (s + h) − W (s))2 = E |W (s + h) − W (s)|2 = h

then on average |W (s + h) − W (s)| ≈ lim

h→0

√ h, and therefore

W (s + h) − W (s) h

10:57:29.

does not exist.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

134

5.5

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

The Stochastic Integral

Of central importance in the remainder of the book will be the calculus of functions defined along a continuous random walk. We have already argued that the derivative of a Wiener process does not exist, but what about the integral? If P is a partition of [0, t] with 0 = t0 < t1 < · · · < tn = t then we define the stochastic integral of a function f (τ ) defined on [0, t] informally as Z t n X f (τ ) dW (τ ) = lim f (tk−1 )(W (tk ) − W (tk−1 )). (5.19) 0

n→∞

k=1

Of course this only makes sense if the limit exists and is the same for all partitions P . Note that the function f is always evaluated at the lefthand endpoint of the subinterval [tk−1 , tk ]. Think of this as ensuring that the function being integrated cannot use any information about the future movements of the random walk as it is being integrated. To give a more concrete example, suppose we are going to place continuous wagers defined by f (τ ) during the interval [0, t] on the outcome of the continuous random walk W (τ ). The stochastic integral given in Eq. (5.19) would represent our net winnings. When convenient the stochastic integral can be written in a form reminiscent of the Fundamental Theorem of Calculus. Z t I(t) − I(0) = f (τ ) dW (τ ) (5.20) 0

The letter “I” is chosen in honor of Itˆo and is sometimes called the Itˆo integral. The stochastic integral is sometimes written in a differential form obtained by informally differentiating Eq. (5.20). dI(t) = f (t) dW (t)

(5.21)

However, beware that we are not allowed to divide both sides of Eq. (5.21) by dt since dW/dt is undefined. In general Eqs. (5.20) and (5.21) are considered to be synonymous. The reader should keep in mind that I(t) is a random variable, not the typical function encountered in elementary calculus. However, the stochastic integral does possess some familiar properties. Theorem 5.7

If f and g are functions defined on [0, t] such that their

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

135

Random Walks and Brownian Motion

stochastic integrals exist and if c is any constant, then Z t Z t Z t (f (τ ) ± g(τ )) dW (τ ) = f (τ ) dW (τ ) ± g(τ ) dW (τ ), 0

0

0

and Z

t

cf (τ ) dW (τ ) = c

0

Z

t

f (τ ) dW (τ ).

0

To address the random variable nature of the stochastic integral consider the following theorem. Theorem 5.8 If f is a deterministic (non-random) function defined on [0, t] for which the stochastic integral exists, then Z t  E f (τ ) dW (τ ) = 0 and (5.22) 0 Z t  Z t V f (τ ) dW (τ ) = (f (τ ))2 dτ. (5.23) 0

0

Proof.

Taking the expected value of both sides of Eq. (5.19) produces " # Z t  n X E f (τ ) dW (τ ) = E lim f (tk−1 )(W (tk ) − W (tk−1 )) n→∞

0

= lim

n→∞

k=1

n X

f (tk−1 )E [W (tk ) − W (tk−1 )] ,

k=1

where we have assumed it is permissible to interchange the order of the operations of taking the limit and finding the expected value. Since E [W (tk ) − W (tk−1 )] = 0, Eq. (5.22) is proved. Proceeding in the same fashion by calculating the variance of both sides of Eq. (5.19) yields ! Z t  n X V f (τ ) dW (τ ) = V lim f (tk−1 )(W (tk ) − W (tk−1 )) n→∞

0

= lim

n→∞

= lim

n→∞

=

10:57:29.

Z

t 0

n X

k=1 n X k=1

k=1

(f (tk−1 ))2 V (W (tk ) − W (tk−1 )) (f (tk−1 ))2 (tk − tk−1 )

(f (τ ))2 dτ

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

136

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

and Eq. (5.23) is established. The reader should note that the assumption that function f is deterministic allows f (tk−1 ) to be “pulled out” of the expected value and variance.  Example 5.4 Let f (τ ) = sin τ and find the mean and variance of the stochastic integral of f for t ≥ 0. According to Eq. (5.22) Z t  E sin τ dW (τ ) = 0, 0

while V

Z

0

t

 Z t t sin τ dW (τ ) = sin2 τ dτ = . 2 0

Since Theorem 5.8 succinctly addressed the mean and variance of stochastic integrals of deterministic functions, our attention now turns to an example in which the integrand is not deterministic. The following example and its analysis are taken from Example 4.3.2 of [Shreve (2004b)] with small modifications to the notation. Suppose the Itˆo integral of W (t) exists then according to the definition of the integral Z t n X W (τ ) dW (τ ) = lim W (tk−1 ) [W (tk ) − W (tk−1 )] , n→∞

0

k=1

where 0 = t0 < t1 < · · · < tn = t is a regular partition of [0, t]. In exercise (18) the reader is asked to verify that n X

k=1

n

W (tk−1 ) [W (tk ) − W (tk−1 )] =

1 2 1X 2 W (t) − (W (tk ) − W (tk−1 )) . 2 2 k=1

Using this allows the Itˆ o integral to be expressed as Z t n X 1 1 W (τ ) dW (τ ) = W 2 (t) − lim (W (tk ) − W (tk−1 ))2 n→∞ 2 2 0 k=1

1 = W 2 (t) − 2 1 2 = W (t) − 2

1 [W, W ] (t) 2 t . 2

The second line of this equation follows from the definition of the quadratic variation of W (t), (see Eq. (5.14)). This example also illustrates a fundamental difference between the Itˆo integral and the Riemann integral of

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

BC8495/Chp. 5

137

elementary calculus. If W (t) is replaced with a continuously differentiable deterministic function f (t) for which f (0) = 0 then Z t Z t Z f (t) 1 f (τ ) df (τ ) = f (τ )f 0 (τ ) dτ = u du = f 2 (t), 2 0 0 f (0) which lacks the final term of the Itˆo integral. The extra term of the form −t/2 is a result of the quadratic variation of the random process W (t). In exercise (19) the reader will demonstrate that the quadratic variation of continuously differentiable functions is zero. 5.6

Continuous Random Walks with Drift

With this informal introduction to stochastic processes completed we can now begin to generalize the discussion. A familiar place to start generalizing is with the deterministic process of exponential growth and decay. Most introductions to calculus (see for example [Stewart (1999)]) contain the mathematical model which states that the rate of change of a non-negative quantity P , is proportional to P . Expressed as a differential equation this statement becomes dP = µP dt

(5.24)

where µ is the proportionality constant and t usually represents time. The proportionality constant is called a “growth rate” if µ > 0 and a “decay rate” if µ < 0. If the value of P is known at a specified value of t (usually at t = 0) then this differential equation has solution P (t) = P (0)eµt (see exercise (9)). This mathematical model is described as deterministic since there is no place for random events to express themselves in the model. Once the initial value of P and the proportionality constant are set, the future evolution of P (t) is completely determined. Equation (5.24) can be rewritten as dP = µ dt, P

(5.25)

and if we make the change of variable Z = ln P then this equation becomes dZ = µ dt.

(5.26)

Now suppose we add a stochastic component to the mix by introducing on the right-hand side of Eq. (5.26) a Wiener process with mean zero and

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

138

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

√ standard deviation σ dt (which of course means the variance is σ 2 dt). We obtain the mathematical model governing the time evolution of Z below. dZ(t) = µ dt + σ dW (t)

(5.27)

A stochastic differential equation of this form is called a generalized Wiener process. Notice it possesses a deterministic part and a stochastic part. The constant µ is called the drift and the constant σ is called the volatility. The integral form of Eq. (5.27) is Z(t) = Z(0) + µt +

Z

t

σ dW (τ ) = Z(0) + µt + σW (t).

0

Do not lose sight of the fact that Z is a random variable. Therefore the Z(t) on the left-hand side of the equation has a mean, variance, and probability distribution. Assuming that it is valid to interchange the order of the limit and the expected value we see that E [Z(t)] = Z(0) + µt + E

Z

t

0

= Z(0) + µt,

 σ dW (τ )

(5.28)

by Theorem 5.8. Meanwhile the variance is calculated as   Z t V (Z(t)) = V Z(0) + µt + σ dW (τ ) 0  Z t σ dW (τ ) =V 0

=

Z

t

σ 2 dt

0

= σ 2 t. Numerically the value of Z(t) can be approximated by choosing a small ∆t = t/n and approximating W (t) =

Z

0

t

dW (τ )

by



∆t

n X

Xj

j=1

where Xj is a standard normal random variable (mean 0 and standard deviation 1). The textbook by Seydel [Seydel (2002)] contains other numerical

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

139

Random Walks and Brownian Motion

results for approximating the solutions to stochastic integrals and for generating random numbers. This approximation is used in the following two examples. Example 5.5 Suppose the drift parameter is µ = 1 and the volatility is σ = 1/4, then the√expected value of the Wiener process is t and the standard deviation is t/4. Plots of the expected value, the expected value plus and minus two standard deviations (the 95% confidence limits), and one realization of the random walk are illustrated in Fig. 5.5. ZHtL 1.5

1.0

0.5

0.2

0.4

0.6

0.8

1.0

t

Fig. 5.5 The behavior of a generalized Wiener process with drift µ = 1 and volatility σ = 1/4.

Example 5.6 Suppose the drift parameter is µ = 1/4 and the volatility is σ = 1, then the expected value of the Wiener process is t/4 and the √ standard deviation is t. Plots of the expected value, the expected value plus and minus two standard deviations (the 95% confidence limits), and one realization of the random walk are illustrated in Fig. 5.6. Continuous random walks such as the Brownian motion with drift possess stopping times just as the discrete symmetric random walk. Enough mathematical background has been established that we may describe the expected value and variance of the stopping time for the generalized Wiener process of Eq. (5.27) under the assumptions that µ > 0 and Z(0) = 0. This derivation follows the line of reasoning outlined in Chap. 10 of [Ross (2003)].

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

140

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

ZHtL 2

1

0.2

0.4

0.6

0.8

1.0

t

-1

Fig. 5.6 σ = 1.

The behavior of a generalized Wiener process with drift µ = 1/4 and volatility

If T ≥ 0 is fixed and we define Y (t) as Y (t) =



W (t) if 0 ≤ t ≤ T , W (T ) if t > T ,

then Y (t) is called a stopped process. The process is stopped at time T , i.e. the process is constant with value Y (T ) for all t > T . Recall that W (t) is a martingale. Likewise the stopped process Y (t) is a martingale. For 0 ≤ t ≤ T , Y (t) = W (t) and therefore is a martingale. For t > T , Y (t) is constant and therefore trivially has no tendency to rise or fall. The following equation is sometimes referred to as the Martingale Stopping Theorem. E [Y (T )] = E [Y (0)]

(5.29)

This result can be understood as stating that the expected value of a stopped martingale process equals the expected value of the process at a fixed time. This is called by some authors Doob’s Optional Stopping Theorem. For a proof of this result the reader may consult [Karatzas (1991)]. Define the stopping time of Z(t) = µt + σW (t) to be the first time Z(t) = z for some fixed z > 0. In order to apply the Martingale Stopping Theorem, the stochastic process must be a martingale, but unfortunately Z(t) is not (see exercise (21)). However W (t) = (Z(t) − µt)/σ is a martingale. Therefore the stopping time can be expressed in two equivalent

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

141

Random Walks and Brownian Motion

ways: 

z − µt T = min{t : Z(t) = z} = min t : W (t) = t≥0 t≥0 σ



.

Now applying Eq. (5.29) we have E [W (T )] = E [W (0)]   z − µT E =0 σ z − µE [T ] =0 σ z E [T ] = . µ Similar to our earlier study of the discrete random walk, let B = {−m, M } where m and M are both positive constants, and let TB = min{Z(t) = −m ∨ Z(t) = M }. t≥0

To be clear, TB is the first time the continuous random process Z(t) = µt+σW (t) achieves either state −m or state M . Now if γ is a constant then γ(Z(t) − µt)/σ = γW (t) is a continuous random process which is normally distributed for each t > 0 with mean zero and variance γ 2 t. Therefore 2 eγW (t) is lognormally distributed with mean eγ t/2 . Thus eγW (t) is not a martingale, but eγW (t)−γ

2

t/2

is a martingale (see exercise (17)). At the stopping time TB then h i h i 2 2 2 E eγW (TB )−γ TB /2 | eγ(0)−γ (0)/2 = E eγW (TB )−γ TB /2 = 1.

(5.30)

If the arbitrary constant γ in Eq. (5.30) is replaced by −2µ/σ we obtain h i 2 E e−2µ(µTB +σW (TB ))/σ = 1. (5.31)

Recall that at time TB , µTB + σW (TB ) = M with probability p, or µTB + σW (TB ) = −m, with probability 1 − p. Equation (5.31) is equivalent to 2

1 = pe−2µM/σ + (1 − p)e2µm/σ 10:57:29.

2

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

142

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

which implies 2

p=

1 − e2µm/σ . e−2µM/σ2 − e2µm/σ2

(5.32)

Thus we have found the probability p that the generalized Wiener process, Z(t) = µt + σW (t), stops at M > 0 before hitting −m < 0. By the Martingale Stopping Theorem, Eq. (5.29), the expected value of the first time the continuous random walk hits the boundary satisfies E [W (TB )] = E [W (0)] = 0 which is equivalent to   Z(TB ) − µTB E [Z(TB )] − µE [TB ] E = = 0, σ σ which in turn implies that E [TB ] =

E [Z(TB )] pM + (1 − p)(−m) = . µ µ

Using the expression for p found in Eq. (5.32), we obtain     2 2 M e2µm/σ − 1 + m e−2µM/σ − 1  E [TB ] = . µ e2µm/σ2 − e−2µM/σ2

(5.33)

A plot of E [TB ] for a fixed choice of m and M is shown in Fig. 5.7. We can generalize the Itˆo integral still further by imagining the drift and volatility are functions of t. For this case the stochastic differential equation has the form dZ(t) = µ(t) dt + σ(t) dW (t). Using the ordinary Riemann integral on the deterministic portion of the right-hand side and the stochastic integral on the random portion we can find Z t Z t Z(t) = Z(0) + µ(τ ) dτ + σ(τ ) dW (τ ). 0

0

If we wish to generalize still further by making µ and σ functions not only of t but of Z (and possibly other variables) then we can no longer integrate in closed form. However, a change of variable may enable us to convert these more general stochastic differential equations into Wiener form. In the next two sections we will develop a version of the chain rule

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

143

Random Walks and Brownian Motion

10 8 E@TB D 6

1.0

4

0.8

2 0.6 Σ

0.2 0.4

0.4 Μ

0.6 0.2

0.8 1.0

Fig. 5.7 For m = M = 1 the expected value of the first time the continuous random process Z(t) hits −m or M is plotted for a range of values of the drift and volatility of the process.

for derivatives which is applicable to functions of stochastic variables and which can help overcome our current limitations in stochastic integration. 5.7

Itˆ o Processes

One class of further generalization of Eq. (5.27) includes the Itˆ o processes of the form dZ = a(Z, t) dt + b(Z, t) dW (t)

(5.34)

where the expressions a and b are each now functions of time t and the random variable Z. In general the stochastic differential equation given in Eq. (5.34) cannot be solved explicitly through integration of both sides. Nevertheless it is an important type of equation for us to use because it will enable us to work with stochastic processes of quantities which depend in turn on other random variables having their own individual stochastic process descriptions. In this way we will develop the fundamentally important Black-Scholes partial differential equation. Before we can do this we must develop an analogue of the chain rule for differentiation found in elementary calculus.

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

144

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

In standard differential calculus if one makes the assignment Z = ln P then it is well understood that dZ = dP/P according to the chain rule. Using this on a generalized Wiener process may lead us to believe that the following equations are equivalent. dZ = µ dt + σ dW (t) dP = µP dt + σP dW (t) First we note that while the first equation is a generalized Wiener process the second is an Itˆ o process. From the point of view of an applied mathematician the second equation has an advantage over the first. As P gets close to zero the “drift-like” quantity µP and the “volatility-like” quantity σP become small as well. This will prevent P from becoming negative (assuming that P (0) > 0). Non-negativity is certainly a property a mathematical model of the price of a financial instrument should possess. The first equation will allow Z to become negative even when the drift is positive. Secondly, if µ and σ represent the mean and standard deviation respectively of random variable dZ (as they did for the stock price data analyzed earlier), does the second equation imply that µ and σ are the mean and standard deviation of the underlying variable P ? The answer is no. What is needed is a valid procedure for changing variables in a stochastic process, in other words, a stochastic calculus version of the chain rule for differentiation. This change of variable differentiation result is known as Itˆ o’s Lemma and is covered in the next section.

5.8

Itˆ o’s Lemma

The proper procedure for changing variables in a stochastic differential equation can be derived using the multivariable version of Taylor’s Theorem. A standard reference containing this result is [Taylor and Mann (1983)]. Here we will give a brief overview of a two-variable version of Taylor’s formula with remainder. We start with the single-variable Taylor’s formula. If f (x) is an (n + 1)-times differentiable function on an open interval containing x0 then the function may be written as f 00 (x0 ) (x − x0 )2 (5.35) 2! f (n) (x0 ) f (n+1) (θ) + ··· + (x − x0 )n + (x − x0 )n+1 n! (n + 1)!

f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) +

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

145

Random Walks and Brownian Motion

The last term above is usually called the Taylor remainder formula and is denoted by Rn+1 . The quantity θ lies between x and x0 . The other terms form a polynomial in x of degree at most n and can be used as an approximation for f (x) in a neighborhood of x0 . This version of Taylor’s formula will be used to derive the two-variable form. Suppose the function F (y, z) has partial derivatives up to order three on an open disk containing the point with coordinates (y0 , z0 ). Define the function f (x) = F (y0 + xh, z0 + xk) where h and k are chosen small enough that (y0 + h, z0 + k) lie within the disk surrounding (y0 , z0 ). Since f is a function of a single variable then we can use the single-variable form of Taylor’s formula in Eq. (5.35) with x0 = 0 and x = 1 to write 1 f (1) = f (0) + f 0 (0) + f 00 (0) + R3 . 2

(5.36)

Using the multivariable chain rule for derivatives we have, upon differentiating f (x) and setting x = 0, f 0 (0) = hFy (y0 , z0 ) + kFz (y0 , z0 ) 00

2

(5.37) 2

f (0) = h Fyy (y0 , z0 ) + 2hkFyz (y0 , z0 ) + k Fzz (y0 , z0 ).

(5.38)

We have made use of the fact that Fyz = Fzy for this function under the smoothness assumptions. The remainder term R3 contains only third order partial derivatives of F evaluated somewhere on the line connecting the points (y0 , z0 ) and (y0 + h, z0 + k). Thus if we substitute Eqs. (5.37) and (5.38) into (5.36) we obtain ∆F = f (1) − f (0)

(5.39)

= F (y0 + h, z0 + k) − F (y0 , z0 )

= hFy (y0 , z0 ) + kFz (y0 , z0 )  1 2 + h Fyy (y0 , z0 ) + 2hkFyz (y0 , z0 ) + k 2 Fzz (y0 , z0 ) + R3 . 2

This last equation can be used to derive Itˆo’s Lemma. Let X be a random variable described by an Itˆo process of the form dX = a(X, t) dt + b(X, t) dW (t)

(5.40)

where dW (t) is a normal random variable and a and b are functions of X and t. Let Y = F (X, t) be another random variable defined as a function of X and t. Given the Itˆ o process which describes X we will now determine the Itˆ o process which describes Y .

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

146

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

Using a Taylor series expansion for Y detailed in (5.39) we find 1 1 ∆Y = FX ∆X + Ft ∆t + FXX (∆X)2 + FXt ∆X∆t + Ftt (∆t)2 + R3 2 2 1 = FX (a∆t + b dW (t)) + Ft ∆t + FXX (a∆t + b dW (t))2 2 1 + FXt (a∆t + b dW (t))∆t + Ftt (∆t)2 + R3 . 2 Upon simplifying, the expression ∆X has been replaced by the discrete version of the Itˆ o process. The dependence of a and b on X and t has been suppressed to simplify the notation. The reader is reminded that the Taylor remainder term R3 contains terms of order (∆t)k where k ≥ 2. Thus as ∆t becomes small ∆Y ≈ FX (a dt + b dW (t)) + Ft dt +

1 FXX b2 (dW (t))2 . 2!

Making use of Eq. (5.18) we can obtain the approximation ∆Y ≈ FX (a dt + b dW (t)) + Ft dt +

1 FXX b2 dt. 2!

(5.41)

The reader should be aware that this is merely an outline of a proof of Itˆo’s Lemma. The interested reader should consult [Neftci (2000), Chap. 10] for a more rigorous proof. To summarize what we have done, the statement of the lemma is given below. Lemma 5.4 (Itˆ o’s Lemma) Suppose that the random variable X is described by the Itˆ o process dX = a(X, t) dt + b(X, t) dW (t)

(5.42)

where W (t) is the standard Wiener process (Brownian motion). Suppose the random variable Y = F (X, t). Then Y is described by the following Itˆ o process.   1 dY = a(X, t)FX + Ft + (b(X, t))2 FXX dt + b(X, t)FX dW (t) (5.43) 2 Now we can return to the question raised earlier regarding the stochastic process followed by a stock price P . If Z = ln P and dZ = µ dt + σ dW (t),

10:57:29.

(5.44)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

BC8495/Chp. 5

147

then we can let F (Z, t) = eZ = P . According to Itˆo’s Lemma then   1 2 Z Z dP = µe + 0 + σ e dt + σeZ dW (t) 2   1 2 = µ + σ P dt + σP dW (t) 2 Note that the coefficient of the deterministic portion of the equation contains the expression 12 σ 2 which would have been absent had the chain rule of deterministic calculus been used instead of Itˆo’s Lemma. This lemma plays a role of central importance in the next chapter in which the Black-Scholes partial differential equation is derived. The next two examples will give the reader some idea of the utility of Itˆ o’s Lemma. Example 5.7 Suppose W (t) is the standard Wiener process. Determine the stochastic process which Y (t) = eW (t) obeys. The random variable W (t) obeys the stochastic differential equation dW (t) = (0) dt + (1) dW (t). Applying Itˆ o’s Lemma with X(t) = W (t), a(X, t) = 0, and b(X, t) = 1 produces   1 1 dY = (0)eW (t) + 0 + eW (t) dt+(1)eW (t) dW (t) = Y (t) dt+Y (t) dW (t). 2 2 Example 5.8 The Ornstein-Uhlenbeck equation (sometimes called the Langevin equation) given below can be thought of as a stochastic extension to the deterministic exponential growth/decay ordinary differential equation. dZ(t) = µZ(t) dt + σ dW (t) Solve this equation assuming Z(0) = Z0 and that µ and σ are constants. The key to solving this equation (and many other similar equations) is to perform the appropriate change of variables and then to use Itˆo’s Lemma. Define Y (t) = e−µt Z(t). According to Itˆ o’s Lemma Y (t) solves the following stochastic differential equation.   1 2 −µt −µt dY = µZ(t)e − µe Z(t) + σ (0) dt + σe−µt dW (t) 2 = σe−µt dW (t)

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

148

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

Using the fact that Y (0) = Z0 and integrating both sides of the previous equation produces Z t Y (t) − Z0 = σe−µs dW (s) 0 Z t µt Z(t) = Z0 e + σeµ(t−s) dW (s). 0

The reader should always keep in mind that the solution to a stochastic differential equation is a random variable. Using Theorem 5.8 we may calculate E [Z(t)] = Z0 eµt Z t  σ 2 2µt V (Z(t)) = σ 2 e2µ(t−s) ds = e −1 . 2µ 0 5.9

Stock Market Example

In Section 5.3 we explored a discrete approximation to a continuous model for generating a random walk which seemed to mimic the fluctuations of the price of a corporation’s stock. The model depended on several parameters which were called µ, σ, and ∆t. In this section we intend to analyze the data associated with the stock price of an actual corporation and determine how to assign values to these parameters which are appropriate to that corporation. The end-of-day closing price of a corporation’s stock is generally available on the World Wide Web. We will use stock data for Sony Corporation. Appendix A contains the raw data analyzed here. The data is graphed in Fig. 5.8. The closing prices of the stock for 248 consecutive trading days are represented in this figure. Since the data is sampled once every trading day (and days on which no trading took place, such as weekends and holidays, are irrelevant) then the value of ∆t appropriate for the model to be developed is ∆t = 1. Simple descriptive statistics can be used to determine µ and σ. The raw stock prices are playing the role of the variable P in our model. To fit the discrete form of the model used in Eq. (5.27) we must take the natural logarithm of the the stock prices, since Z = ln P . Then we must form ∆Z by subtracting consecutive days logarithmic prices. If the values of ∆Z just calculated fit the model then they will appear to be normally distributed. The mean of this data will be our estimate of µ and the standard deviation will be an estimate of σ, the volatility. A histogram of the values of ∆Z

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

149

Random Walks and Brownian Motion

P 60 50 40 30 20 10 50

100

150

200

250

t

Fig. 5.8 The closing prices of Sony Corporation stock between August 13, 2001 (day 1) and August 12, 2002 (day 248).

15 12.5 10 7.5 5 2.5 -0.1 Fig. 5.9

-0.05

0

0.05

0.1

A histogram of the values of the ln P (ti+1 ) − ln P (ti ) for i = 1, 2, . . . , 247.

calculated from the data in Appendix A appears in Fig. 5.9. It is certainly plausible that the values of ∆Z summarized there are normally distributed. The mean of data is µ ≈ −0.000555 day−1 and the standard deviation is σ ≈ 0.028139 day−1 . Thus, despite the apparent randomness of the closing 10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

150

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

prices there is some structure and organization hidden in them. The estimates of drift and volatility hold for the random variable Z. Thus as before, Z t Z(t) = Z(0) + µt + σ dW (τ ) 0

which implies that the random motion of the stock prices obeys the formula P (t) = P (0)eµt+σW (t) .

(5.45)

Since P is a lognormal random variable, Lemma 3.1 implies its mean and variance are respectively E [P (t)] = P (0)e(µ+σ

2

/2)t

V (P (t)) = (P (0))2 e(2µ+σ

2

and  2  )t eσ t − 1 .

The discrete approximation of the stochastic Eq. (5.27) evidently matches well the behavior of the stock price of an actual company. In the next chapters we will begin to make use of the brief information presented in this chapter on stochastic processes. A reader interested in a more rigorous introduction to the mathematics and statistics of stochastic processes and stochastic differential equations should consult one or more of the references such as [Durrett (1996)] or [Mikosch (1998)]. In this book we are merely taking an intuitive, non-rigorous approach to their use as tools for modeling prices of stocks and their derivatives. 5.10

Exercises

(1) Prove Theorem 5.2. (2) Another example of a discrete martingale can be explored by considering an experiment called Polya’s urn. Suppose initially an urn contains one black marble and one white marble. During each iteration of the experiment a marble will be randomly drawn, its color noted, and replaced along with a new marble of the same color. Show that the fraction of black marbles in the urn is a martingale. (3) Another, less formal, method for obtaining the probability that an unbiased random walk which begins in state S(0) = i > 0 and finishes in state S = A while avoiding an absorbing boundary at 0 is to treat the Laplacian Eq. (5.6) as a continuous boundary value problem. Find

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

BC8495/Chp. 5

151

all the functions f (x) which satisfy f 00 (x) = 0 subject to the conditions that f (0) = 0 and f (A) = 1. (4) Another, less formal, method for obtaining the stopping time for an unbiased random walk which begins in state 0 ≤ S(0) = i ≤ A and finishes in either state S = 0 or S = A is to treat the Poisson Eq. (5.7) as a continuous boundary value problem. Find all the functions g(x) which satisfy g 00 (x) = −2 subject to the conditions that g(0) = 0 and g(A) = 0. (5) Suppose f (x) is four times continuously differentiable at x = x0 . Use Taylor’s Theorem [Stewart (1999)] to expand f (x) about x = x0 . Then by using f (x0 + h) and f (x0 − h) show that f 00 (x0 ) ≈

f (x0 + h) − 2f (x0 ) + f (x0 − h) . h2

(6) If a stock price can change (up or down) by only one dollar per day and the probability of a unit increase is 1/2 what is the probability that the stock price will increase by $25 before decreasing from the present value by $50? (7) What is the expected number of days that will elapse before the stock described in exercise (6) increases in price by $25 or decreases by $50? (8) What is the expected number of days that will elapse before the stock described in exercise (6) increases in price by $25 while avoiding a decrease of $50? (9) Assuming that µ and P (0) are constants, verify by differentiation and substitution that P (0)eµt solves Eq. (5.24). (10) Using the growth constant and volatility of the Sony Corporation stock found earlier, use Eq. (5.45) to generate a realization of the random walk of stock prices for another year’s duration. (11) Investigate the random walk of another company’s stock. Collect, if possible, a year’s worth of closing prices (P (t1 ), P (t2 ), . . . , P (tn )) for the stock of your favorite company. Plot a histogram of ∆Z = ln P (ti+1 ) − ln P (ti ) and also calculate the mean and standard deviation of this random variable. Does the distribution of ∆Z appear to be normal? (12) Using the growth constant and volatility of the stock data you found in exercise (11), use Eq. (5.45) to generate a realization of the random walk of stock prices for another year’s duration. (13) Using the chain rule for derivatives from ordinary calculus (not the stochastic version governed by Itˆo’s Lemma) verify the expressions

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

152

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

found for f 0 (0) and f 00 (0) in Eqs. (5.37) and (5.38). Find an expression for the Taylor remainder R3 . (14) Suppose that P is governed by the stochastic process dP = µP dt + σP dW (t) and that Y = P n . Find the process which governs Y . (15) Suppose that P is governed by the stochastic process dP = µP dt + σP dW (t) and that Y = ln P . Find the process which governs Y . (16) Suppose that X is a normally distributed variable with mean  4  random 2 4 µ = 0 and variance σ . Show that E X = 3σ . (17) Let W (t) be the standard Wiener process. Show that the exponential random variable defined as Z(t) = eW (t)−t/2 is a martingale. (18) Let 0 = t0 < t1 < · · · < tn = t be a regular partition of [0, t] and let W (s) be the continuous symmetric random walk for which W (0) = 0. Then show that n X

k=1

n

W (tk−1 ) [W (tk ) − W (tk−1 )] =

1 2 1X 2 W (t)− (W (tk ) − W (tk−1 )) . 2 2 k=1

(19) Suppose the function f (t) is continuous and has a continuous derivative on the interval [0, T ]. Show that the quadratic variation of f is zero. (20) Find the mean and variance of the stochastic integral Z 2 t3 dW (t) 0

(21) Let Z(t) = µt + σW (t) where W (t) is the standard Wiener process. Show that if µ > 0, Z(t) is not a martingale. (22) Find the expected value of TB for µ = 0 (no drift) from Eq. (5.33). (23) Suppose W (t) is the standard Wiener process and define Y (t) = W (t) . Determine the stochastic differential equation solved by Y (t). 1 + t2 (24) The mean reverting Ornstein-Uhlenbeck stochastic differential equation [Øksendal (2003)] can be thought of as a stochastic version

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Random Walks and Brownian Motion

BC8495/Chp. 5

153

of Newton’s Law of Cooling. It takes the form dX(t) = k(µ − X(t)) dt + σ dW (t) where µ represents the constant temperature of the environment and k > 0 and σ > 0 are constants. (a) If Z(t) = ekt (µ − X(t)), find the stochastic differential equation solved by Z(t). (b) Solve the stochastic process found in exercise (24a) for Z(t). (c) Find E [X(t)] and V (X(t)). (25) Stochastic integrals have many properties similar to the properties of n+1 Riemann integrals. Using Itˆo’s Lemma with Y (t) = (W (t)) and the Wiener process dW (t) = (0) dt + 1 dW (t) establish the reduction of order formula Z t Z 1 n t n n+1 n−1 (W (s)) dW (s) = (W (t)) − (W (s)) ds. n+1 2 0 0 (26) Suppose f (t) is a continuous deterministic function and X(t) obeys the stochastic process 1 dX = − [f (t)]2 dt + f (t) dW (t). 2 Define Z(t) = e−X(t) and find the stochastic differential equation solved by Z(t). (27) Suppose the stochastic processes X1 (t) and X2 (t) have the following respective stochastic differentials. dX1 = a1 (t)X1 dt + b1 (t)X1 dW (t) dX2 = a2 (t)X2 dt + b2 (t)X2 dW (t) If Y (t) = X1 (t)X2 (t), use the product rule for stochastic processes d(X1 X2 ) = X2 dX1 + X1 dX2 + dX1 dX2 to find the stochastic differential for Y . (28) An extension of the logistic ordinary differential equation for bounded population growth to a stochastic differential equation has the form dP = rP (K − P ) dt + αP dW (t)

P (0) = P0 .

10:57:29.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

154

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 5

An Undergraduate Introduction to Financial Mathematics

The constant K is the carrying capacity of the environment. The quantity rK is the per capita reproductive rate of the population at low population density. The constant α is related to the size of the perturbation in population size due to random events in the environment. (a) Define Y = −1/P and show the stochastic differential for Y [Gard (1988)] is dY = [(α2 − rK)Y − r] dt − αY dW (t). (b) Consider the stochastic differential equation dX = (α2 − rK)X dt − αX dW (t), with initial condition X(0) = 1. Show the solution to this initial value problem is 2

X(t) = e(α

/2−rK)t−αW (t)

.

Hint: let Z = ln X. 2 (c) Define Y0 (t) = 1/X(t) = e(rK−α /2)t+αW (t) and show the stochastic differential for Y0 is dY0 = rKY0 dt + αY0 dW (t). (d) Notice that in Ex. (28b) we solved a stochastic differential equation which is the same is the one in Ex. (28a) save for one term on the right-hand side. Assume that Y (t) = X(t)Z(t) where Y (t) is the solution to Ex. (28a), X(t) is the solution to Ex. (28b), and Z(t) is an unknown stochastic process. Use the product rule for differentials of stochastic processes and the result of Ex. (28c) to show that dZ = −rY0 dt. (e) Since Y0 (t) has been found, we can write Z t Z(t) = Z(0) − r Y0 (s) ds. 0

Use this to show that the solution to the stochastic logistic equation can be written as 2

P (t) =

10:57:29.

P0 e(rK−α /2)t+αW (t) . Rt 1 + rP0 0 e(rK−α2 /2)s+αW (s) ds

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

Chapter 6

Forwards and Futures

This chapter introduces some of the concepts and terminology associated with the buying and selling of securities such as stocks. The main issue discussed will be the pricing of forward contracts and futures. The present chapter will give the reader the opportunity to apply the theory of interest, the arbitrage principle, and some elementary stochastic processes to the problem of pricing two commonly traded financial derivatives. The term “derivative” is used because the values of these financial instruments is “derived” from underlying securities or commodities. The material of this chapter can also be treated as a “warm-up” exercise for the later chapters on options and the development of the Black-Scholes option pricing formula. 6.1

Definition of a Forward Contract

At its essence, a forward is an agreement between two agents (which we will usually call the “party” and the “counter-party”) to buy or sell a specified quantity of a commodity at a specified price on a specified date in the future. The forward is an obligation to buy or sell at the agreed upon quantity, price, and time. If the party or counter-party breaks the agreement, they may face legal and financial consequences. Consider the situation of a manufacturer producing portable MP3 players. Their product depends upon an adequate supply of solid state memory to match the manufacturing output. If the manufacturer is concerned that the readily available supply of memory may fall short of their needs three months from now, they may enter into a forward contract with a memory supplier to sell them, say, 100, 000 units of memory for one million dollars in three months. Once the forward contract is established, the memory supplier must come up with the 100, 000 units of memory in three months and must sell it for

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

156

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

An Undergraduate Introduction to Financial Mathematics

one million dollars even if another buyer would be willing to pay more for it. The MP3 manufacturer must buy the 100, 000 units of memory in three months for one million dollars even if another supplier is willing to sell it for less. While there are many reasons an institution may buy or sell a forward contract, the chief reason is that the forward reduces the risk of market prices moving against the party or counter-party. The MP3 manufacturer may fear the business disruption which a shortage of solid state memory at affordable prices may cause, while the memory supplier may worry that increased supply or improved technology may decrease the price of memory three months hence. While we will not usually concern ourselves with the issue, the party or counter-party of a forward contract can effectively “cancel” a forward contract by entering into an opposing forward contract. For example, suppose the MP3 manufacturer after setting up the original forward contract decides they will not need the solid state memory. They can enter into a forward contract with another memory buyer. The MP3 manufacturer becomes a seller of its unneeded memory. A forward contract can be set up between two agreeing agents and can be customized to the precise needs of the buyer and seller. In contrast securities such as stocks (and later we will see futures) are often traded in a market. A stock market is created to increase the efficiency of the process of buying and selling stocks. Creating and maintaining this efficient environment for trading stocks is the business of market makers. Like any (for profit) business the market makers expect to earn money from the operation of the market. There are a number of ways this may happen. The market makers may charge buyers and sellers fees or commissions. Money can also be earned from the difference in the buying and selling price of a stock. In a stock market there may be several owners of a particular corporation’s stock who are willing to sell and a number of buyers wishing to add that stock to their portfolio, and investors, both buyers and sellers may have different prices for the stock. A price a buyer is willing to pay is called a bid price. A price a seller of a stock is willing to accept is known as an ask price. The ask price is known also as the offer price. Typically the ask prices exceed the bid prices. At a given moment in the market the difference between the lowest ask price and the highest bid price is called the bid/ask spread or bid/offer spread. The market maker earns the bid/ask spread on stock trades as reward for operating the market. Commissions, fees, and the bid/ask spread are examples of transaction costs associated with stock trading.

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Forwards and Futures

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

157

Example 6.1 Suppose the lowest ask price of a share of stock is $50.10 and the highest bid price for the stock is $50.00. The bid/ask spread is therefore $0.10 per share. A stock buyer who issues a buy order for 1000 shares will pay $50, 100. The seller will receive $50, 000 and the market maker will earn $100 on the trade (plus any other fees or commissions charged). Without market makers and the markets for trading they create and operate, buyers and sellers of stocks and commodities would be responsible for locating each other. Anyone who has sold a used car knows that there are costs associated with finding potential buyers. When trading in a market buyers and sellers also benefit from knowing the last price at which a stock traded and from the price competition among traders. 6.2

Pricing a Forward Contract

Consumers are accustomed to an instantaneous transfer of ownership when purchasing most items. An MP3 player for instance becomes the buyer’s property the instant the buyer transfers the appropriate amount of money to the seller. To describe the process of purchasing a share of stock we must generalize and will assume that events or actions take place at different times separated by a finite interval. To make the initial scenario simple we will assume that events may take place at time t = 0 and also later at time t = T > 0. There are three components or actions involved in the simplest description of a stock purchase: (1) fixing or agreeing on the price for the stock, (2) making payment for the stock, and (3) transferring ownership of the stock from the seller to the buyer. Logically the price is fixed before payment is made so it is assumed that the price of the stock is fixed at t = 0. However, the remaining two actions may occur at either t = 0 or t = T depending on the arrangement made between buyer and seller. The traditional buyer/seller mode of purchase in which all three events occur simultaneously is called an outright purchase. The situation in which the buyer receives ownership of the stock at time t = 0 but pays for the purchase at the later time t = T is called a fully leveraged purchase. A fully leveraged purchase is equivalent to buying the stock on credit or with borrowed money. A buyer who pays for the stock at time t = 0 but does not receive ownership until time t = T is said to have purchased the security using a prepaid forward contract. The prepaid forward contract allows the seller to retain certain rights associated with ownership

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

158

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

An Undergraduate Introduction to Financial Mathematics

of the stock until t = T . For example, the seller would retain voting rights inherent in ownership of the stock until t = T . Another important reserved right for the seller in the prepaid forward contract situation is the right to receive any dividend payments associated with the stock. Lastly if payment for the stock and transfer of ownership both take place at t = T , it is said that the stock is purchased using a forward contract. The timing and coordination of the payment for the purchase and the transfer of ownership have implications for the amount which should be paid for the contract and security. The simplest case to analyze is the case of outright purchase of the stock. If the stock is worth S(0) at time t = 0 and payment and transfer of ownership will take place at time t = 0 then the amount paid should be S(0). To pay any other amount would create an arbitrage opportunity for either the buyer or the seller (see exercise 2). Since the fully leveraged purchase is equivalent to the buyer purchasing the stock with borrowed money, then if the continuously compounded interest rate for borrowing is rb , the cost of the fully leveraged stock purchase is S(0)erb T . Determining the price of a prepaid forward or forward contract requires a more detailed (though still elementary) argument. To make the argument rigorous we must introduce some new terms and establish some assumptions. We will assume the stock pays no dividends. Dividends are periodic payments paid by a corporation to the stock holders. The funds for the dividends are paid out of a portion of the profit made by the corporation. We will also assume the stock obeys a Wiener process of the form dS = µS dt + σS dW (t). Thus if the stock is worth S(0) at time t = 0 then at time t = T , the expected value of the price of the stock will be E [S(T )] = S(0)eµT according to Lemma 3.1. To develop the pricing argument we will make use of the terms long and short. While the implications of long and short positions can seem foreign (or perhaps even illegal) to novices, keep the following simple definitions in mind. An investor who owns (rather than merely has possession of) a stock, security, or other commodity is said to be in a long position relative to the item. If an investor outright purchases a stock, they have entered into a long position with regards to the stock. An investor who sells a stock, security, or commodity (perhaps by borrowing it on short-term loan from the owner) and must re-purchase it later is in a short position relative to the item. Typically an investor will short a stock if they believe the value of the stock will decrease in the near future.

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Forwards and Futures

BC8495/Chp. 6

159

They borrow the stock, sell it while the price is still relatively high, repurchase it after the price has fallen, return the stock to the lender, and keep the difference as profit. A more detailed description of the practice of shorting stocks is contained in [McDonald (2006)]. The final assumption is that pricing is done within a “no-arbitrage” framework. This means that it is not possible for any party to make a risk-free positive profit. Now we may justify the price of a prepaid forward contract. Theorem 6.1 The price F of a prepaid forward contract on a nondividend paying stock initially worth S(0) at time t = 0 for which ownership of the stock will be transferred to the buyer at time t = T > 0 is F = S(0). Proof. Suppose F < S(0). The buyer can purchase the forward and sell the security (the buyer has entered into a long position on the forward and a short position on the security). Since S(0) − F > 0 the buyer has a positive cash flow at t = 0. At t = T , the buyer receives ownership of the security and immediately closes their short position in the security. The cash flow at t = T is therefore zero. Thus the total cash flow at t = 0 and t = T is S(0) − F > 0. There is no risk in obtaining this positive profit since the forward obligates the seller to deliver the security to the buyer so that the buyer’s short position in the security can be closed out. Hence if F < S(0) arbitrage is present. Suppose F > S(0). The buyer can purchase the security at time t = 0 and sell a prepaid forward (the buyer has entered into a long position on the security and a short position on the forward). Since F − S(0) > 0 the buyer has a positive cash flow at t = 0. At t = T , the buyer must transfer ownership of the security to the party who purchased the forward. The cash flow at t = T is therefore zero. Thus the total cash flow at t = 0 and t = T is F − S(0) > 0. There is no risk in this situation since the buyer owns the security at time t = 0 and thus will with certainty be able to transfer ownership at t = T . Hence if F > S(0) arbitrage is present. Consequently F = S(0).  An alternative proof of the pricing formula for the prepaid forward contract makes use of a present value argument. Since the stock is worth S(0) at time t = 0 and has an expected value of E [S(T )] = S(0)eµT at time t = T , then the continuously compounded rate of return for the stock is µ. Consequently the present value of the prepaid forward is F = E [S(T )] e−µT = S(0).

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

160

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

An Undergraduate Introduction to Financial Mathematics

To summarize the prepaid forward contract, we now understand that for a non-dividend paying stock the value of the prepaid forward contract is the same as the present value of the stock. The reader may question the fairness of paying “full price” for a stock for which transfer of ownership will not be made until a future time. However, we have shown above this is the correct no-arbitrage price. To settle any misgivings a reader might have about this, conduct the following thought experiment. As long as the stock pays no dividends it can be thought of as inert during the interval [0, T ]. Think of buying the stock through the mail. Even though the mail delivery will occur a few days after the purchase, the only behavior of the stock of interest is the market fluctuation of its value, which occurs whether or not the stock is physically in the buyer’s possession. It is more common that an investor will purchase a forward contract rather than a prepaid forward. Recall that a forward contract is similar to a prepaid forward except that the payment for the forward and the transfer of the ownership of the security take place simultaneously at t = T > 0 while the price F of the forward contract is determined at time t = 0. In this section we will assume that the risk-free interest rate (which may be the interest rate on US Treasury Bonds) is denoted by r and this interest is compounded continuously. The reader can easily modify the arguments given below to other interest compounding schedules. We will assume that parties can borrow and lend money at rate r. To determine the price of a forward contract the we can adopt a quick intuitive approach that since only the timing of the payment is different between the forward contract and the prepaid forward, the no-arbitrage price of a forward contract should be the future value of the prepaid forward, all other conditions being the same. Theorem 6.2 Suppose a share of a non-dividend paying stock is worth S(0) at time t = 0 and that the continuously compounded risk-free interest rate is r, then the price of the forward contract is F = S(0)erT . Proof. Suppose F < S(0)erT . The buyer can purchase the forward (which they will not have to pay for until t = T ) and sell the security at time t = 0. The value of the security is S(0) which is lent out at the risk-free rate compounded continuously. Thus the net cash flow at time t = 0 is S(0) − S(0) = 0. At t = T , when the borrower repays the loan, the buyer’s cash balance is S(0)erT . The buyer pays F for the forward in

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Forwards and Futures

BC8495/Chp. 6

161

order to receive the security which is then used to close out the forward position. The cash flow at t = T is therefore −F . Thus the total cash flow at t = 0 and t = T is S(0)erT − F > 0. There is no risk in obtaining this positive profit since the forward obligates the seller to deliver the security to the buyer so that the buyer’s short position in the security can be closed out. Hence if F < S(0)erT arbitrage is present. Suppose F > S(0)erT . The buyer can sell a forward contract which will be paid for at time t = T and borrow S(0) to purchase the security at time t = 0. Thus the net cash flow at time t = 0 is S(0) − S(0) = 0. At t = T , the buyer must repay the loan of S(0)erT and will sell the security for F . The cash flow at t = T is therefore F − S(0)erT > 0. Thus the total cash flow at t = 0 and t = T is F −S(0)erT > 0. There is no risk in this situation since the buyer owns the security at time t = 0 and thus will with certainty be able to transfer ownership at t = T . Hence if F > S(0)erT arbitrage is present. Therefore the no-arbitrage price of the forward contract on the non-dividend paying stock is F = S(0)erT .  Once a prepaid forward or forward contract has been priced and purchased, an investor will be interested in the profit from the transaction. The value of the stock at time t = T is a random variable and may differ from E [S(T )]. The profit from the forward contract is defined as the difference between the price of the contract and the value of the stock when t = T . This is the amount of money the investor would make if they immediately sold the stock at time t = T , the time at which ownership is transferred to the investor. Mathematically we may express this as profit = S(T ) − S(0)erT . Example 6.2 Suppose a share of stock is currently trading for $25 and the risk-free interest rate is 4.65% per annum. The price of a two-month forward contract is F = 25e0.0465(2/12) ≈ 25.1945. A plot of the profit curve is shown in Fig. 6.1. A positive profit is made if at time t = 2/12 the stock is trading above F . In the following section the pricing formulas developed above will be generalized to include the effects of transaction costs and stocks that pay dividends.

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

162

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

An Undergraduate Introduction to Financial Mathematics

Profit 4

2

22

24

26

28

30

SHTL

-2

-4

Fig. 6.1 The profit on a long forward contract is a linear function of the stock price at time t = T .

6.3

Dividends and Pricing

Dividends are paid to owners of stock shares from the profits earned by the corporation issuing the shares. Not all shares earn dividends. Dividends on shares from individual corporations may be paid annually or semi-annually. Occasionally it will be convenient to think of dividends on large and diverse collections of shares as paid continuously. Note that dividends are paid to the shareholders, not to the owners of prepaid forwards or forward contracts. When pricing a prepaid forward or forward contract we must carefully consider the effect of any dividends that will be paid to the shareholders between t = 0 and t = T , since the owner of the forward will not receive these disbursements. The value of the forward must be decreased by the present value of the dividends paid during the interval [0, T ]. If the risk-free interest rate is r, then the present value of an amount D paid at time t where 0 ≤ t ≤ T is De−rt . Consequently if n dividends in the amounts {D1 , D2 , . . . , Dn } are paid at times {t1 , t2 , . . . , tn } in the interval [0, T ], the price of a prepaid forward on a stock currently valued at S(0) becomes

F = S(0) −

n X

Di e−rti .

i=1

Using the idea that the value of a forward contract is the future value of the prepaid forward we have the value of the forward contract for a stock

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Forwards and Futures

BC8495/Chp. 6

163

paying dividends at discrete times expressed as F = S(0)erT −

n X

Di er(T −ti ) .

i=1

Example 6.3 Suppose the risk-free interest rate is 4.75%. A share of stock whose current value is $121 per share will pay a dividend in six months of $3 and another in twelve months of $4. An investor is pricing a one-year forward contract and one-year prepaid forward on the stock assuming that transfer of ownership will take place immediately after the second dividend is paid. The value of the prepaid forward is F = 121 − 3e−0.0475(6/12) − 4e−0.0475(12/12) ≈ 114.256. The value of a forward contract on the dividend paying stock is F = 114.256e0.0475(12/12) ≈ 119.814. If a forward is being purchased on a portfolio of stocks paying dividends at many times, it may be convenient to think of the investment as paying dividends continuously. In this case let the dividend rate be denoted by δ. Only simple modifications are needed in the pricing formulas above. The value of a prepaid forward on a stock currently worth S(0) becomes F = S(0)e−δT . The value of a forward contract on the same security is F = S(0)e(r−δ)T . Example 6.4 An investment valued at $117 pays dividends continuously at the annual rate of 2.55%. The risk-free interest rate is 3.95%. Therefore the price of a four-month prepaid forward on the investment is F = 117e−0.0255(4/12) ≈ 116.01. The value of a four-month forward contract on the investment is F = 117e(0.0395−0.0255)(4/12) ≈ 117.547. 6.4

Incorporating Transaction Costs

In the previous sections we proved there is a single no-arbitrage price for a prepaid forward and a forward contract. This ignores the possibility

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

164

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

An Undergraduate Introduction to Financial Mathematics

of there being transaction costs associated with the buying and selling of the security and the forward. In this section we will generalize the specifications surrounding the forward contract and determine an interval of no-arbitrage prices for the forward rather than a single no-arbitrage price. In this section we will make use of the following definitions and notation for the components of the transactions. S a : the time t = 0 ask price at which the security can be bought. S b : the time t = 0 bid price at which the security can be sold. In general Sb < Sa. rb : the continuously compounded interest rate at which money may be borrowed. rl : the continuously compounded interest rate at which money may be lent. In general rl < rb . k: the cost per transaction for executing a purchase or sale. In this section we will derive an interval of forward contract prices of the form [F − , F + ] for which no arbitrage is possible when F − ≤ F ≤ F + . Outside of this interval arbitrage may be possible. Suppose the forward contract has value F . We will show below that in the absence of arbitrage F ≤ (S a + 2k)er

b

T

≡ F +.

For the sake of contradiction assume that F > F + . At time t = 0 the buyer may borrow amount S a + 2k to purchase the security and sell the forward contract. Since a transaction cost of k is incurred for both the purchase of the security and the sale of the forward the amount 2k, in addition to the ask price S a of the security at time t = 0, must be borrowed. As before the payment for the forward will be made at time t = T > 0. Thus the net cash flow at time t = 0 is zero. At time t = T the loan must be repaid in b the amount of (S a + 2k)er T and the buyer receives F for the forward. The total cash flow for times t = 0 and t = T is therefore F − (S a + 2k)er

b

T

= F − F + > 0.

Hence when F > F + arbitrage results. Now we will show that in the absence of arbitrage F ≥ (S b − 2k)er 10:57:37.

l

T

≡ F −.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

165

Forwards and Futures

For the sake of contradiction assume that F < F − . At time t = 0 the buyer can purchase the forward contract (for which F will be paid at time t = T > 0) and sell short the security for S b . A transaction cost of k is paid at time t = 0 for the forward contract and another transaction cost of k is incurred during the short sale the net proceeds from the sale are S b − 2k. This amount is lent out at interest rate rl until time t = T . At time t = T l the buyer’s cash balance is (S b − 2k)er T . Also at this time the buyer pays F for the forward contract and closes out the short position in the security. Thus the total cash flow at times t = 0 and t = T is (S b − 2k)er

l

T

− F = F − − F > 0.

Hence when F < F − arbitrage is possible. To summarize, for the situation when transaction costs are included in the mathematical model, the arbitrage-free forward contract price must satisfy the inequality (S b − 2k)er

l

T

≤ F ≤ (S a + 2k)er

b

T

.

(6.1)

Example 6.5 Suppose the asking price for a certain stock is $55 per share, the bid price is $54.50 per share, the fee for buying or selling a share or a forward contract is $1.50 per transaction, the continuously compounded lending rate is 2.5% per year, and the continuously compounded borrowing rate is 5.5% per year. The price of a three-month forward contract on the stock would fall in the interval (54.50 − 2(1.50))e0.025(3/12) ≤ F ≤ (55 + 2(1.50))e0.055(3/12) 51.7223 ≤ F ≤ 58.8030

Throughout the remainder of this book, unless specifically mentioned, we will ignore the complications introduced by transaction costs and dividends. 6.5

Futures

A forward contract is very “customizable” in that the terms of the contract can be arranged to the satisfaction of the parties involved. The date of maturity of the forward, the volume of stocks or of a commodity to be exchanged at the maturity of the contract, and any necessary collateral to be held to reduce the risk of default by one or more of the parties may all be decided by the parties engaged in the contract. The parties may even decide that at maturity they will only exchange the net amount of

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

166

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

An Undergraduate Introduction to Financial Mathematics

profit earned by the parties on the transaction instead of actually selling or buying the underlying security or commodity. Futures are like this last type of “cash-settled” forward contract with some additional differences. Futures are generally traded in a more structured exchange market. Futures have standardized maturity dates (typically a few months in the future) and standardized volumes of the underlying security or quantity. There are other important differences between forward and futures contracts. There is generally less risk of a party involved in a futures contract defaulting since daily adjustments to futures contracts take place and are managed by the clearinghouse associated with a futures exchange. The clearinghouse will require a deposit from both the party and the counter-party to the futures contract. This deposit is called a margin. The margin protects each party to the futures contract against default by the other party. The clearinghouse will then, based on subsequent changes in the futures price, require additional deposits to the margin so as to protect both parties from default. The process of adjusting the financial amounts owed to the parties in the futures contract is called marking-to-market. In contrast, recall that forward contracts are settled on the date of maturity of the contract. A futures exchange will generally have rules governing the practice of trading depending on changes in the price of the contract traded. For example trading on a particular futures contract may be temporarily halted if the price suddenly moves downward by a specified threshold proportion. The last difference we will mention is that due to the standardization and trading infrastructure provided by a futures exchange, futures are easily traded. An investor wishing to rid themselves of a particular obligation implied by a contract may easily purchase an offsetting opposite contract with the same date of maturity as the original contract. If the risk-free interest rate is constant for the life of the contract, the price of a futures contract is the same as the price of a forward contract. This will be our assumption for the remainder of this chapter. We will also present an extended example of the process of marking-to-market. For the purpose of the example assume an investor is purchasing a 7-day futures contract whose initial price is $1000. The price may change daily until maturity. A volume of 5000 futures contracts will be purchased. The clearinghouse will require that a minimum margin of 10% of the current value of the futures contract be maintained until maturity. The margin will earn

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

167

Forwards and Futures Table 6.1 The daily values of the futures price and the margin balance for a hypothetical seven-day futures contract. The column headed “Margin Balance” shows the margin balance after interest is credited but before the amount in the “Margin Call” column is added to the margin balance. Day 0 1 2 3 4 5 6 7

No. of Contracts 5000 5000 5000 5000 5000 5000 5000 5000

Futures Price 1000.00 987.90 987.97 990.53 988.37 973.89 968.70 980.82

Price Change — −12.10 0.07 2.56 −2.16 −14.48 −5.19 12.12

Margin Balance 500, 000.00 439, 691.82 494, 489.50 507, 479.20 496, 873.89 424, 664.51 461, 181.81 545, 135.81

Margin Call — 54, 258.18 0.00 0.00 0.00 62, 280.49 23, 168.19 0.00

interest at the risk-free rate of 14% per annum compounded continuously.1 The ultimate profit to the investor will be the difference between the final margin balance and the future value of the initial margin balance. To start the investor deposits a margin of (5000)(1000)(0.10) = 500, 000. Suppose that on the next day the price of the futures contract has fallen to $987.90. The change in futures price is ∆F = −12.10. Multiplied by the 5000 futures contracts the investor owns, the wealth of the investor has changed by −60, 500. This loss will be taken from the margin balance. The initial margin balance has earned a day’s interest, so its new post-loss balance is 500, 000e0.14/365 − 60, 500 ≈ 439, 691.82. Minding the minimum 10% margin requirement, we see the margin should be at least (5000)(987.90)(0.10) = 493, 950, therefore the owner of the futures contract must add 493, 950 − 439, 691.82 = 54, 258.18 to the margin. The request of the additional margin deposit is called a margin call. After bringing the margin up to the minimum level (sometimes called the maintenance margin) the next day’s adjustments can be made. This daily process of marking-to-market continues until the futures contract matures. The daily values of the futures price and the margin balance are shown in Table 6.1. The profit to the holder of the long position in the futures contract is the difference between the final margin balance 1 The interest rate is set artificially high to magnify the daily changes in the margin balance.

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

168

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

An Undergraduate Introduction to Financial Mathematics

and the future value of the initial margin. In this example 545, 135.81 − 500, 000e7(0.14)/365 = 43, 791.50. 6.6

Exercises

(1) Suppose the bid/ask spread for a share of a particular stock is $0.25. An investor buys 1000 shares and then immediately sells them (before the bid and ask prices can change). What is the total transaction cost (the so-called round trip cost) to the investor? (2) Show that in the absence of arbitrage, the price paid for outright purchase of a stock should be S(0), the price of the stock at the time of outright purchase. (3) Suppose the continuously compounded interest rate for borrowing is 5.05% per year. What is the cost to the buyer for a fully leveraged purchase of a stock worth now $17 and for which payment will be made in one month? (4) Suppose the continuously compounded risk-free interest rate is 4.75% per year. What is the cost of a three-month forward contract on a non-dividend paying stock whose value currently is $23? (5) Suppose the risk-free interest rate is 3.65%. A share of stock whose current value is $97 per share will pay a dividend in six months of $2.50 and another in twelve months of $2.75. Find the values of one-year forward contract and one-year prepaid forward on the stock assuming that transfer of ownership will take place immediately after the second dividend is paid. (6) An investment valued at $195 pays dividends continuously at the annual rate of 1.95%. The risk-free interest rate is 4.55%. Find the prices of a three-month prepaid forward and a three-month forward contract on the investment. (7) A security is currently priced at $1000. The continuously compounded risk-free interest rate is 5.05% annually. The price of a six-month forward contract for the security is $990. If the security pays dividends continuously at rate r, find r expressed as an annual percentage. (8) Suppose the asking price for a certain stock is $75 per share, the bid price is $74 per share, the fee for buying or selling a share or a forward contract is $2 per transaction, the continuously compounded lending rate is 3% per year, and the continuously compounded borrowing rate is 4% per year. Find the interval of no-arbitrage prices for a six-month

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Forwards and Futures

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

169

forward contract on the stock. (9) A $80 stock pays a $2 dividend every 3 months, with the first dividend coming 3 months from today. The continuously compounded risk-free interest rate is 4.5%. (a) What is the price of a prepaid forward contract that expires 12 months from today immediately after the fourth dividend payment? (b) What is the price of a forward contract that expires 12 months from today immediately after the fourth dividend payment? (10) A $75 stock pays a continuous dividend at the rate of 8%. The continuously compounded risk-free interest rate is 4.75%. (a) What is the price of a prepaid forward contract that expires 12 months from today? (b) What is the price of a forward contract that expires 12 months from today? (11) Suppose the price of a stock is $85 and the continuously compounded interest rate is 6.25%. If the price for a 12-month forward contract on the stock is $89, what is the continuous dividend yield on the stock? (12) Suppose an investor buys 1000 futures contracts for $875 each. The continuously compounded annual interest rate is 5.5% and the futures contracts will be marked to market weekly. The initial margin is equal to 25% of the value of the contracts purchased and margin equal to 80% of the initial margin must be maintained. What is the greatest price of the contract one week later which will trigger a margin call? (13) A security is currently priced at $950 per share. An investor purchases a 6-month futures contract on 100 shares of the security. The continuously compounded risk-free interest rate is 6%. The initial margin on the futures contract is 12.5% and the futures position will be marked to market monthly. The maintenance margin is likewise 12.5%. What is the highest security price for which a margin call will be made after one month? (14) Suppose a stock pays no dividends and the risk-free rate of continuously compounded interest is r. An investor has the choice between the following investments: (a) Purchasing one share of the stock at time t = 0 and selling it at time t = T > 0, (b) Entering into a long forward contract on the stock at time t = 0 and lending the present value of the long forward contract until

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

170

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 6

An Undergraduate Introduction to Financial Mathematics

time t = T . Show that the payoffs of the two investments are the same. (15) An investor is purchasing a 10-day futures contract whose initial price is $850. The price of the contract changes daily following the path described in the table below. Day 1 2 3 4 5

Price 774.67 779.39 778.42 749.56 742.87

Day 6 7 8 9 10

Price 735.64 741.59 759.88 766.25 805.36

A volume of 1500 futures contracts will be purchased. The clearinghouse will require that a minimum margin of 15% of the current value of the futures contract be maintained until maturity. The margin will earn interest at the risk-free rate of 10% per annum compounded continuously. Create a table for the daily accounting of marking to market for this futures contract similar to Table 6.1. What is the profit on the futures contract to the investor? (16) Suppose the price of copper is $4.10 per ounce. Copper purchased as an investment must be stored until it is sold. The storage cost must be paid monthly in advance and is $0.10 per ounce per month. Assume the risk-free interest rate is 3.5% compounded continuously. Find the price of a forward contract for an ounce of copper due to be delivered to the buyer in 6 months. (17) At the beginning of June, forward prices on copper were as listed in the table below. The storage cost of copper is $0.0053 per ounce per month payable at the beginning of the month. The forwards mature at the ends of the given months. Month July August September December

F 4.0995 4.0945 4.1215 4.1235

Estimate the price of an ounce of copper on at the beginning of June and estimate the continuously compounded interest rate. (18) Suppose at the beginning of June a company owns a copper mine which will produce the amounts of copper shown in the following table

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Forwards and Futures

BC8495/Chp. 6

171

at the end of each listed month. Month July August September December

Copper Production (ounces) 1500 1575 1625 1500

The cost to mine an ounce of copper is 1.25. Assume the risk-free interest rate is 3% compounded continuously. Using the forward prices for copper in Ex. 17 find the present value of the mining operation at the beginning of June. (19) A bond is currently selling for $1050. An investor owns a forward contract on this bond. The maturity date for the forward is one year from now and the delivery price of the contract is $1000. The bond pays an $50 dividend every six months. One dividend will be paid six months from now and a second dividend will be paid one year from now just prior to the maturity of the forward contract. Find the current value of the forward contract if the interest rate for six months is 2.75% and the interest rate for one year is 3.25% compounded semiannually. (20) Currency forwards are sometimes used by banks to eliminate risk due to uncertainties in future currency exchange rates when debts must be paid in a foreign currency. The currency forward allows the spot exchange rate to be “locked in” even though the debt will not be repaid until sometime in the future (generally one to six months hence [Cuthbertson and Nitzsche (2004)]). Suppose the spot exchange rate of euros for dollars is S (i.e. S euros will currently buy $1). Let the annual continuously compounded interest rate for dollars be r and the annual continuously compounded interest rate for euros be re . Let F be the forward exchange rate (again in euros for dollars). Show that in the absence of arbitrage F = Se(r−re )T where T is the time to maturity of the forward.

10:57:37.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

This page intentionally left blank

10:57:37.

172

BC8495/Chp. 6

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

Chapter 7

Options

In the present world of finance there are many types of financial instruments which go by the name of options. At its simplest, an option is the right, but not the obligation, to buy or sell a security such as a stock for an agreed upon price at some time in the future. The agreed upon price for buying or selling the security is known as the strike price. Options come with a time limit at which (or prior to) they must be exercised or else they expire and become worthless. The deadline by which they must be exercised is known as the exercise time, strike time, or expiry date. Often the exercise time will simply be called expiry. In the remainder of the text the three terms will be used interchangeably. An option to buy a security in the future is called a call option. An option to sell is known as a put option. Types of options can also be distinguished by their handling of the expiry date. The European option can only be exercised at maturity, while an American option can be exercised at or before expiry. Of the two types, the European option is simpler to treat mathematically, and will be the focus of much of the rest of this book. However, in practice, American options are more commonly traded. There is a mathematical price to be paid in terms of complexity for the added flexibility of the American-style option. Suppose stock in a certain company is selling today for $100 per share. An investor may not want to buy this stock today, but may want to own it in the future. To reduce the risk of financial loss due to a potential large increase in the price of the stock during the next three-month period, they buy a European call option on the stock with a three-month strike time and a strike price of $110. At the expiry date, if the price of the stock is above $110 and the investor now wishes to buy the stock, they have the right as holder of the call option to purchase it for $110, even if the market

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

174

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

value of the stock is higher. Otherwise, if the value of the stock is below $110 and the investor still wishes to buy the stock, they will let the call option expire without exercising it, and purchase the stock at its market price. Call options allow investors to protect themselves against paying an unexpectedly high price in the future for a stock which they are considering purchasing. We have used the language of “buying” an option. Thus options themselves have a value (a price), just as the securities which underlie the options have a value. An important issue to consider is, how are values assigned to options? In light of the Arbitrage Theorem of Chapter 4, if the option is mis-priced relative to the security, arbitrage opportunities may be created. In this chapter we will explore some of the relationships between option and security prices. We will derive the Black-Scholes partial differential equation together with the boundary and final conditions which govern the prices of European style options. During the explanations of many of the concepts in this chapter, we will refer to buying or selling a financial instrument. In the financial markets it is often possible to sell something which we do not yet own by borrowing the object from a true owner. For example, investor A may borrow 100 shares of stock from investor B and sell them with the understanding that by some time in the future investor A will purchase 100 shares of the same stock and return them to investor B. Normally one buys an object first and sells it later, but in the financial arena one can often sell first and then buy later. Borrowing and selling an object with the agreement to re-purchase and return it later is called adopting a short position in the object, or sometimes shorting the object. Adopting a short position can be a profitable transaction if the investor believes the price of the object is going to decrease. Purchasing an object first and then selling it in the future is called adopting a long position.

7.1

Properties of Options

There are many relationships between the values of options, their underlying securities, and other market parameters. The maintenance of most of these relationships is necessary to eliminate the possibility of arbitrage. We will cover some of these relationships here and develop methods and strategies for proving these relationships must hold in an arbitrage-free setting. Throughout this section we will use the following definitions.

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Options

BC8495/Chp. 7

175

C a : value of an American-style call option C e : value of a European-style call option K: strike price of an option P a : value of an American-style put option P e : value of a European-style put option r: continuously compounded, risk-free interest rate δ: continuously compounded, dividend yield rate S: price of a share of a security T : exercise time or expiry of an option t: current time, generally with 0 ≤ t ≤ T We have used the term “risk-free” to describe the interest rate r. It is assumed that this is the rate of return for an investment which carries no risk. While the reader may wish to debate whether a complete absence of risk can ever be achieved, there are some investments which carry very little risk, for example, U.S. Treasury Bonds. For the initial exploration of option properties, we will also assume that the security pays no dividends. Once the basic relationships have been established, the effects of dividends are straightforward to accommodate. Consider the cost of an American option compared with a European option. In the absence of arbitrage, an American-style option must be worth at least as much as its European counterpart, i.e. C a ≥ C e and P a ≥ P e . Naturally we are assuming that the strike prices, strike times, and underlying securities are the same. On the contrary, suppose that and American-style option is worth less than its corresponding European-style option, i.e. C a < C e . Taking the position of an informed investor, knowing that the American-style option has all the characteristics of the European option and in addition has the increased flexibility that it may be exercised early, no investor would purchase a European-style call option if it cost more than the American-style option. To formulate a more mathematical proof, suppose an investor sells the European-style call option and purchases the American-style option. Since C e − C a > 0, the investor may purchase a risk-free bond paying interest at rate r compounded continuously. At the expiry date, the bond will have value (C e − C a )erT . If the owner of the European option wishes to exercise the option, the investor insures this is possible by exercising their own American option. If the owner of the European option allows it to expire unused, then the investor can do the same with the American option. Thus in both cases the investor makes a risk-free profit of (C e − C a )erT which is an example of arbitrage. 14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

176

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

It is also the case that C e ≥ S − Ke−rT or arbitrage is present. The reader should practice interpreting these inequalities in financial terms. The last inequality can be understood as that a European call option on a non-dividend paying stock must be worth at least as much as the difference between the current value of the stock and the present value of the strike price associated with the option. To establish the inequality (and most of the option properties to come) we use the technique of proof by contradiction. Assuming an absence of arbitrage and that C e < S − Ke−rT , an investor can purchase the call option and sell the security at time t = 0. Since 0 < Ke−rT < S−C e , the investor can invest S−C e in a risk-free bond paying interest rate r. At the strike time the bond is worth (S − C e )erT which is greater than K. Thus if the investor chooses to exercise the option (because S at time T is greater than K), the bond can be cashed out, the security purchased for K, and there is still capital left over. In other words there is risk-free profit. On the other hand, if S at time T is less than K, the short position on the security is eliminated using the bond and purchasing the stock for S (which is less than K). Again there will still be capital left over. Once again there is a risk-free profit. There is strict arbitrage-free relationship between the value of a European call and put with the same strike price and expiry date on the same underlying security. This is the important Put-Call Parity Formula of Eq. (7.1). P e + S(0) = C e + Ke−rT

(7.1)

In financial terms, the value of a European put and the stock equals the value of a European call and the present value of the strike price (with the assumptions that the underlying stocks for both options are the same, the stock pays no dividends, and the expiry dates of the options are the same). To see why this relationship must be true, imagine portfolio A represents the left-hand side of Eq. (7.1) while the right-hand side is represented by portfolio B. The Put-Call Parity formula implies that in an arbitrage-free setting these portfolios must have the same value. Suppose portfolio A is worth less than portfolio B, i.e. P e + S < C e + Ke−rT .

(7.2)

An investor can borrow at interest rate r an amount equal to P e + S − C e . This would allow the investor to purchase the security, the European put option, and to sell the European call option. At the strike time of the two

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Options

BC8495/Chp. 7

177

options, the investor must repay principal and interest in the amount of (P e + S − C e )erT . If the security is worth more than K at time T , the put expires worthless and the call will be exercised by its owner. The investor must sell the security for K. Thus the net proceeds of this transaction are K − (P e + S − C e )erT > 0

(7.3)

since inequality (7.3) is equivalent to the one in (7.2). If the security is worth less than K at time T , the call expires worthless and the put will be exercised by the investor. Again the investor will sell the security for K. The net proceeds of this transaction are the same as in the previous case. Thus there is a risk-free profit to be realized if portfolio A is worth less than portfolio B. Now suppose portfolio A is worth more than portfolio B, i.e. P e + S > C e + Ke−rT .

(7.4)

An investor can sell the security and the European put option and buy the call option. This generates an initial positive flow of capital in the amount of S + P e − C e . This amount will be invested in a risk-free bond earning interest at rate r until the expiry date arrives. At that time the investor will have (S +P e −C e )erT . If the security is worth more than K at time T , then the put option is worthless and investor will exercise the call option. The investor purchases the security for K (thus canceling their short position). This leaves the investor with a net gain of (P e + S − C e )erT − K > 0

(7.5)

since inequality (7.5) is equivalent to the one in (7.4). If the security is worth less than K at time T , the call option expires unused and the owner of the put option will exercise it. Thus the investor will clear their short position by buying the security for K and their net gain is as before. Consequently, there exists an arbitrage opportunity if portfolio B is worth less than portfolio A. Therefore the two portfolios must have the same value, i.e. the Put-Call Parity formula must be true. Assuming that the prices of European call and put options are nonnegative (a mild assumption), we can derive two corollary inequalities to the Put-Call Parity formula. C e ≥ S − Ke−rT

P e ≥ −S + Ke−rT 14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

178

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

The first of these has already been proved directly. While none of the formulas or inequalities developed so far enable us to price options directly, they are useful checks for arbitrage opportunities. 7.2

Including the Effects of Dividends

Now that the Put-Call Parity formula (7.1) for non-dividend paying stocks has been developed, incorporating the effects of paid dividends is a matter of understanding how dividends affect the value of the underlying security. In this section we will explore how discrete and continuous dividend payments alter the value of a security and modify the Put-Call Parity formula to account for these changes. Suppose a stock will pay a dividend in the amount of δS(t) (that is, a dividend proportional to the value of the stock S(t)) at time td > 0. What happens to the value of the stock at the instant the dividend is paid? Since something of value is leaving the corporation issuing the stock, the value of the company (and hence its stock) should reflect this loss. Consider the values of the stock immediately before and after the payment of the dividend. lim S(t) = S(t− d)

t→t− d

lim S(t) = S(t+ d)

t→t+ d

The before and after dividend stock values are related by the following equation. − S(t+ d ) = S(td )(1 − δ)

(7.6)

This equation states that the value of the stock must diminish by the amount of the dividend payment. A “no arbitrage” argument will establish − this rigorously. If S(t+ d ) > S(td )(1 − δ), then an investor may purchase the stock immediately before the dividend is paid for S(t− d ), collect a dividend − in the amount of δS(td ), and sell the stock immediately after the dividend payment for S(t+ d ), producing a net profit of − − + − S(t+ d ) + δS(td ) − S(td ) = S(td ) − S(td )(1 − δ) > 0. − If S(t+ d ) < S(td )(1 − δ), then an investor may short the stock immediately before the dividend is paid for S(t− d ) and purchase the stock immediately

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Options

BC8495/Chp. 7

179

after the dividend payment for S(t+ d ), producing a net profit of + − + S(t− d ) − S(td ) > (1 − δ)S(td ) − S(td ) > 0.

In either case arbitrage is present. Thus for a single discrete dividend payment the value of the stock drops by the amount of the dividend. This affects the left-hand side of the PutCall Parity formula (7.1). We do not simply subtract the dividend payment from the left-hand side since this would ignore the time value of the dividend, instead we should subtract the present value of the dividend payment. This idea is generalized to a finite number of discrete dividend payments made between the sale of the options and expiry. If n dividend payments − of the form δS(t− i ) will be made at times ti for i = 1, 2, . . . , n then the Put-Call Parity formula for discrete dividend payments can be expressed as e

P + S(0) − δ

n X

−rti S(t− = C e + Ke−rT . i )e

(7.7)

i=1

In some situations it is mathematically convenient to think of the dividends as a continuous stream of payments at a rate δ per unit time. For example, if the security under consideration is an index, a financial instrument made up of hundreds or perhaps thousands of stocks each of which may be paying dividends on its own schedule. In this context δ is called the dividend yield. If the dividend yield is constant then the dividend paid over a short time interval [t, t + ∆t] is approximately δS(t)∆t. Therefore S(t + ∆t) − S(t) = −δS(t) ∆t dS = −δS(t) (as ∆t → 0) dt S(t) = S(0)e−δt . Consequently for European options on securities which pay dividends at a continuous, constant dividend yield δ, the Put-Call Parity formula takes on the form P e + S(0)e−δT = C e + Ke−rT .

14:29:39.

(7.8)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

180

7.3

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

Pricing an Option Using a Binary Model

In this section we will examine a very simple option/security situation and determine the arbitrage-free price for the security. The analysis done here is too simple to be applied directly to any real-world example of a security; however, it provides an example of an application of the Arbitrage Theorem (Theorem 4.5). This type of analysis can be generalized and extended to more relevant real-world examples and thus provides a good starting point for the goal of developing option pricing formulae. The analysis performed here is based on a binomial model, so named since the security is assumed to take on one of two values in the future. Suppose the price of a share of stock is $100 currently. After a single, indivisible unit of time T , the price of the share will either be $200 or $50. The stock will be worth $200 after time T with probability p or will be worth $50 with probability 1 − p. Currently an investor can purchase a European call option whose value is C. The exercise time and strike price of the option are respectively T and $150. In the absence of arbitrage what is the value of C? The investor could buy either the option or the stock at the beginning of the time interval. In the absence of arbitrage, the expected value of the investor’s profit from either course of action should be zero regardless of the direction in which the price of the stock moves. Since the option or stock may be purchased at the beginning of the time interval but the profit (if any) arrives after time T has elapsed, we must calculate the present value of any potential profit. We will assume the interest rate per T unit of time is r. Suppose the investor purchases the stock initially. At time t = 0 their net gain is −100 + 200(1 + r)−1 with probability p or −100 + 50(1 + r)−1 with probability 1 − p. In an arbitrage-free setting the expected value of this gain is zero.   0 = −100 + 200(1 + r)−1 p + −100 + 50(1 + r)−1 (1 − p) 0 = −100(1 + r) + 150p + 50 1 + 2r p= 3

(7.9)

On the other hand, if the investor purchases the option initially, at time t = 0 their net gain is −C + (200 − 150)(1 + r)−1 with probability p or −C with probability 1 − p. Again, the expected value of this gain will be zero in the absence of arbitrage.

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

181

Options

 0 = −C + 50(1 + r)−1 p + (−C)(1 − p)

0 = 50p(1 + r)−1 − C 50(1 + 2r) C= 3(1 + r)

(7.10)

Equation (7.10) was found using Eq. (7.9). The arbitrage-free probability and the option cost both depend on the interest rate. Figure 7.1 shows a parametric plot of the points (p(r), C(r)). If the probability p and the call option price deviate from the curve shown, then a risk-free profit can be made.

24

22 C 20

18

0.4

0.5

0.6

0.7

0.8

0.9

1.0

p

Fig. 7.1 rate.

A parametric plot of probability and call option cost as a function of interest

Example 7.1 at time T = 1

Suppose the current value of the stock is S(0) = $100 and

S(1) =



$200 with probability p = 11/30, $50 with probability 1 − p = 19/30.

Suppose further that the risk-free interest rate is r = 5% and that a European call option for the stock can be purchased for C = $17.50. Note that by Eq. (7.10) an arbitrage-free option should be priced at $17.46. The strike price of the option is K = $150. A risk-free profit can be generated in

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

182

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

this situation since the option has been mis-priced according to Eqs. (7.10) and (7.9). Suppose an investor borrows sufficient cash to purchase x shares of stock and y call options. The portfolio is originally worth 100x + 17.50y and they must pay back e0.05 (100x + 17.50y) at time T = 1. If S(1) = $200 then the call option will be exercised. Liquidating the portfolio yields 200x + (200 − 150)y in revenue. If S(1) = $50 the call option should not be exercised. Cashing out the portfolio produces 50x in revenue. Thus if x and y can be found so that the following set of linear inequalities are simultaneously satisfied then a risk free profit can be realized. 200x + 50y > e0.05 (100x + 17.50y) 50x > e0.05 (100x + 17.50y) One such solution is the point with coordinates (x, y) = (1000, −3000). Therefore if the investor takes a long position in 1000 shares of the stock and shorts 3000 call options on the stock, their risk-free profit will be $64.62.

7.4

Black-Scholes Partial Differential Equation

In this section we will derive the fundamental equation governing the pricing of options, the famous Black-Scholes partial differential equation. The ideas of arbitrage, stochastic processes, and present value converge at this point in the study of financial mathematics. We begin by supposing that S is the current value of a security and that it obeys a stochastic process of the form dS = µS dt + σS dW (t)

(7.11)

If F (S, t) is the value of any type of option (more generally called a financial derivative), then according to Itˆo’s Lemma (Lemma 5.4), F obeys the following stochastic process.   1 2 2 dF = µSFS + σ S FSS + Ft dt + σSFS dW (t) (7.12) 2 Suppose a portfolio of value P is created by selling the option and buying ∆ units of the security. Thus the value of the portfolio is P = F − ∆S. The notation ∆ for the number of units of the security purchased is standard in the derivation and analysis of the Black-Scholes equation. The reader

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Options

BC8495/Chp. 7

183

should not let the notation ∆S suggest “change in S”, it merely means ∆ multiplied by S. Since the portfolio is a linear combination of the the option and the security then the stochastic process governing the portfolio is dP = d(F − ∆S)

= dF − ∆dS   1 = µSFS + σ 2 S 2 FSS + Ft dt + σSFS dW (t) 2 − ∆ (µS dt + σS dW (t))   1 2 2 = µS [FS − ∆] + σ S FSS + Ft dt 2 + σS (FS − ∆) dW (t)

(7.13)

Notice the coefficient of the normal random variable, dW (t), contains the factor FS − ∆. Equation (7.13) can be simplified if we assume ∆ = FS . This has the beneficial effect of reducing the number of stochastic terms in Eq. (7.13). Randomness is not completely eliminated since the value of the security S remains and is stochastic. Under the assumption that ∆ = FS , Eq. (7.13) becomes   1 2 2 σ S FSS + Ft dt. (7.14) dP = 2 In an arbitrage-free setting the difference in the returns from investing in the portfolio described above or investing an equal amount of capital in a risk-free bond paying interest at rate r should be zero. Thus the following equations are true. 0 = rP dt − dP   1 2 2 = rP dt − σ S FSS + Ft dt (from Eq. (7.14)) 2 1 2 2 0 = σ S FSS + Ft − rP 2 1 0 = σ 2 S 2 FSS + Ft − r(F − ∆S) 2 1 0 = Ft + rSFS + σ 2 S 2 FSS − rF 2

(7.15)

Notice that ∆ was replaced by FS to obtain the final form of the equation. This is the well-known Black-Scholes equation for pricing financial derivatives. Equation (7.15) is an example of a partial differential equation

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

184

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

or PDE, for short. While the general theory of solving PDEs is beyond the scope of this book, we will briefly mention a few concepts related to the Black-Scholes PDE. PDEs are often described by their order, type, and linearity properties. The Black-Scholes PDE is a second order equation since the highest order derivative of the unknown function F present in the equation is the second derivative. This PDE is of parabolic type. The best known example of the parabolic PDE is the heat equation which is used to describe the distribution of temperature along an object, for example a metal rod. For an introductory treatment of the heat equation see [Boyce and DiPrima (2001)]. Since the coefficients of Ft and FSS have the same algebraic signs, the Black-Scholes PDE is sometimes referred to as a backwards parabolic equation. In [Wilmott et al. (1995)] the BlackScholes PDE is solved by appropriate changes of variables until it becomes the heat equation. The heat equation can be solved by several independent techniques which all naturally, ultimately yield the same solution. One elementary solution technique is the method known as separation of variables. Lastly, the Black-Scholes PDE is an example of a linear partial differential equation since if F1 and F2 are two solutions to the equation then c1 F1 + c2 F2 is also a solution where c1 and c2 are any constants (see exercise 10). Financial derivative products of many types obey the Black-Scholes equation. Different solutions correspond to different initial/final and boundary side conditions imposed while solving the equation. In the next section we will describe in more detail the initial and boundary conditions relevant to the Black-Scholes PDE. In Sec. 8.4 we will derive a solution to the Black-Scholes equation for the price of a European style call option.

7.5

Boundary and Initial Conditions

Without the imposition of side conditions in the form of initial or final conditions and boundary conditions, a partial differential equation can have many different solutions. Mathematicians refer to a differential equation with many possible solutions as “ill-posed”. In this section we will discuss the appropriate conditions to impose in order to obtain a solution to Eq. (7.15) which describes the value of a European call option. The domain of the unknown function F of Eq. (7.15) is a region in (S, t)space. We will refer to this region as Ω ⊂ (S, t)-space. For a European call option we are only interested in the value of the option during the time

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

185

Options

interval [0, T ], where T is the exercise time of the option. During this time interval the price of the security underlying the option may be any non-negative value. Thus the domain of the solution to Eq. (7.15) is Ω = {(S, t) | 0 ≤ S < ∞ and 0 ≤ t ≤ T }. At time t = T the value of the security will either exceed the strike price (in which case the call option will be exercised generating an income flow of S(T ) − K > 0) or the security will have a value less than or equal to the strike price (in which case the call option expires unused and has value 0). When S(T ) > K the call option is said to be in the money. When S(T ) ≤ K the call option is said to be out of the money. Thus the terminal value of a European call option is (S(T ) − K)+ = max{S(T ) − K, 0} where S(T ) is the value of the underlying security at the exercise time and K is the strike price of the option. Graphically the payoff of the portfolio is said to resemble a “hockey stick”. See Fig. 7.2. Hence if F represents HSHTL-KL+

K

Fig. 7.2

SHTL

The piecewise linear curve representing the payoff of a European call option.

a European call option, we can use the following equation as the final condition for the Black-Scholes PDE. F (S, T ) = (S(T ) − K)+

(7.16)

We see from Eq. (7.11) that if S = 0, then dS = 0, i.e. S never changes

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

186

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

and hence remains zero. The boundary at S = 0 is said to be invariant. Thus on the portion of the boundary of Ω where S = 0, the call option would never be exercised and hence must have zero value. Thus we have derived one boundary condition, namely F (0, t) = 0.

(7.17)

Now if we suppose that S is approaching infinity, it becomes increasingly likely that the call option will be exercised, since as S → ∞, S will exceed any finite value of K. Likewise as S → ∞ a put option would never be exercised. Thus according to the Put-Call Parity formula in Eq. (7.1), as S → ∞, C → S − Ke−r(T −t) . Also as S → ∞, the difference S − K ≈ S and hence we have the second boundary condition F (S, t) = S − Ke−r(T −t) ,

as S → ∞.

(7.18)

Thus to summarize the Black-Scholes equation and its final and boundary conditions for a European call option we have the following set of equations. 1 rF = Ft + rSFS + σ 2 S 2 FSS for (S, t) in Ω, 2 F (S, T ) = (S(T ) − K)+ for S > 0, F (0, t) = 0 for 0 ≤ t < T ,

F (S, t) = S − Ke−r(T −t)

as S → ∞.

In the next chapter, a method for determining the solution to this initial, boundary value problem will be described. 7.6

Option Strategies

Earlier in this chapter some of the properties of put and call options were discussed. The most important of these is the Put-Call Parity formula, Eq. (7.1). In this section common uses of options will be explored to reinforce the idea that options function as a form of insurance against changes in the value of an asset (or the lack of change in the value of an asset). The strategies to follow will include cases of an investor being in long as well as short position in an asset and situations in which options are purchased from or sold to others. The simplest case to consider is analogous to the purchase of homeowner’s insurance. Suppose an investor has purchased a stock for price

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

187

Options

S(0) and wishes to insure against the loss of value by the stock. The investor may purchase a put option which establishes the minimum price, the strike price K, for which the stock may be sold in the future. Homeowner’s purchase insurance on their dwellings to make certain that regardless of the occurrence of damages to the home due to natural events (within limits laid out in the insurance policy), the home maintains a minimum value, usually sufficient to replace the structure in the case of complete loss. The strategy of purchasing a put to insure a long position in an asset is called a floor since the asset holder is placing a “floor” under the value of the asset. This is apparent from the profit diagram of the strategy. Assume the investor borrows funds at the risk-free rate to finance the portfolio. If the asset is purchased for S(0) and the put for P while the strike price is K, the continuously compounded, risk-free interest rate is r, then at time t with 0 ≤ t ≤ T , the expiry time the value of the portfolio is max{S(t), K} − (S(0) + P )ert . Since the investor purchased a put with strike price K, the stock can always be sold for a minimum of K. At a fixed time t the graph of the profit resembles that in Figure 7.3. Profit

K

SHtL

Fig. 7.3 The profit diagram for a portfolio consisting of a long position in an asset and a put option on that asset.

For the case of an investor employing the floor strategy, it is plausible that the investor believes the asset will increase in value and wants to insure against the possibility of it losing value. In contrast if the investor believes

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

188

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

the asset will not increase in value, they may want to sell a call option on the asset to another investor. Since the investor who owns the asset is creating and selling the call option, this is sometimes called writing a covered call. If the asset is purchased for S(0), the covered call with strike price K is sold for C, and the continuously compounded, risk-free interest rate is r, then at time t with 0 ≤ t ≤ T the profit from the covered call strategy is min{S(t), K} + (C − S(0))ert . If the asset does not increase in value the call writer keeps the premium charged for the call and hence the call writer may earn a positive profit even when the asset does not appreciate in value. See Figure 7.4. Profit

K

SHtL

Fig. 7.4 The profit diagram for a portfolio consisting of a long position in an asset and sold covered call.

The option strategies available to the investor in a short position in an asset mirror those of the investor in a long position. A short position in an asset must be cleared at a time in the future and the investor may want to insure against the price of the asset rising too high. In this case the investor may want to use an option strategy known as a cap, so called because a “cap” is placed on the amount the investor will have to pay to clear the short position. The cap consists of purchasing a call option on the asset. If the cash generated by the short sale of the asset is S(0), the price of the call is C, the continuously compounded, risk-free interest rate is r, and the strike price of the call is K, then at the time the call is exercised to clear

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

189

Options

the short position, the profit takes the form (S(0) − C)ert − min{K, S(t)}. The net cash flow generated setting up the portfolio earns interest at the risk-free rate. The investor pays either the strike price or the time t price for the asset (naturally whichever is lower) to close out the short position. Figure 7.5 illustrates the shape of the profit diagram. Profit

K

SHtL

Fig. 7.5 The profit diagram for a portfolio consisting of a short position in an asset and a call option on that asset.

An alternative available to the investor who has shorted an asset it to sell a put option. This strategy is known as writing a covered put. Since the investor has sold the put, they are obligated to buy the asset at the strike price of the put, if the option is exercised. This may mean buying the asset at a loss if the current price is below the strike price (keep in mind the investor has shorted the stock and thus does not own it). The investor will have to buy the asset to close the short position in any case, thus the proceeds from the sale of the covered put may provide some extra, positive cash flow to offset the extra cost of the asset if it increases in value. If the put is not exercised, the investor earns the premium charged for the put. The profit of the covered put strategy is given by the following. (S(0) + P )ert − max{S(t), K} The proceeds from the short sale and the sale of the covered put earn interest at the risk-free rate. At time t that the put is exercised, the investor

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

190

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

purchases the stock for the strike price K (in this case S(t) < K) or for the current price S(t) (in this case the option expires unused). See Figure 7.6. Profit

K

SHtL

Fig. 7.6 The profit diagram for a portfolio consisting of a short position in an asset and sold covered put.

The strategies considered so far have each involved an asset and a single option (either purchased or sold). More complicated strategies can be formulated with multiple options. The next category of strategies to be discussed involves two options of the same type (two calls or two puts) and are called spreads. When an investor purchases an option a premium is paid, i.e., there is a cost. This cost decreases the future profit of the investment (but also lowers the risk). Some investors offset some of the cost of the purchased option by selling an otherwise identical option with a different strike price. In exercise 12 the reader will show that calls are non-increasing functions of the the strike price while puts are non-decreasing functions of the strike price. Thus an investor may offset the cost of a call option with strike price K1 by selling a call option with strike price K2 > K1 . Some out of pocket premium must still be paid, but the amount is reduced compared to the simple cap strategy. In the discussion to follow it will be assumed that the out of pocket costs are borrowed at the continuously compounded, risk-free interest rate. The reduced cost does potentially lower the profit inherent in the investment strategy. An investor buying a call option has the belief that the underlying asset will increase in value before expiry. The investor views the call as a way to make a profit equal to the difference between the asset price (at the time

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

191

Options

of exercise) and the strike price of the option, minus the premium charged for the call. The investor may create a bull spread by selling a call option with a higher strike price, but otherwise identical to the one purchased. Let C(K1 ) and C(K2 ) be the premiums charged for the purchased and sold calls respectively. Since K1 < K2 then the initial cash flow is C(K2 ) − C(K1 ) ≤ 0. The profit will depend on where the asset price falls relative to the two strike prices. To keep the explanation simple, assume the options are of European type so that they must be exercised (if at all) at the same time. If S(T ) < K1 < K2 then both options expire unused and the profit is (C(K2 ) − C(K1 ))erT . If K1 < S(T ) < K2 then the purchased option is exercised and the sold option expires unused. In this case the profit is (C(K2 ) − C(K1 ))erT + S(T ) − K1 . Finally if K1 < K2 < S(T ) then both options will be exercised and the profit is (C(K2 ) − C(K1 ))erT + K2 − K1 . The graph of the profit resembles the piecewise linear plot illustrated in Figure 7.7. The investor limits their potential profit by selling the higher strike call. Profit

K1

Fig. 7.7

K2

SHTL

The profit diagram for a typical bull spread.

A bear spread is constructed by purchasing a put with strike price K2 and selling a put with strike price K1 < K2 . An investor employing this option strategy may believe the underlying asset will decrease in value before expiry. Since P (K1 ) ≤ P (K2 ) the investor pays a premium to set up the bear spread; however, the premium is smaller than what would be paid to set up the floor option strategy. In much the same way as the profit for the bull spread was determined we see that if S(T ) < K1 < K2 , both puts

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

192

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

are exercised and the profit to the investor is (P (K1 )−P (K2 ))erT +K2 −K1 . The investor sells the asset for the higher strike price K2 and must guarantee the owner of the lower strike option can sell the asset at price K1 . In effect the investor sells for K2 and buys back for K1 . If K1 < S(T ) < K2 the lower strike put is not exercised and the investor sells the asset for K2 . In this case the profit is (P (K1 ) − P (K2 ))erT + K2 − S(T ). Finally if K1 < K2 < S(T ) then both puts expire unused and the profit is (P (K1 ) − P (K2 ))erT ≤ 0. Investors can reduce risk by entering into a bull spread option strategy (if they believe asset prices will rise) or a bear spread strategy (if they believe asset prices will decline). Some investors may believe that asset prices will change, but are unable to commit to the prediction that they will increase or decrease. This type of investor believes that asset prices are volatile and that in the future the asset price is likely to be different than it is today. To reduce investment risk in a volatile asset, an investor may wish to purchase a call and a put with the same expiration date. The call generates a profit if asset prices rise while the put generates a profit if the asset price falls. If the purchased call and put have the same strike price, which would generally be near the current price of the asset so that both options are at-the-money, this strategy is known as a straddle. At-themoney options are generally more expensive since they are more likely to be exercised. Thus a straddle may have an unacceptably high premium for some investors. A lower premium may be found when purchased out-of-themoney calls and puts (again with the same expiry). This strategy is known as a strangle. The trade-off with the strangle is that asset prices may have to move further to generate a profit than they would with a straddle. Example 7.2 Suppose the current price of a security is $100. An investor can purchase 3-month call and put options with a strike price of $100 for a total of $13.89 to create a straddle. As an alternative the investor can purchase a 3-month call with a strike price of $110 for $3.75 and a 3-month put with a strike price of $90 for $2.46. The premium cost to set up the strangle is $6.21. The profit earned by these alternatives depends on the price of the security at the time of exercise. See Figure 7.8. The profit generated by the straddle is higher than that produced by the strangle if the security price moves further from the common strike price of the straddle’s put and call. However, the strangle generates a smaller loss than the strangle if the security price remains nearly the same over the lives of the options. The extra “insurance” provided by the strangle comes at the price of lower potential profit than that which could be earned with the straddle.

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

193

Options

profit

40

30

20

straddle

strangle

10

60

80

100

120

140

160

SHtL

-10

Fig. 7.8

Comparison of the profits generated by a straddle and a strangle.

An investor holding contrary beliefs, namely that the price of an asset will exhibit less volatility than other investors predict may wish to write or sell a straddle. The profit from a written straddle has a graph which is a reflection of the profit curve of the purchased straddle of which an example is shown in Figure 7.8. The straddle seller makes a positive profit if the asset’s future price remains close to the strike price of the at-the-money written call and put options. However, the losses suffered by this option strategy can be large if the asset price changes significantly. In this case the straddle writer bet that asset prices would not change much, but in reality they did. Thus a cautious investor may wish to insure the written straddle against large losses by purchasing a strangle. The option strategy of combining a written straddle with a purchased strangle is known as a butterfly spread. As described the butterfly spread would involve two calls (one sold and one purchased) and two puts (again, one sold and one purchased). Surprisingly an equivalent payoff and profit can be achieved using two purchased calls (or puts) and two sold calls (or puts). Suppose an investor purchases two call options, one with strike price K1 and one with strike price K3 and sells two calls both with strike price K2 . Assume that K1 < K2 < K3 and generally that the K1 -strike and K3 -strike options are out-of-the-money. All four calls have the same expiration date. For the sake of simplicity assume as well that K2 = (K1 + K3 )/2, the average of the purchased calls. At the time the options are exercised the payoff of the

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

194

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics Table 7.1 Payoff from a butterfly spread created by buying two out-of-the– money calls and selling two at-the-money calls. Asset Price S(t) < K1 K1 < S(t) < K2 K2 < S(t) < K3 K3 < S(t)

Payoff from Purchased Call 0 S(t) − K1 S(t) − K1 S(t) − K1

Payoff from Purchased Call 0 0 0 S(t) − K3

Payoff from Sold Calls 0 0 −2(S(t) − K2 ) −2(S(t) − K2 )

Total Payoff 0 S(t) − K1 K3 − S(t) 0

butterfly spread can be determined from Table 7.1. The profit obtained from the butterfly spread is the payoff minus the net cost of the premium required to set up the option strategy. In general the profit curve resembles that shown in Figure 7.9. profit

K1

Fig. 7.9

K2

K3

SHtL

Profit curve for a butterfly spread with K2 = (K1 + K3 )/2.

The opportunities and methods for combining options are limited only by the imaginations of the investors. The interested reader is urged to explore more option strategies in the texts by [Hull (2000)] and [McDonald (2006)]. 7.7

Exercises

(1) Show that in the absence of arbitrage P a ≥ P e where the underlying security, exercise time, and strike price for both options are the same. (2) Consider a European call option with a strike price of $60 which costs

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Options

(3)

(4)

(5)

(6)

(7)

(8)

(9)

(10)

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

195

$10. Draw a graph illustrating the net payoff of the option for stock prices in the interval [0, 100]. Show that in the absence of arbitrage the value of a call option, either European or American, never exceeds the value of the underlying security. What is the minimum price of a European style call option with an exercise time of three months and a strike price of $26 for a security whose current value is $29 while the continuously compounded interest rate is 6%? If a share of a security is currently selling for $31, a three-month European call option is $3 with a strike price of $31, and the risk-free interest rate is 10%, what is the arbitrage-free European put option price for the security? If a share of a security is currently selling for $31, a three-month European call option is $3, a three-month European put option is $2.25, and the risk-free interest rate is 10%, what is the arbitrage-free strike price for the security? What is the minimum price of a European put option with an exercise time of two months and a strike price of $14 for a stock whose value is $11 while the continuously compounded interest rate is 7%? If a share of a security is currently selling for $31, a three-month European call option is $3, a three-month European put option is $1, the strike price for both options is $30, and the risk-free interest rate is 10%, what arbitrage opportunities are open to investors? A stock has a current value of $36. A four-month call option with a strike price of $38 will cost an investor $2.25. If the continuously compounded interest rate is 4.75%, find the price of the four-month put option with a strike price of $38. Suppose the functions f1 (S, t) and f2 (S, t) each solve the Black-Scholes Eq. (7.15). Show that if c1 and c2 are constants then the function f (S, t) = c1 f1 (S, t) + c2 f2 (S, t) also solves the Black-Scholes equation.

(11) What final and boundary conditions are appropriate for the BlackScholes Eq. (7.15) when F represents a European-style put option? (12) Suppose there are options on the same stock with two different strike prices, K1 < K2 . (a) Show that C(K1 ) ≥ C(K2 ). (b) Show that P (K1 ) ≤ P (K2 ). 14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

196

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

(c) Show that C(K1 ) − C(K2 ) ≤ K2 − K1 . (d) Show that P (K2 ) − P (K1 ) ≤ K2 − K1 .

(13) Consider three call options C(K1 ), C(K2 ), and C(K3 ) and three put options P (K1 ), P (K2 ), and P (K3 ) with corresponding strike prices K1 < K2 < K3 . C(K2 ) − C(K1 ) C(K3 ) − C(K2 ) (a) Show that ≤ . K2 − K1 K3 − K2 P (K2 ) − P (K1 ) P (K3 ) − P (K2 ) (b) Show that ≤ . K2 − K1 K3 − K2 (14) Suppose the continuously compounded, risk-free interest rate is 3.25%, an investor borrows money to purchase a stock for $500 and a sixmonth American put option for $40. The strike price of the put is $495. (a) If after two months the price of the stock is $450 and the investor decides to exercise the put option, determine the amount of profit for this floor strategy. (b) If after two months the price of the stock is $550 and the investor decides to sell the stock, determine the amount of profit for this floor strategy. (15) Suppose the continuously compounded, risk-free interest rate is 3.25%, an investor borrows money to purchase a stock for $498 and sells a six-month American covered call for $35. The strike price of the call is $500. (a) At expiry the price of the stock is $490 and the investor decides to sells the stock. Determine the amount of profit for this covered call strategy. (b) At expiry the price of the stock is $510 and the owner of the call decides to exercise it. Determine the amount of profit to the option writer for this covered call strategy. (16) Suppose the continuously compounded, risk-free interest rate is 2.95%, an investor shorts a stock for $525 and purchases a four-month American call option for $50. The strike price of the put is $530. (a) If after two months the price of the stock is $500 and the investor decides to close the short position, determine the amount of profit for this cap strategy. (b) If after two months the price of the stock is $555 and the investor decides to close the short position, determine the amount of profit for this cap strategy.

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

197

Options

(17) Suppose the continuously compounded, risk-free interest rate is 2.95%, an investor shorts a stock for $475 and sells a three-month American covered put for $45. The strike price of the call is $495. (a) At expiry the price of the stock is $485 and the investor clears the short position. Determine the amount of profit for this covered put strategy. (b) At expiry the price of the stock is $515 and the owner of the put decides to exercise it. Determine the amount of profit to the option writer for this covered put strategy. (18) Suppose the continuously compounded, risk-free interest rate is 3.75%, an investor creates a bull spread on a stock by purchasing a two-month European call option with a strike price of $100 for $7.57 and selling a two-month European call option with a strike price of $110 for $4.75. Find the profit to the investor if at expiry the stock is worth (a) $98, (b) $107, (c) $115. (19) Suppose the continuously compounded, risk-free interest rate is 4.25%, an investor creates a bear spread on a stock by purchasing a threemonth European put option with a strike price of $425 for $15.75 and selling a three-month European put option with a strike price of $400 for $10.25. Find the profit to the investor if at expiry the stock is worth (a) $375, (b) $410, (c) $450. (20) A stock is currently trading for $50. An investor unsure about the future direction of the stock price may purchase the following options. Option Type Call Call Put Put

Strike Price $50 $55 $50 $45

(a) What is the premium cost of a straddle? (b) What is the premium cost of a strangle?

14:29:39.

Premium $3.64 $1.70 $2.72 $0.94

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

198

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 7

An Undergraduate Introduction to Financial Mathematics

(c) In what stock price intervals does the straddle produce a higher profit than the strangle? (21) An investor creates a butterfly spread using puts. Puts with strike prices of $45 and $55 are purchased while two puts with strike prices of $50 are sold. The premiums (per option) for the puts are shown in the table below. Strike Price $45 $50 $55

Premium $0.94 $2.72 $5.68

Determine the profit to the investor if the underlying asset at expiry has the following values. (a) (b) (c) (d)

$40 $47 $52 $57

(22) It was mentioned that the butterfly spread could be created by buying two out-of-the-money calls and selling two at-the-money calls or by buying a strangle and selling a straddle. Show that the payoff from buying a straddle consisting of a put with strike price K1 and a call with strike price K3 and selling a call and a put with strike price K2 = (K1 + K3 )/2 is the same as that shown in Table 7.1.

14:29:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

Chapter 8

Solution of the Black-Scholes Equation

In Chapter 7 the Black-Scholes partial differential equation was derived and summarized in Eq. (7.15). Every European style option satisfies this PDE. The differences in the options are due to different boundary and payoff conditions. In the present chapter we will solve the Black-Scholes equation with boundary and final conditions appropriate for a European style call option. Several methods are available to solve this equation. Some take the limit of a discrete time binomial model of security prices (see [Ross (1999)]). In [Wilmott et al. (1995)] the Black-Scholes PDE is transformed through a sequence of changes of variables to an ordinary differential equation which is solved by elementary means. Their change of variable approach will be mimicked here, but the Fourier Transform will be used to solve the equation. For readers unfamiliar with the Fourier Transform, a brief introduction to the relevant operations are included in this chapter. The Fourier Transform is a standard mathematical device used to convert certain types of partial differential equations into ordinary differential equations. A more complete introduction to Fourier Transforms can be found in either [Greenberg (1998)] or [Jeffrey (2002)]. If the boundary and final conditions of the Black-Scholes PDE are changed from those of the European calls and puts (examples of so-called vanilla options) to more esoteric conditions (of the exotic options), more sophisticated solution techniques or even numerical approximations of the solution may be required. 8.1

Fourier Transforms

Suppose the function f (x) is defined for all real numbers and that the Fourier Transform of f is denoted by F {f (x)} or often as fˆ(ω). By defini14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

200

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

tion the Fourier Transform of f is F{f (x)} = fˆ(ω) =

Z



f (x)e−iωx dx,

(8.1)

−∞

√ where i = −1 and ω is a parameter. The Fourier Transform maps the original function f which is a function of x to a new function fˆ which depends on ω which is often thought of as a frequency. The Fourier Transform of f is meaningful if and only if the improper integral converges. There are many theorems in the literature on the Fourier Transform which list conditions under which the integral of the Fourier Transform converges. It is possible to state very general but technical conditions in such a convergence theorem. For our purposes we will adopt a less technical theorem which states that if the domain of f is all real numbers and if • f and f 0 are piecewise continuous on every interval of the form [−M, M ] for arbitrary M > 0, and Z •



−∞

|f (x)| dx converges,

then the Fourier Transform of f exists, see Theorem 17.9.1 of [Greenberg (1998)]. Example 8.1

Consider the piecewise-defined function  1 if |x| ≤ 1 f (x) = 0 otherwise.

Its Fourier Transform is readily found through the following calculation. Z ∞ fˆ(ω) = f (x)e−iωx dx −∞ 1

=

Z

e−iωx dx

−1

=− =

 1 −iω e − eiω iω

2 sin ω ω

In the previous example we carried out integration of a complex exponential as if it followed the same rules of integration as real-valued functions. This is in fact true. The Euler Identity, eiθ = cos θ + i sin θ was also used to simplify the result [Churchill et al. (1976)].

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solution of the Black-Scholes Equation

BC8495/Chp. 8

201

The Fourier Transform also has an interesting effect on derivatives of functions. Suppose the Fourier Transform of f exists and that f is differentiable for all real numbers. Z ∞ 0 F{f (x)} = f 0 (x)e−iωx dx −∞ Z ∞ −iωx ∞ = f (x)e − f (x)(−iω)e−iωx dx −∞ −∞ Z ∞ = iω f (x)e−iωx dx −∞

ˆ = iω f(ω)

(8.2)

The technique of integration by parts was used to shift the derivative from f to the exponential function. Since f is assumed to be Fourier transformable, f vanishes as |x| → ∞, which implies the leading expression is zero. This property of the Fourier Transform can be extended to higher order derivatives. Theorem 8.1 If f (x), f 0 (x), . . . , f (n−1) (x) are all Fourier transformable and if f (n) (x) exists (where n ∈ N) then F {f (n) (x)} = (iω)n fˆ(ω). Proof. The result has already been shown in Eq. (8.2) for the case where n = 1. The technique of proof by induction is used to establish this result for all higher order derivatives. Suppose the result has been proved for n = k. Z ∞ F{f (k+1) (x)} = f (k+1) (x)e−iωx dx −∞ Z ∞ ∞ = f (k) (x)e−iωx − f (k) (x)(−iω)e−iωx dx −∞ −∞ Z ∞ (k) −iωx = iω f (x)e dx −∞

= iω(iω)k fˆ(ω) (induction hypothesis) = (iω)k+1 fˆ(ω)

Thus the result is true for n = k + 1 and by the principle of mathematical induction, hence for all n ∈ N.  Another Fourier Transform result which we will have need for later concerns the transform of the convolution of two functions. By definition

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

202

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

the Fourier Convolution of two functions f and g is Z ∞ (f ∗ g)(x) = f (x − z)g(z) dz.

(8.3)

−∞

The Fourier Convolution is not the same as the Fourier Transform. The convolution involves no complex integration, merely the integration of the product of two real-valued functions. It is possible to calculate the Fourier Transform of this convolution. Theorem 8.2 (Fourier Convolution Theorem) F {(f ∗ g)(x)} = fˆ(ω)ˆ g(ω), in other words the Fourier Transform of the Fourier Convolution of f and g is the product of the Fourier Transforms of f and g. Proof. We begin by applying the definitions of Fourier Transform and Convolution.  Z ∞ Z ∞ F{(f ∗ g)(x)} = f (x − z)g(z) dz e−iωx dx −∞

−∞

Fubini’s Theorem (see [Bleecker and Csordas (1996)]) allows the order of integration to be switched, and thus the Fourier Transform of the Convolution may be written as  Z ∞ Z ∞ −iωx f (x − z)e dx dz. g(z) F{(f ∗ g)(x)} = −∞

−∞

Now the change of variable x = u + z is used to rewrite the interior integral above. Z ∞  Z ∞ −iω(u+z) F{(f ∗ g)(x)} = g(z) f (u)e du dz −∞ −∞ Z ∞  Z ∞ = g(z)e−iωz f (u)e−iωu du dz −∞

= fˆ(ω)ˆ g (ω)

8.2

−∞



Inverse Fourier Transforms

So far we have seen that the Fourier Transform can be used to map functions of x, their derivatives, and their convolutions to other functions which depend on a frequency-like variable w. We intend to apply this transformation to the Black-Scholes partial differential equation and its associated side conditions. However, any useful solution to the Black-Scholes equation

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solution of the Black-Scholes Equation

BC8495/Chp. 8

203

should depend on the variables and constants in the original equation and not on the frequency variable w. It is necessary to have a method for transforming a function of w back into a function of x. This is the purpose of the Inverse Fourier Transform. By definition the inverse Fourier Transform of fˆ(ω) given by Z ∞ 1 F −1 {fˆ(ω)} = fˆ(ω)eiωx dω. (8.4) 2π −∞ Example 8.2 Suppose a is a positive constant, then the inverse Fourier Transform of e−a|w| is Z ∞ 1 −1 −a|w| F {e }= e−a|w| eiωx dω 2π −∞ Z 0 Z ∞ 1 1 (a+ix)ω = e dω + e(−a+ix)ω dω 2π −∞ 2π 0 1 1 = + 2π(a + ix) 2π(a − ix) a = π(a2 + x2 ) Readers with some prior exposure to the Fourier Transform may prefer more symmetric definitions of the transform and the inverse transform. Variations on the formulas abound, though they are all equivalent and compatible with the versions given above in Eqs. (8.1) and (8.4). Some people prefer each √ of the forward and inverse transforms to be scaled by a factor involving 2π as in the following. Z ∞ 1 fˆ(ω) = √ f (x)e−iωx dx 2π −∞ Z ∞ 1 f (x) = √ fˆ(ω)eiωx dω 2π −∞ 8.3

Changing Variables in the Black-Scholes PDE

To recap the partial differential equation and its side conditions for a European style call option, we have the following three equations 1 rF = Ft + σ 2 S 2 FSS + rSFS for t ∈ (0, T ), S ∈ (0, ∞) 2 F (S, T ) = (S − K)+ for S ∈ (0, ∞) 14:29:51.

(8.5) (8.6)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

204

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

F (0, t) = 0

(8.7)

F (S, t) → S − Ke

−r(T −t)

as S → ∞ for t ∈ [0, T ).

(8.8)

The condition given in Eq. (8.6) states that when the expiry date for the option arrives, the call option will be worth nothing if the value of the stock is less than the strike price. Otherwise the call option is worth the excess of the stock’s value over the strike price. The boundary conditions are specified in Eqs. (8.7) and (8.8). The first condition implies that if the stock itself becomes worthless before maturity, the call option is also worthless. An investor could buy the stock for nothing and then let the call option expire unused. The second part of this condition follows from the European call option property discussed in Chapter 7, namely C e ≥ S − Ke−rT . By the Put-Call Parity formula in Eq. (7.1), as S → ∞ a put option becomes worthless and the value of a call option approaches becomes S − Ke−r(T −t) asymptotically. A judicious change of variables for the Black-Scholes equation can simplify it. In fact, in only a few steps we can convert Eq. (8.5) into a more well known partial differential equation, namely the heat equation. Suppose F , S, and t are defined in terms of the new variables v, x, and τ as in the following equations. S = Kex ⇐⇒ x = ln

S K

2τ σ2 ⇐⇒ τ = (T − t) 2 σ 2 F (S, t) = Kv(x, τ )

t=T−

(8.9) (8.10) (8.11)

How are the Black-Scholes equation and its side conditions altered by this change of variables? The multivariable form of the chain rule can be used to determine new expressions for the derivatives present in Eq. (8.5).   1 FS = (Kv(x, τ ))S = K (vx xS + vτ τS ) = K vx + 0 = e−x vx (8.12) S The reader will be asked to verify that Ft = − FSS =

14:29:51.

Kσ 2 vτ 2

and

e−2x (vxx − vx ) K

(8.13) (8.14)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

205

Solution of the Black-Scholes Equation

in exercises (7) and (8). Substituting the results in Eqs. (8.11)–(8.14) in the Black-Scholes Eq. (8.5) and simplifying produces the equation vτ = vxx + (k − 1)vx − kv

(8.15)

where k = 2r/σ 2 (see exercise (9)). The final condition for F is converted by this change of variables into an initial condition since when t = T , τ = 0. The initial condition v(x, 0) is then found to be Kv(x, 0) = F (S, T ) = (S − K)+

= K(ex − 1)+

v(x, 0) = (ex − 1)+

(8.16)

Since limS→0+ x = −∞, then 0 = lim F (S, t) = lim Kv(x, τ ) =⇒ lim v(x, τ ) = 0. x→−∞

S→0+

x→−∞

Likewise since as limS→∞ x = ∞, lim F (S, t) = S − Ke−r(T −t) = lim Kv(x, τ ) x→∞

S→∞

ex − e−kτ = lim v(x, τ ) x→∞

Thus we have derived a pair of boundary conditions for the partial differential equation in (8.15). So the original Black-Scholes initial, boundary value problem can be recast in the form of the following. vτ = vxx + (k − 1)vx − kv

v(x, 0) = (ex − 1)+

v(x, τ ) → 0

x

for x ∈ (−∞, ∞)

as x → −∞ and

v(x, τ ) → e − e

−kτ

2

for x ∈ (−∞, ∞), τ ∈ (0, T 2σ ) (8.17)

as x → ∞, τ ∈ (0,

(8.18)

(8.19) T σ2 2 )

(8.20)

The independent variable x can be thought of as corresponding to a spatial variable. The reader should note that where previously the price of the security S was assumed to take on only non-negative values, now x can be any real number.

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

206

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

Now another change of variables is needed. If α and β are constants then we can introduce the new dependent variable u. v(x, τ ) = eαx+βτ u(x, τ ) vx = e

αx+βτ

vxx = e

αx+βτ

vτ = e

αx+βτ

(8.21)

(αu(x, τ ) + ux )

(8.22)

2

α u(x, τ ) + 2αux + uxx (βu(x, τ ) + uτ )



(8.23) (8.24)

Substituting the expressions found in (8.21)–(8.24) into Eq. (8.17) produces uτ = (α2 + (k − 1)α − k − β)u + (2α + k − 1)ux + uxx

(8.25)

Since α and β are arbitrary constants they can now be chosen appropriately to simplify Eq. (8.25). Ideally the coefficients of ux and u would be zero. Solving the two equations: 0 = α2 + (k − 1)α − k − β

0 = 2α + k − 1

yields α = (1 − k)/2 and β = −(k + 1)2 /4. The initial condition for u can be derived from the initial condition given in (8.18). v(x, 0) = (ex − 1)+

u(x, 0) = e(k−1)x/2 (ex − 1)+  x e − 1 if x > 0, (k−1)x/2 =e 0 if x ≤ 0.  (k+1)x/2 e − e(k−1)x/2 if x > 0, = 0 if x ≤ 0. = (e(k+1)x/2 − e(k−1)x/2 )+

Likewise we can derive boundary conditions at x = ±∞ for u from Eqs. (8.19) and (8.20). To summarize then the original Black-Scholes partial differential equation, initial, and boundary conditions have been converted to the following system of equations. uτ = uxx u(x, 0) = (e

for x ∈ (−∞, ∞) and τ ∈ (0, T σ 2 /2)

(k+1)x/2

−e

(k−1)x/2 +

)

for x ∈ (−∞, ∞) 2

(8.26) (8.27)

u(x, τ ) → 0 as x → −∞ for τ ∈ (0, T σ /2)

(8.28)

u(x, τ ) → e

(8.29)

(k+1) [x+(k+1)τ /2] 2

14:29:51.

−e

(k−1) [x+(k−1)τ /2] 2

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solution of the Black-Scholes Equation

BC8495/Chp. 8

207

as x → ∞ for τ ∈ (0, T σ 2 /2). The PDE of Eq. (8.26) is the well-known heat equation of mathematical physics. The Fourier Transform technique introduced earlier will now be used to solve this initial boundary value problem. 8.4

Solving the Black-Scholes Equation

We begin by taking the Fourier transform of both sides of the heat equation (since the Black-Scholes equation has been through two changes of variables and has been converted into the heat equation on the real number line) in Eq. (8.26).

d dτ

Z



F {uτ } = F {uxx} Z ∞ uτ e−iωx dx = uxx e−iωx dx −∞ −∞ Z ∞ u(x, τ )e−iωx dx = (iω)2 u(x, τ )e−iωx dx

Z

−∞



dˆ u = −ω 2 u ˆ dτ

−∞

(8.30)

where u ˆ is the Fourier transform of u(x, τ ). The reader should note that in deriving Eq. (8.30) Theorem 8.1 is used. This last equation is an ordinary differential equation of the type used to model exponential decay. Separating variables (see Sec. 5.4, especially Eq. (5.24) and the paragraphs which follow it) and solving this equation produces a solution of the form 2

uˆ(w, τ ) = De−ω τ . The expression represented by D is any quantity which is constant with respect to τ . Even though this equation was solved using ordinary differential equations techniques, the quantity u ˆ is a function of both w and τ . To evaluate D we can set τ = 0 and determine that u ˆ(w, 0) = D. Thus D is merely the Fourier transform of the initial condition in Eq. (8.27). For simplicity of notation we will write D = fˆ(ω). Thus the Fourier transformed solution to the heat equation is 2

uˆ(w, τ ) = fˆ(ω)e−ω τ .

(8.31)

Now this solution must be inverse Fourier transformed and then have its variables changed back to the original variables of the Black-Scholes

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

208

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

equation. 2 F −1 {ˆ u(w, τ )} = F −1 {fˆ(ω)e−ω τ }

2 1 u(x, τ ) = (e(k+1)x/2 − e(k−1)x/2 )+ ∗ √ e−x /(4τ ) 2 πτ Z ∞ (x−z)2 1 z z = √ (e(k+1) 2 − e(k−1) 2 )+ e− 4τ dz (8.32) 2 πτ −∞

The reader should note that the Fourier Convolution Theorem (8.2) was used above. In exercise (5) the reader will also provide the details of the essential result that 2 2 1 F −1 {e−ω τ } = √ e−x /(4τ ) . 2 πτ

If the substitution z = x +

√ 2τ y is made then Eq. (8.32) becomes

u(x, τ ) Z ∞ + √ √ 2 1 = √ e(k+1)(x+ 2τ y)/2 − e(k−1)(x+ 2τ y)/2 e−y /2 dy 2π −∞ (k+1)x/2 (k+1)2 τ /4 Z ∞ √ 2 e e 1 √ = e−(y− 2 (k+1) 2τ ) /2 dy (8.33) √ 2π −x/ 2τ Z 2 √ 2 1 e(k−1)x/2 e(k−1) τ /4 ∞ √ − e−(y− 2 (k−1) 2τ ) /2 dy √ 2π −x/ 2τ The reader will be asked to fill in the details of this derivation in exercises (11) and (12). The two improper integrals of Eq. (8.33) can be expressed in terms of the cumulative distribution function for a normal ran√ 1 dom variable. We will make the substitution w = y − (k + 1) 2τ in the 2 √ first integral and w0 = y − 12 (k − 1) 2τ in the second. The first integral of Eq. (8.33) equals the following expression Z

√ √ x/ 2τ + 21 (k+1) 2τ

e

−w 2 /2

−∞

√ dw = 2πΦ



√ x 1 √ + (k + 1) 2τ 2 2τ



where Φ is the cumulative distribution function for a normal random variable with mean zero and standard deviation one. The reader should verify this calculation (exercise (13)) and a similar calculation for the second improper integral of Eq. (8.33).

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solution of the Black-Scholes Equation

BC8495/Chp. 8

209

Thus far the solution to the initial boundary value problem in Eqs. (8.26)–(8.29) is given by   √ x 1 (k+1)x/2+(k+1)2 τ /4 u(x, τ ) = e Φ √ + (k + 1) 2τ (8.34) 2 2τ   √ 2 x 1 − e(k−1)x/2+(k−1) τ /4 Φ √ + (k − 1) 2τ . 2 2τ The reader should take a moment to check that the expression for u(x, τ ) given in Eq. (8.34) satisfies the boundary conditions as x → ±∞ specified in Eqs. (8.28) and (8.29). Now we begin the task of reconverting variables to those of the BlackScholes initial boundary value problem as stated in Eqs. (8.5)–(8.8). Using the change of variables in Eq. (8.21) this solution can be re-written in terms of the function v(x, τ ), where 2

v(x, τ ) = e−(k−1)x/2−(k+1) τ /4 u(x, τ )   √ 1 x = ex Φ √ + (k + 1) 2τ 2 2τ   √ x 1 − e−kτ Φ √ + (k − 1) 2τ . 2 2τ

(8.35)

Now using the change of variables described in Eqs. (8.9) and (8.10) the reader can show in exercise (16) that √ ln(S/K) + (r + σ 2 /2)(T − t) x 1 √ w = √ + (k + 1) 2τ = 2 σ T −t 2τ √ √ 1 x w0 = √ + (k − 1) 2τ = w − σ T − t. 2 2τ

(8.36) (8.37)

Substituting w and w0 in Eq. (8.35) yields v(x, τ ) =

  √ S Φ (w) − e−r(T −t) Φ w − σ T − t K

which upon using Eq. (8.11) produces the sought after Black-Scholes European call option pricing formula. For the sake of simplicity and meaningful notation the value of the European call will be denoted C.   √ C(S, t) = SΦ (w) − Ke−r(T −t) Φ w − σ T − t (8.38) A surface plot of the value of the European call option as a function of S and t is shown in Fig. 8.1.

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

210

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

T

C

t

K S

A surface plot of C e as a function of S and t.

Fig. 8.1

The value of a European put option could be found via a similar set of calculations or the already determined value of the European call could be used along with the Put-Call Parity formula (7.1) to derive  √  P (S, t) = Ke−r(T −t)Φ σ T − t − w − SΦ (−w) . (8.39) A surface plot of the value of the European put option as a function of S and t is shown in Fig. 8.2.

Example 8.3 Suppose the current price of a security is $62 per share. The continuously compounded interest rate is 10% per year. The volatility of the price of the security is σ = 20% per year. The cost of a five-month European call option with a strike price of $60 per share can be found using Eqs. (8.36) and (8.38). If we summarize the quantities we know, in the notation of this section, we have: T = 5/12,

t = 0,

r = 0.10,

σ = 0.20,

S = 62,

and K = 60.

Thus we have w ≈ 0.641287 and therefore the price of the European call option is C = $5.80. Example 8.4 Suppose the current price of a security is $97 per share. The continuously compounded interest rate is 8% per year. The volatility

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

211

Solution of the Black-Scholes Equation

T

P

t

K S

Fig. 8.2

A surface plot of P e as a function of S and t.

of the price of the security is σ = 45% per year. The cost of a three-month European put option with a strike price of $95 per share can be found using Eqs. (8.36), and (8.39). If we summarize the quantities we know, in the notation of this section, we have: T = 1/4,

t = 0,

r = 0.08,

σ = 0.45,

S = 97,

and K = 95.

Thus we have w ≈ 0.293985 and consequently the put option price is P = $6.71. 8.5

Binomial Model (Optional)

As an alternative to the partial differential equation solution to the BlackScholes equation, we may derive the same formulas from a simple, discrete model. Thus for readers uncomfortable with the formality and technicalities of the previous approach or those readers wanting to understand an alternative model which forms the basis of numerical methods for evaluating options, this section describes what is known as the binomial model. This derivation was initially developed by Cox, Ross, and Rubinstein [Cox et. al. (1979)] and is sometimes referred to as the Cox-Ross-Rubinstein model. We will see that the value of the option given by the binomial

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

212

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

model will, in the sense of a limit, equal the previously described value given by the Black-Scholes option pricing formula, Eq. (8.38). We will restrict our attention to European Call options in this section. The value of a European Put option is easily found either through the Put-Call Parity formula, Eq. (7.1) or via a simple modification to the following derivation. We will use the following assumptions in our derivation: • • • • •

Strike price of the call option is K. Exercise time of the call option is T . Initial price of the security is S(0). Continuously compounded risk-free interest rate is r. Price of the security follows a geometric Brownian motion with drift µ and volatility σ.

The binomial model is often called the lattice model for reasons which will soon become apparent. Suppose the time interval [0, T ] is partitioned into n equal subintervals of length ∆t = T /n. At the start of the first subinterval [0, ∆t] the value of the security is S(0). The binomial model assumes that the value of the security will evolve to uS(0) where u > 1 with probability p or to dS(0) where 0 < d < 1 with probability 1 − p. Since value of the security may only increase by the factor u or decrease by the factor d, the name “binomial model” is appropriate. A graph of the lattice model for a single time step is shown in Fig. 8.3.

SHTL=u SH0L p

SH0L

1-p SHTL=d SH0L Fig. 8.3

A single time step of size ∆t is shown for the binomial model of security values.

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

213

Solution of the Black-Scholes Equation

The assumption that the value of the security follows a geometric Brownian motion implies that dS = µS dt + σS dW (t) S(t) = S(0)e(µ−σ E [S(t)] = S(0)e

2

/2)t+σW (t)

(exercise (22))

µt

 2  2 V (S(t)) = (S(0)) e2µt eσ t − 1 .

The expressions for the expected value and variance are a result of Lemma 3.1. The parameters of the binomial lattice model must be selected so that the discrete model of the security has the same expected value and variance as the continuous model. Hence if the security can take on only values uS(0) and dS(0) with probabilities p and 1 − p respectively then after a time step of length ∆t we have puS(0) + (1 − p)dS(0) = S(0)eµ∆t pu + (1 − p)d = eµ∆t .

(8.40)

Another assumption of the binomial model is that the variance in the value of the security at time ∆t generated by the discrete model must agree with the variance of the continuous random variable S(∆t). Therefore  2  2 (S(0)) e2µ∆t eσ ∆t − 1 = p(uS(0))2 + (1 − p)(dS(0))2 

e2µ∆t eσ

2

∆t



− [p(uS(0)) + (1 − p)(dS(0))]

2

− 1 = pu2 + (1 − p)d2 − (eµ∆t )2

e(2µ+σ

2

)∆t

= pu2 + (1 − p)d2

(8.41)

So far we have written the probability p as a function of µ and ∆t and now we have Eq. (8.41) which relates u and d to µ, σ, and ∆t. If there existed another equation relating u and d we could express p, u, and d as functions of µ, σ, and ∆t. We are at liberty to define any useful relationship between u and d for this last desired equation. A simple equation whose significance will be explained shortly is ud = 1. We may solve Eq. (8.40) for p to yield p=

14:29:51.

ueµ∆t − 1 u2 − 1

(8.42)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

214

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

where d has been replaced by 1/u. Performing the same replacement in Eq. (8.41) and inserting the newly found value for p we then find the value for u as   q 2 2 1 −µ∆t u= e + e(µ+σ )∆t + e−µ∆t + e(µ+σ2 )∆t − 4 . (8.43) 2 For the sake of completeness, an expression for d is   q 2 2 1 −µ∆t e−µ∆t + e(µ+σ2 )∆t − 4 . d= e + e(µ+σ )∆t − 2

(8.44)

Just as we did for the continuous model we must impose a “no-arbitrage” condition on the binomial model. Π will denote the initial value of a portfolio consisting of a short position in the call option and a long position in the underlying security S. For every option sold, ∆ units of the underlying are bought. As usual we will assume the underlying security is infinitely divisible. The value of ∆ must be chosen so as to make the portfolio Delta neutral, in other words whether the price of the security rises or falls, the portfolio value must remain the same. We will introduce the symbols Cu and Cd to represent the value of the option for the cases in which the security is worth at time ∆t, uS(0) and dS(0) respectively. The Delta neutrality of the portfolio can be expressed as Cu − ∆uS(0) = Cd − ∆dS(0)

which implies ∆=

Cu − Cd . S(0)(u − d)

(8.45)

Now at time ∆t the portfolio is worth Π(∆t) = C(∆t) − ∆S(∆t). At time ∆t the underlying security will be worth uS(0) (with probability p) or dS(0) (with probability 1 − p). If no arbitrage is permitted the value of the portfolio at time ∆t should be equal to the future value of Π earning interest at the risk-free rate r. Thus er∆t Π = Π(∆t) er∆t (C − ∆S(0)) = Cu − ∆uS(0)   Cu − Cd u(Cu − Cd ) r∆t C− = Cu − e u−d u−d 14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solution of the Black-Scholes Equation

BC8495/Chp. 8

215

which can be solved for C to produce C=

1 − de−r∆t ue−r∆t − 1 Cu + Cd . u−d u−d

(8.46)

Another way of interpreting Eq. (8.46) is as giving the future value of the option as a function of its two potential binomial values at the next time step. er∆t C =

er∆t − d u − er∆t Cu + Cd u−d u−d

Note that the coefficients of Cu and Cd sum to one. Thus setting uer∆t − 1 er∆t − d = u−d u2 − 1

(8.47)

er∆t C = p0 Cu + (1 − p0 )Cd .

(8.48)

p0 = we may write

In other words the value of the option at time t is the present value of the expected value of the option at time t + ∆t. The reader should note that the probability of an upward movement in the price of the security (refer to Eq. (8.42)) and the probability of an upward movement in the value of the call option (see Eq. (8.47)) are closely related. The main difference is that the drift parameter of the security has been replaced with the riskfree, continuously compounded interest rate. The no-arbitrage condition then imposes the restriction that the binomial probability and the increase and decrease factors (u and d respectively) must be calculated using r in place of σ. When calculating the value of options we will use the symbols p, u, and d but the reader should keep in mind that these quantities are computed using r in place of µ. So far a great deal of work has gone into constructing a discrete model of a single time step in the evolution of the value of the security. Once the model parameters p, u, and d are known a lattice of n = T /∆t steps can be constructed. At each vertex of the lattice the value of the security may proceed up by a multiplicative factor u with probability p or down by a multiplicative factor d with probability 1 − p. The parameters are assumed to remain constant across the time steps. For the sake of illustration a lattice of three time steps is shown in Fig. 8.4. Since we chose ud = 1 then we see that udS(0) = duS(0) = S(0) and thus bifurcations in the lattice come together after two time steps during which one upward movement

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

216

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

SH0L u3

SH0L u2

SH0L d u2

SH0L u

SH0L

SH0L d u

SH0L d2 u

SH0L d

SH0L d2

SH0L d3

Fig. 8.4

A binomial model of three time steps.

and one downward movement occurred. For this reason the graphs in this discrete model are sometimes called recombining trees. Now we may use the lattice model to approximate the value of a European call option. Recall that the payoff value of a European call option is (S(T ) − K)+ . Thus C(T ) = (S(T )−K)+ . In an arbitrage-free setting the value of the call at any time must be the same as the present value of the expected payoff at expiry. Working backward in time from expiry, at t = T − ∆t the call option is worth   C(T − ∆t) = e−r∆t E (u2Y −1 − K)+ where Y ∈ {0, 1}. The assignment Y = 0 implies the security decreases in price over the next time step, while Y = 1 indicates it increases in price. To generalize the discussion to the situation of an n-step binomial lat-

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solution of the Black-Scholes Equation

BC8495/Chp. 8

217

tice, define Y to be the random variable denoting the number of upward movements in the value of the security since t = 0 (at which time the value of the security was S(0)). Note that Y is a binomial random variable in the sample space {0, 1, . . . , n}. Thus if C represents the value of the call at time t = 0, then   C = e−rT E (uY dn−Y S(0) − K)+   = e−rT E (u2Y −n S(0) − K)+ h i = e−rT E (u2Y −T /∆t S(0) − K)+ .

(8.49)

Example 8.5 Now we may approximate the value of a European call option and compare the values generated by the Black-Scholes option pricing formula in Eq. (8.38) and the lattice method. Assume the current price of a security is $62 per share, the continuously compounded interest rate is 10% per year, the volatility of the price of the security is σ = 20% per year. If the strike price of the option is $60 per share with an expiry of 5 months, then C = $5.789 according to example 8.3. To approximate the value of the option we must create a recombining tree of values of the security. Since the option expires in 5 months, it is convenient for the lattice to have 5 steps. The time step will be one month, so ∆t = 1/12. The parameters of the binomial lattice can be found using ∆t, r = 0.10, σ = 0.20 and Eqs. (8.42)–(8.44). u = 1.06036,

d = 0.943073,

and p = 0.556697.

The lattice of security values is shown in Fig. 8.5. Since the option is of European type on the values for S in the right-most column are important for the determination of the value of the option. The value of the European call option can be approximated by calculating the present value of the expected payoff of the option. The table below summarizes the most important information. S 83.1122 73.9189 65.7425 58.4705 52.0029 46.2507

14:29:51.

(S − K)+ 23.1122 13.9189 5.7425 0 0 0

Probability 0.0534682 0.212886 0.339046 0.269985 0.107496 0.0171199

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

218

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

83.1122

78.3809

73.9189

69.7109

65.7425

62.

73.9189

69.7109

65.7425

62.

58.4705

65.7425

62.

58.4705

55.142

58.4705

55.142

52.0029

52.0029

49.0426

46.2507

Fig. 8.5

A binomial lattice of security prices for example 8.5.

The last column of the table contains the probability that the security price shown in the column labeled S is achieved. These values are calculated using the binomial probability density formula of Eq. (2.3). If we think of the columns labeled (S − K)+ and probability as vectors then the approximate value of the option is the present value of their dot product. Therefore we find that (23.1122)(0.0534682) + (13.9189)(0.212886) + (5.7425)(0.339046) e(0.10)(5/12) 6.14588 = = 5.89506. 1.04255

C≈

This compares favorably to the value given by the Black-Scholes option pricing formula for European call options.

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

219

Solution of the Black-Scholes Equation

If the reader has access to spreadsheet software or a computer algebra system they may verify that if ∆t for the discrete model is decreased from one month to one day, the binomial model yields a call option value of C ≈ 5.80305 which provides numerical evidence that the discrete price is converging to the continuous price. The reader should observe that u2Y −T /∆t = e(2Y −T /∆t) ln u and since Y is a binomial random variable, as n becomes large (or equivalently as ∆t becomes small) Y can be approximated by a normal random variable. Consequently (2Y − T /∆t) ln u is a linear function of a normal random variable (for large n) and is itself a normal random variable. We saw earlier that the expected value of a binomial random variable is np where here n = T /∆t is the number of trials and the probability of success on a single trial is p. Thus   T lim E [(2Y − T /∆t) ln u] = lim (2p − 1) ln u ∆t→0 ∆t→0 ∆t (2p − 1) ln u = lim ∆t→0 ∆t/T which is indeterminate of the form 0/0. Applying l’Hˆopital’s Rule and using the values of p and u found in Eqs. (8.42) and (8.43) respectively, we have   σ2 lim E [(2Y − T /∆t) ln u] = r − T. (8.50) ∆t→0 2 In a similar manner we find that the limit of the variance of (2Y −T /∆t) ln u is given by 2

lim V ((2Y − T /∆t) ln u) = lim (2 ln u) p(1 − p)

∆t→0

∆t→0

T ∆t

(ln u)2 ∆t→0 ∆t

= T lim

since from Eq. (8.42) we know that p → 1/2 as ∆t → 0. The remaining limit is again indeterminate of the form 0/0. By l’Hˆopital’s Rule we obtain lim V ((2Y − T /∆t) ln u) = σ 2 T.

∆t→0

(8.51)

Thus we have shown that we can write the cost of the call option as   C = e−rT E (S(0)eX − K)+ , (8.52) 14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

220

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

where X is a normal random variable with   σ2 E [X] = r − T and 2

V (X) = σ 2 T,

and probability density function 2 2 2 1 g(x) = √ √ e−(x−T (r−σ /2)) /(2σ T ) . σ T 2π

Define a new variable Z to be Z=

X − (r − σ 2 /2)T √ . σ T

Since Z is a linear function of a normal random variable X, then Z is also a normal random variable. We note that   X − (r − σ 2 /2)T √ E [Z] = E σ T E [X] (r − σ 2 /2)T √ = √ − σ T σ T (r − σ 2 /2)T (r − σ 2 /2)T √ √ = − σ T σ T = 0, and furthermore that 

X − T (r − σ 2 /2) √ σ T V (X) = 2 σ T σ2 T = 2 σ T = 1.

V (Z) = V



Thus Z is a standard normal random variable. We know its probability distribution is given by the function, 2 1 h(z) = √ e−z /2 . 2π

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

221

Solution of the Black-Scholes Equation

Returning to the expected value to be calculated in Eq. (8.52), we have   E (S(0)eX − K)+ Z ∞ = (S(0)ex − K)+ g(x) dx −∞ Z ∞ = (S(0)ex − K)g(x) dx = =

ln(K/S(0)) ∞

Z

(S(0)ewσ

Zw∞

T +(r− σ2 )T



T +(r− σ2 )T

2



Zw∞ ∗

=



w∗

(S(0)ewσ

√ √ σ2 )T )σ T dw − K)g(wσ T + (r − 2

2

− K)h(ω) dw Z ∞ √ 2 wσ T +(r− σ2 )T S(0)e h(ω) dw − Kh(ω) dw w∗

where ln(K/S(0)) − T (r − σ 2 /2) √ . σ T

w∗ =

Consider the expression contained in the first integrand above, ewσ

√ T +T (r−σ2 /2)

√ 2 2 1 h(ω) = √ ewσ T +T (r−σ /2) e−w /2 2π √ 2 2 1 = √ e−(w −2σw T +σ T )/2+rT 2π 1 (−(w−σ√T )2 )/2+rT = √ e 2π 1 rT (−(w−σ√T )2 )/2 = √ e e 2π √ rT = e h(w − σ T ).

Now we will substitute this expression in Eq. (8.53) and find that   E (S(0)eX − K)+ Z ∞ Z ∞ √ rT = S(0)e h(w − σ T ) dw − K h(ω) dw w∗ w∗ ! Z w∗ √ rT = S(0)e 1− h(w − σ T ) dw −∞

−K

1−

Z

14:29:51.

w∗

−∞

h(ω) dw

!

(8.53)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

222

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

= S(0)e −K

rT

1−

1−

Z

Z

√ w ∗ −σ T

−∞

w∗

−∞

h(ω) dw

!

h(x) dx !

  √  − K (1 − Φ (w∗ )) , = S(0)erT 1 − Φ w∗ − σ T

(8.54)

where Φ (x) is the probability that a standard normal random√variable is less than x. For convenience sake we will set w = −(w∗ − σ T ). Since Φ (−x) = 1 − Φ (x), then Eq. (8.54) can be rewritten as  √    E (S(0)eX − K)+ = S(0)erT Φ (w) − KΦ w − σ T .

(8.55)

Now we can substitute this expression into Eq. (8.52) and finally we can write the Black-Scholes Option Pricing Formula as   √  C = e−rT S(0)erT Φ (w) − KΦ w − σ T  √  = S(0)Φ (w) − Ke−rT Φ w − σ T .

(8.56)

If we would like to know that value of the call option at any time 0 ≤ t ≤ T then a simple shift in the time variable can be used in Eq. (8.56). This agrees with the formula derived in Eq. (8.38) and repeated below for completeness.   √ C(S, t) = SΦ (w) − Ke−r(T −t) Φ w − σ T − t where w=

ln(S(0)/K) + (r + σ 2 /2)(T − t) √ σ T −t

In conclusion, whether we derive the value of the European call option from the continuous model via the solution of the Black-Scholes partial differential equation or by developing a multi-time step binomial model and taking its limit as the time step becomes infinitesimally small, the formula remains the same.

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solution of the Black-Scholes Equation

8.6

BC8495/Chp. 8

223

Exercises

(1) Let the function f (x) be defined as f (x) =



e−ax if x ≥ 0 0 otherwise

where a is a positive constant. Show that the Fourier Transform of f 1 is fˆ(ω) = a+iω . (2) Suppose f (x) is the function f (x) =



sin x if |x| ≤ π, 0 if |x| > π

then find fˆ(ω). 2 (3) Find the Fourier Transform of f (x) = xe−x . (4) Let the function fˆ(ω) be defined as fˆ(ω) =



e−aω if w ≥ 0 0 otherwise

where a is a positive constant. Show that the inverse Fourier Trans1 form of fˆ is f (x) = 2π(a−ix) . (5) If a is a positive constant verify that the inverse Fourier Transform of 2 2 e−aω is 2√1πa e−x /4a . (6) If a is a positive constant find the inverse Fourier Transform of fˆ(ω) =



−ie−aω if ω ≥ 0, ieaω if ω < 0.

(7) Using the new variables described in Eqs. (8.9), (8.10), (8.11), and the multivariable chain rule, verify the identity shown in Eq. (8.13). (8) Using the new variables described in Eqs. (8.9), (8.10), (8.11), the expression for FS shown in (8.12), and the multivariable chain rule, verify the identity shown in Eq. (8.14). (9) Verify that the Black-Scholes partial differential Eq. (8.5) can be simplified to the form shown in Eq. (8.15) by using the variables described in (8.9), (8.10), and (8.11) and the partial derivatives found in (8.12), (8.13), and (8.14).

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

224

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 8

An Undergraduate Introduction to Financial Mathematics

(10) Fill in some of the details in the derivation of Eqs. (8.26)–(8.29) by showing that u(x, τ ) → e

(k+1) [x+(k+1)τ /2] 2

−e

(k−1) [x+(k−1)τ /2] 2

as x → ∞. (11) Verify that the expression 

e(k+1)(x+

√ 2τ y)/2

− e(k−1)(x+



2τ y)/2

+

√ is non-zero whenever y > −x/ 2τ . (12) Use completion of the square in the exponents of the integrand in Eq. (8.33) to derive Eq. (8.33). √ (13) Using the change of variable w = −y + 12 (k + 1) 2τ in the first improper integral of Eq. (8.33) verify that Z ∞ √ 2 1 1 √ e−(y− 2 (k+1) 2τ ) /2 dy √ 2π −x/ 2τ Z x/√2τ + 12 (k+1)√2τ w2 1 = √ e− 2 dw 2π −∞   √ √ 1 = Φ x/ 2τ + (k + 1) 2τ 2 where Φ is the cumulative distribution function for a normal random variable with mean zero and standard deviation one. √ (14) Using the change of variable w0 = −y + 12 (k − 1) 2τ in the second improper integral of Eq. (8.33) verify that Z ∞ √ 2 1 1 √ e−(y− 2 (k−1) 2τ ) /2 dy √ 2π −x/ 2τ Z x/√2τ + 12 (k−1)√2τ w02 1 e− 2 dw0 = √ 2π −∞   √ √ 1 = Φ x/ 2τ + (k − 1) 2τ 2 where Φ is the cumulative distribution function for a normal random variable with mean zero and standard deviation one. (15) Verify that the expression for u(x, τ ) given in Eq. (8.34) satisfies the initial condition as specified in Eq. (8.27).

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solution of the Black-Scholes Equation

BC8495/Chp. 8

225

(16) Verify using the change of variable formulas in Eqs. (8.9) and (8.10) and the fact that k = 2r/σ 2 that √ √ 1 ln(S/K) + (r + σ 2 /2)(T − t) √ x/ 2τ + (k + 1) 2τ = 2 σ T −t √ √ √ 1 ln(S/K) + (r + σ 2 /2)(T − t) √ x/ 2τ + (k − 1) 2τ = − σ T − t. 2 σ T −t (17) What is the price of a European call option on a stock when the stock price is $52, the strike price is $50, the interest rate is 12%, the stock’s volatility is 30%, and the exercise time is three months? (18) What is the price of a European put option on a stock when the stock price is $69, the strike price is $70, the interest rate is 5%, the stock’s volatility is 35%, and the exercise time is six months? (19) A European call option on a stock has a market value of $2.50. The stock price is $15, the strike price is $13, the interest rate is 5%, and the exercise time is three months. What is the volatility of the stock? (20) Show by differentiation and substitution in Eq. (8.5) that the price of the security itself, i.e. f (S, t) = S solves the Black-Scholes equation. (21) Show by differentiation and substitution in Eq. (8.5) that money earning the risk-free interest rate r compounded continuously solves the Black-Scholes equation. (22) (optional) Show that if S obeys a stochastic process of the form dS = µS dt + σS dW (t) with initial value S(0) then for t ≥ 0 S(t) = S(0)e(µ−σ

2

/2)t+σW (t)

(23) (optional) What is the price of a European call option on a stock when the stock price is $43, the strike price is $42, the risk-free interest rate is 11%, the stock’s volatility is 25%, and the exercise time is four months? Use the binomial lattice model with a time step of one month to evaluate the option. (24) (optional) What is the price of a European put option on a stock when the stock price is $96, the strike price is $100, the risk-free interest rate is 6%, the stock’s volatility is 33%, and the exercise time is three months? Use the binomial lattice model with a time step of one month to evaluate the option.

14:29:51.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

This page intentionally left blank

14:29:51.

226

BC8495/Chp. 8

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 9

Chapter 9

Derivatives of Black-Scholes Option Prices

Now that the solution to the Black-Scholes equation is known we can investigate the sensitivity of the solution to changes in the independent variables. Knowledge of these sensitivities allows a portfolio manager to minimize changes in the value of a portfolio when underlying variables such as the risk-free interest rate change. The material of the present chapter will be used extensively in Chapter 10. An understanding of the sensitivity of an option’s value to changes in its independent variables will also provide a new way of interpreting the Black-Scholes equation itself. To mathematicians the word derivative means an instantaneous rate of change in a quantity. To a quantitative analyst a derivative is a financial entity whose value is derived from the value of some underlying asset. Hence a European call option is a derivative (in the quantitative analytical sense) whose value is a function of (among other things) the value of the security underlying the option. In this chapter we explore derivatives from the mathematician’s perspective. In Chapter 10 we will use these derivatives in the manner of a quantitative analyst. Members of the quantitative financial profession refer to the subject matter of this chapter as the “Greeks” since a Greek letter is used to name each derivative (except for one which we will meet in due time). Due to the typically large volume of options written in a contract, very accurate calculations of the Greeks are necessary to avoid round-off errors. Motivation for accuracy and accurate numerical techniques may be found in [Chawla (2006)] and [Chawla and Evans (2005)]. 9.1

Theta

The quantity Theta, Θ, is defined to be the rate of change of the value of an option with respect to t. We will calculate Θ for the European call

14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

228

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 9

An Undergraduate Introduction to Financial Mathematics

and put options derived in Chapter 8. Since the valuation of Europeanstyle options depends heavily on the cumulative distribution function of the standard normal random variable and a composite variable w, we will begin by investigating their partial derivatives. The derivative with respect to t of quantity w defined in Eq. (8.36) is given by   ∂w 1 ln(S/K) 2 √ = − r − σ /2 . (9.1) ∂t T −t 2σ T − t By the Fundamental Theorem of Calculus [Smith and Minton (2002), Chap. 4] the cumulative distribution function for the standard normal distribution has derivative 2 1 φ (x) = √ e−x /2 . 2π

(9.2)

Hence using the Chain Rule [Stewart (1999), Chap. 3] for derivatives we can see that the time rate of change in the value of a European call option (defined in Eq. (8.38)) is    √ ∂C ∂w = Sφ (w) − Ke−r(T −t) rΦ w − σ T − t ∂t ∂t     ∂w √ σ +φ w−σ T −t + √ ∂t 2 T −t    ∂w √ = Sφ (w) − Ke−r(T −t)φ w − σ T − t ∂t      √ √ σ −r(T −t) − Ke φ w−σ T −t rΦ w − σ T − t + √ 2 T −t √   −w 2 /2 −r(T −t)−(w−σ T −t)2 /2 Se ln(S/K) − Ke 2 p ΘC = − r − σ /2 T −t 2σ 2π(T − t) !   σe−(w−σ√T −t)2 /2 √ −r(T −t) p − Ke . (9.3) rΦ w − σ T − t + 2 2π(T − t)

To take an example, consider a European call option on a stock whose current price is $250. The strike price is $245, the annual risk-free interest rate is 2.5%, the volatility of the stock price is 20%, and the strike time is four months. Thus substituting S = 250, K = 245, r = 0.025, σ = 0.20, T = 1/3, and t = 0 into Eq. (9.3) produces Θ=

14:30:05.

∂C = −19.9836. ∂t

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Derivatives of Black-Scholes Option Prices

BC8495/Chp. 9

229

The case that Θ is negative for an option is typical since options lose value as the expiry date approaches. The calculation of Θ for a European put option is carried out in a similar fashion to that for the call option. The reader will be asked to provide the details of the following derivation. ΘP =

∂P ∂t

(9.4) 2

=

Se−w /2 p 2σ 2π(T − t)





ln(S/K) − r − σ 2 /2 T −t   √ + Kre−r(T −t) Φ σ T − t − w √   2 Ke−r(T −t)−(w−σ T −t) /2 ln(S/K) p − − r + σ 2 /2 T −t 2σ 2π(T − t)

(9.5)

Equation (9.5) will be put to use in the following example. Consider a European put option on a stock whose current price is $325. The strike price is $330, the annual risk-free interest rate is 3.5%, the volatility of the stock price is 27%, and the strike time is three months. Thus substituting S = 325, K = 330, r = 0.035, σ = 0.27, T = 1/4, and t = 0 into Eq. (9.5) produces Θ=

∂P = −28.7484. ∂t

The derivative Θ is unique among the derivatives covered in this chapter. The passage of time t is the only non-stochastic variable among the list of independent variables upon which C or P depend. 9.2

Delta

The value of an option is also sensitive to changes in the price of the underlying stock. This rate of change is called Delta, ∆. The astute reader may recall that ∆ was introduced in Sec. 7.4 during the derivation of the Black-Scholes partial differential equation. Delta was used to eliminate a stochastic term from Eq. (7.13). The clever choice of ∆ left us with a deterministic partial differential equation to solve rather than a stochastic differential equation. The starting point for deriving ∆ is again Eq. (8.38).    ∂w √ ∂C = Φ (w) + Sφ (w) − Ke−r(T −t) φ w − σ T − t ∂S ∂S 14:30:05.

(9.6)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

230

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 9

An Undergraduate Introduction to Financial Mathematics

We can make use of Eq. (9.2) and the following result, ∂w 1 √ = . ∂S σS T − t

(9.7)

Substituting Eqs. (9.2) and (9.7) into Eq. (9.6) produces ! √ 2 2 Se−w /2 Ke−r(T −t) e−(w−σ T −t) /2 1 √ √ √ − σS T − t 2π 2π   −r(T −t) −w 2 /2 √ Ke e −(σ2 (T −t)−2σw T −t)/2 1− = Φ (w) + p e S σ 2π(T − t)

∂C = Φ (w) + ∂S

2

e−w /2 = Φ (w) + p × σ 2π(T − t)   Ke−r(T −t) −(σ2 (T −t)−2(r+σ2 /2)(T −t)−2 ln(K/S))/2 1− e S   2 e−w /2 Ke−r(T −t) r(T −t)+ln(S/K) = Φ (w) + p e 1− S σ 2π(T − t)   −w 2 /2 e K = Φ (w) + p 1 − eln(S/K) S σ 2π(T − t) = Φ (w) .

(9.8)

We can put this formula to use in the following example. Suppose the price of a security is $100, the risk-free interest rate is 4% per annum, the annual volatility of the stock price is 23%, the strike price is $110, and the strike time is six months. Under these conditions w = −0.381747 and ∆C =

∂C = 0.351325. ∂S

Delta for a European put option could be calculated directly from the definition for P ; however, the Put-Call Parity formula (Eq. (7.1)) provides a convenient shortcut. ∂ ∂ (P + S) = (C + Ke−r(T −t) ) ∂S ∂S ∂P ∂C +1 = ∂S ∂S ∂P ∆P = = Φ (w) − 1 ∂S 14:30:05.

(9.9)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Derivatives of Black-Scholes Option Prices

BC8495/Chp. 9

231

Thus using the same parameter values as in the previous example, ∂P = 0.351325 − 1 = −0.648675. ∂S So, as the European call option increases in value with an increase in the value of the underlying security, the put option decreases in value.

9.3

Gamma

The Gamma, Γ, of an option or collection of options is defined to be the second derivative of the option with respect to S, the price of the underlying security. Hence Γ is the partial derivative of ∆ with respect to S. Thus for a European call option Γ is ∂2C ∂w = φ (w) ∂S 2 ∂S 2 e−w /2 p = . σS 2π(T − t)

(9.10)

We can readily see from Eq. (9.9) that the Γ for a European put is the same as that for a European call 2

Γ=

∂2P ∂2C e−w /2 p = = . 2 2 ∂S ∂S σS 2π(T − t)

(9.11)

If all independent variables are held fixed except for S, then Γ measures the concavity of the value of the option as a function of S. Suppose we consider a European call option on a security whose value is $295. The strike price is $290, the annual risk-free interest rate is 4.25%, the volatility of the security is 25% per annum, and the expiry date is two months. We can calculate the price of the option, its ∆, and its Γ as follows. C = 15.7173,

∆C = 0.613297,

Γ = 0.0127122.

Thus the value of the option will increase with the price of the underlying security at an increasing rate. This will have important implications for the practice known as hedging which will be explored in Chapter 10. In terms of elementary calculus, the graph of C as a function of S (holding all other variables fixed) is increasing and concave upward on an interval containing S = 295. See Fig. 9.1.

14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

232

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 9

An Undergraduate Introduction to Financial Mathematics

C 60 50 40 30 20 10

280

300

320

340

S

Fig. 9.1 Since the function C(S) is increasing and concave upward on an interval containing S = 295 the price of the European call option increases at an increasing rate when the price of the underlying security is increased from $295.

9.4

Vega

The Vega, V, of an option is the change in the value of the option as a function of the volatility of the underlying security. Among the partial derivatives of the value of an option, Vega distinguishes itself as the only partial derivative given a non-Greek letter name. Vega comes to us through medieval Latin from Arabic. For a European call option defined as in Eq. (8.38),    ∂w √  √ ∂C ∂w = Sφ (w) − Ke−r(T −t)φ w − σ T − t − T − t , (9.12) ∂σ ∂σ ∂σ where, according to exercise (12), ∂w √ = T − t − w/σ. ∂σ Substituting Eq. (9.13) into Eq. (9.12) produces ∂C ∂σ

  h√ i √ w √ − Ke−r(T −t) φ w − σ T − t T −t− − T −t σ √  w √ w −r(T −t) = Sφ (w) T −t− + Ke φ w−σ T −t σ σ   √ √ w = S T − tφ (w) − Sφ (w) − Ke−r(T −t)φ w − σ T − t σ 14:30:05.

(9.13)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Derivatives of Black-Scholes Option Prices

BC8495/Chp. 9

233

 √ √ 2 w  −w2 /2 = S T − tφ (w) − √ Se − Ke−r(T −t)e−(w−σ T −t) /2 σ 2π √ = S T − tφ (w)   √ 2 2 w − √ e−w /2 S − Ke−r(T −t)+wσ T −t−σ (T −t)/2 σ 2π   √ 2 w = S T − tφ (w) − √ e−w /2 S − Keln(S/K) σ 2π √ S T − t −w2 /2 = √ e . (9.14) 2π Note that the value of the European call option increases with increasing volatility in the underlying security. It is easily seen from the Put-Call Parity formula (Eq. (7.1)) that √ ∂P S T − t −w2 /2 e , (9.15) = √ ∂σ 2π in other words the rate of change in the value of a European put option with respect to volatility is the same as the rate of change in the value of the corresponding European call option with respect to volatility. Thus we may unambiguously define Vega for a European style option to be √ S T − t −w2 /2 e . (9.16) V= √ 2π For example the Vega for a European style option (either call or put) on a security whose current value is $160, whose strike price is $150, and whose expiry date is five months can be calculated using Eq. (9.16). Suppose the volatility of the security is 20% per year and the risk-free interest rate is 2.75% per year. Then w = 0.653219 and V = 33.2866. 9.5

Rho

The Rho, ρ, of an option is the rate of change in the value of the option as a function of the risk-free interest rate r. Once again the European call option will be used to demonstrate the calculation of Rho. Using Eq. (8.38) it is seen that    ∂w √ ∂C = Sφ (w) − Ke−r(T −t)φ w − σ T − t ∂r   ∂r √ + K(T − t)e−r(T −t) Φ w − σ T − t . (9.17) 14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

234

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 9

An Undergraduate Introduction to Financial Mathematics

Calling on the definition of w given in Eq. (8.36) we see that ∂w = ∂r

√ T −t . σ

(9.18)

Substituting Eq. (9.18) into Eq. (9.17) yields  √T − t   √ ∂C = Sφ (w) − Ke−r(T −t) φ w − σ T − t ∂r σ   √ −r(T −t) + K(T − t)e Φ w−σ T −t  √T − t  √ 2 2 √ = Se−w /2 − Ke−r(T −t) e−(w−σ T −t) /2 σ 2π   √ + K(T − t)e−r(T −t) Φ w − σ T − t   e−w2 /2 √T − t √ 2 √ = S − Ke−r(T −t)+wσ T −t−(T −t)σ /2 σ 2π   √ + K(T − t)e−r(T −t) Φ w − σ T − t   e−w2 /2 √T − t √ = S − Keln(S/K) σ 2π   √ + K(T − t)e−r(T −t) Φ w − σ T − t   √ ρC = K(T − t)e−r(T −t) Φ w − σ T − t .

(9.19)

Note that for ρC > 0 implying that the value of a European call option increases with increasing interest rate. Consider the situation of a European call option on a security whose current value is $225. The strike price is $230, while the strike time is four months. The volatility of the price of the security is 27% annually and the risk-free interest rate is 3.25%. Under these conditions, w = 0.0064434 and ∂C = 33.4156. ∂r The Rho for a European put option can be determined from the Put-Call Parity formula (Eq. (7.1)) and Eq. (9.19). ∂P ∂C = − K(T − t)e−r(T −t) ∂r ∂r 

 √ = K(T − t)e−r(T −t) Φ w − σ T − t − K(T − t)e−r(T −t) 14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Derivatives of Black-Scholes Option Prices

   √ = −K(T − t)e−r(T −t) 1 − Φ w − σ T − t  √  ρP = −K(T − t)e−r(T −t) Φ σ T − t − w .

BC8495/Chp. 9

235

(9.20)

In the last equation above we made use of the fact that Φ (x) = 1 − Φ (−x) for the cumulative distribution function of a normally distributed random variable. Note that for ρP < 0 implying that the value of a European put option decreases with increasing interest rate. To put Eq. (9.20) to work in an example, consider the situation of a European put option on a security whose underlying value is $150. The strike price is $152, the volatility is 15% per annum, the expiry date is three months, and the risk-free interest rate is 2.5%. Then w = −0.0557697 and ∂P = −20.8461. ∂r 9.6

Relationships Between ∆, Θ, and Γ

The value F of an option on a security S must satisfy the Black-Scholes partial differential Eq. (7.15). This equation is repeated below for convenience. 1 rF = Ft + rSFS + σ 2 S 2 FSS 2 This equation assumes the risk-free interest rate is r and the volatility of the security is σ. In this chapter we have given names to the partial derivatives Ft , FS , and FSS , namely Θ = Ft ,

∆ = FS ,

Γ = FSS .

Thus the Black-Scholes partial differential equation can be re-written as 1 rF = Θ + rS∆ + σ 2 S 2 Γ. 2

(9.21)

The value of the option is a linear combination of ∆, Θ, and Γ. When the value of the option is insensitive to the passage of time Θ = 0 and Eq. (9.21) becomes   σ2 S F =S ∆+ Γ . 2r 14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

236

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 9

An Undergraduate Introduction to Financial Mathematics

If ∆ should be of large magnitude then Γ must be large and of opposite sign. Returning to Eq. (9.21), if the rate of change in the value of the option with respect to the value of the underlying security is zero (or at least very small) then the Black-Scholes equation becomes 1 rF = Θ + σ 2 S 2 Γ. 2 Thus Θ and Γ tend to be of opposite algebraic signs. In practice, calculation of Θ can be used to approximate Γ. Table 9.1 summarizes the option value derivatives discussed in this chapter and lists some of their relevant properties. This table is provided as a convenient place to find the definitions of these quantities in the future. Table 9.1 Listing of the derivatives of European option values and some of their properties. Name Theta Delta Gamma Vega Rho

Symbol ΘC ΘP ∆C ∆P Γ V ρC ρP

Definition Eq. (9.3) Eq. (9.5) Φ (w) Φ (w) − 1 Eq. (9.10) Eq. (9.16) Eq. (9.19) Eq. (9.20)

Property

∆C > 0 ∆P < 0 Γ>0 V>0 ρC > 0 ρP < 0

In the next chapter the practice of hedging will be described. Hedging is performed on a portfolio of securities, a collection which may include stocks, put and call options, cash in savings accounts, etc. The mathematical aspect of hedging is made possible by the linearity property of the Black-Scholes partial differential equation. An operator L[·] is linear if L[aX + Y ] = aL[X] + L[Y ] for all scalars a and vectors X, Y in the domain of L. In the case of the Black-Scholes PDE the vectors are solution functions. If we define the operator L[·] to be

L[X] =

∂X ∂X 1 ∂2X + rS + σ2 S 2 − rX, ∂t ∂S 2 ∂S 2

then any solution X, to the Black-Scholes partial differential equation, Eq. (7.15), will satisfy L[X] = 0.

14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Derivatives of Black-Scholes Option Prices

9.7

BC8495/Chp. 9

237

Exercises

(1) Find Θ for a European call option on a stock whose current value is $300. The annual risk-free interest rate is 3%. The strike price is $310 while the strike time is three months. The volatility of the stock is 25% annually. (2) Consider the Θ for a European call option on a stock whose current value is $100. The annual risk-free interest rate is 4.12%. The strike price of the option is $103 and expiry occurs in twelve months. The volatility of the stock price is 30% annually. Plot Θ as a function of the stock price at twelve months to expiry, six months to expiry, and one month to expiry. (3) Carefully work through the details of the derivation of Θ for a European put option. Confirm that you obtain the result in Eq. (9.5). (4) Find Θ for a European put option on a stock whose current value is $275. The annual risk-free interest rate is 2%. The strike price is $265 while the expiry date is four months. The volatility of the stock is 20% annually. (5) Find the partial derivative of w (as defined in Eq. (8.36)) with respect to S. (6) Find ∆ for a European call option on a stock whose current value is $150. The annual risk-free interest rate is 2.5%. The strike price is $165 while the strike time is five months. The volatility of the stock is 22% annually. (7) Find ∆ for a European call option on a stock whose current value is $50. The annual risk-free interest rate is 3.25%. The strike price is $55 while the strike time is four months. The volatility of the stock is 27% annually. (8) Find ∂P/∂S for a European put option on a stock whose current value is $125. The annual risk-free interest rate is 5.5%. The strike price is $140 while the expiry date is eight months. The volatility of the stock is 15% annually. (9) Find ∆ for a European put option on a stock whose current value is $75. The annual risk-free interest rate is 6.5%. The strike price is $81 while the expiry date is seven months. The volatility of the stock is 33% annually. (10) Calculate Γ for a European put option for a security whose current price is $180. The strike price is $175, the annual risk-free interest rate

14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

238

(11)

(12) (13)

(14)

(15) (16)

(17)

(18)

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 9

An Undergraduate Introduction to Financial Mathematics

is 3.75%, the volatility in the price of the security is 30% annually, and the strike time is four months. Calculate Γ for a European call option for a security whose current price is $205. The strike price is $195, the annual risk-free interest rate is 4.65%, the volatility in the price of the security is 45% annually, and the strike time is two months. Using partial differentiation derive Eq. (9.13) from Eq. (8.36). Calculate the V of a European style option on a security whose current value is $300, whose strike price is $305, and whose expiry date is six months. Suppose the volatility of the security is 25% per year and the risk-free interest rate is 4.75% per year. Calculate the V of a European style option on a security whose current value is $123, whose strike price is $125, and whose expiry date is three months. Suppose the volatility of the security is 35% per year and the risk-free interest rate is 5.15% per year. Estimate the change in the value of the option in exercise (14) if the volatility of the security is 30% per year. Calculate the Rho of a European style put option on a security whose current value is $270, whose strike price is $272, and whose strike time is two months. Suppose the volatility of the security is 15% per year and the risk-free interest rate is 3.75% per year. Calculate the Rho of a European style call option on a security whose current value is $305, whose strike price is $325, and whose expiry date is four months. Suppose the volatility of the security is 35% per year and the risk-free interest rate is 2.55% per year. Option elasticity is the ratio of the percentage change in the price of an option to the percentage change in the underlying price of the stock. It can be denoted and calculated for a call option as Ω=

(S)(∆) . C

Calculate the option elasticity of a European style call option on a security whose current value is $425, whose strike price is $435, and whose expiry date is six months. Suppose the volatility of the security is 17% per year and the risk-free interest rate is 5.25% per year. (19) Calculate the option elasticity of a European style put option on a security whose current value is $315, whose strike price is $310, and whose expiry date is four months. Suppose the volatility of the security is 24% per year and the risk-free interest rate is 4.75% per year.

14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Derivatives of Black-Scholes Option Prices

BC8495/Chp. 9

239

(20) The volatility of an option is the product of the absolute value of the elasticity of the option multiplied by the volatility of the underlying security. Find the volatility of the European call option described in exercise (18).

14:30:05.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

This page intentionally left blank

14:30:05.

240

BC8495/Chp. 9

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

Chapter 10

Hedging

Hedging is the practice of making a portfolio of investments less sensitive to changes in market variables such as the prices of securities and interest rates. If F (S, t) is a solution to the Black-Scholes partial differential equation Eq. (7.15), the quantity ∆ = FS is central to hedging a portfolio. In fact this same quantity ∆ was used to derive the Black-Scholes PDE. The term “Delta” may be used in two senses in this chapter. In some cases ∆ will refer to the instantaneous rate of change of a financial derivative with respect to the value of the underlying security. This is the mathematical Delta. The term may also be used as part of the phrase “Delta neutral” used to describe a portfolio consisting of the underlying security and financial derivatives. The portfolio P made up of one option F and ∆ shares of the underlying security S is Delta neutral if PS = 0. This is the quantitative analytical sense of “Delta”. The Black-Scholes equation can be thought of as the statement that when the ∆ of a portfolio is zero, the rate of return from a portfolio consisting of ownership of the underlying security and sale of the option (or the opposite positions) should be the same as the rate of return at the risk-free interest rate on the same net amount of cash. In Chapter 9 the partial derivatives, also known as the “Greeks”, of the values of European put and call options were calculated. In this chapter they will be used to hedge portfolios of investments. 10.1

General Principles

Before launching into a discussion of hedging, let us pause and recall what happens when one entity (say, a bank) sells a call option on a stock to another entity (say, an investor). The bank has promised the investor that

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

242

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

An Undergraduate Introduction to Financial Mathematics

they will be able to purchase the stock at a predetermined price (the strike price), at a predetermined time (the expiry date) in the future. The stock must be available to the investor at the strike price even if the market value of the stock exceeds the strike price. The bank is in a favorable position if, at the strike time, the market price of the stock is below the strike price set when the call option was sold to the investor. In this case the call option will not be exercised and the bank keeps the premium it received when it sold the option. Conversely, if the market price of the stock exceeds the strike price of the option, the bank must ensure the investor can find a seller of the stock who will agree to accept the strike price. One way to accomplish this would be for the bank itself to be the seller of the stock to the investor. Consider this strategy: the bank creates a European call option on a stock whose current price is $50 while the strike price is $52, the risk-free interest rate is 2.5% per annum, the expiry date is four months, and the volatility of the stock price is 22.5% per annum. According to the BlackScholes option pricing formula, the value of this European call option is $1.91965. So that we deal with whole numbers, suppose an investor purchases one million of these call options. The bank accepts revenue equal to $1, 919, 650. At two extremes, the bank could wait until the expiry date to purchase the one million shares of the stock, or they could purchase one million shares at the time the investor bought the million call options. The former strategy is called a naked position since the bank’s potential responsibility for providing the stock at the strike price is exposed to the movements in the market until expiry. The latter strategy is called a covered position since the bank is now protected against increases in the price of the security prior to the expiration of the option. Suppose the bank takes the covered position, they will expend $50M, considerably more than the revenue generated by the sale of the call options. If S represents the price per share of the stock at the strike time then the net revenue generated by the transaction obeys the formula 1919650 + (min{52, S(T )}e−0.025/3 − 50) · 106 . Notice that the present value of the stock price must be used rather than just S(T ). The reader can check that when S(T ) ≈ 48.4827 the net revenue is zero. Figure 10.1 shows if the stock is below that price the bank loses money (at S(T ) = 46 the loss comes to more than $2.4M), and if the stock price is higher the bank profits (at S(T ) = 52 the profit is nearly $3.5M).

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

243

Hedging

*106 4 3 2 1 48

50

52

54

S

-1 -2 -3 -4 Fig. 10.1 The line shows the profit or loss to the seller of a European call option who adopts a covered position by purchasing the underlying security at the time they sell the option.

Since the strike price is $52, the maximum net revenue for the bank occurs at that price. If the bank adopts a naked position it will purchase the stock at expiry and instantaneously sell it to the investor at the strike price. In this scenario the net revenue generated by this sequence of transactions is 1919650 + min(0, 52 − S(T ))e−0.025/3 · 106 . The net revenue is zero when S(T ) is approximately $53.9357. As long as the stock price remains below the strike price the bank keeps its revenue from the sale of the options. However the bank’s losses could be dramatic if the price of the stock rises before the expiry date. At a price of $56 per share the bank’s net loss would be approximately $2.0M. Figure 10.2 illustrates the profit or loss to the seller of the European call option. Neither of these schemes are practical for hedging portfolios in the financial world due to their potentially large costs. The naked and the covered positions represent two extremes, one in which no stock is held and the other in which all the potentially needed shares are held from the moment the option is sold until expiry. A better strategy may be a compromise between these extremes. Perhaps some fraction of the total, potential number of necessary shares should be held by the bank. The optimal number would

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

244

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

An Undergraduate Introduction to Financial Mathematics

*106 2

1

50

52

54

56

S

-1

-2 Fig. 10.2 The line shows the profit or loss to the seller of a European call option who adopts a naked position by purchasing the underlying security at the expiry date of the option.

be the number which reduces the potential loss to the bank. In the next section a simple method for hedging is explored.

10.2

Delta Hedging

The purpose of hedging is to eliminate or at least minimize the change in the value of a portion of an investor’s or an institution’s portfolio of investments as conditions change in the financial markets. The hedged portion of a portfolio may represent the value of some future financial obligation, for example a pension intended to fund retirement. In general, people and institutions invest in stocks to earn a greater rate of return per dollar invested than can be earned in low risk government-issued bonds. The higher rate of return can be thought of a payment for accepting the greater risk. Hedging is a method of reducing the risk at the cost of some of the reward. The value of a portfolio containing stocks and options may change due to changes in the value of the stock underlying the option. The reader will recall from the previous chapter that Delta is the partial derivative of the value of an option with respect to the value of the underlying security. In Sec. 9.2 expressions for Delta for European call and put options were derived. One can think of Delta in the following way: for every unit change in

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Hedging

BC8495/Chp. 10

245

the value of the underlying security, the value of the option changes by ∆. Delta is the marginal option value of the underlying security. A portfolio consisting of securities and options is called Delta neutral if for every option sold, ∆ units of the security are bought. Example 10.1 Consider the case of a security whose price is $100, while the risk-free interest rate is 4% per annum, the annual volatility of the stock price is 23%, the strike price is set at $105, and the expiry date is 3 months. Under these conditions w = −0.279806, the value of a European call option is C = 2.96155 and Delta for the option is ∆=

∂C = 0.389813. ∂S

Thus if a firm sold an investor European call options on ten thousand shares of the security1 , the firm would receive $29615.50 and purchase $389813 = (10000)(0.389813)(100) worth of the security, most likely with borrowed money. The firm may choose to do nothing further until the strike time of the option. In that case, this is referred to as a “hedge and forget” scheme. On the other hand, since the price of the security is dynamic, the firm may choose to make periodic adjustments to the number of shares of the security it holds. This strategy is known as rebalancing the portfolio. The example begun above can be extended to include weekly rebalancing. Assume that the value of the security follows the random walk illustrated in Fig. 10.3. In this case the European call option will be exercised since the price of the security at the expiry date exceeds the strike price of $105. At the end of the first week the value of the security has declined to $98.79. Upon re-computing, ∆ = 0.339811. Thus the investment firm would adjust its security holdings so that it now owned 3398.11 shares of the security2 with a total value of $335699. Also at the end of the first week the firm will have incurred costs in the form of interest on the money initially borrowed to purchase shares of the security. This cost amounts to (389813)(e0.04/52 − 1) = $299.9717. 1 Typically an option is keyed to 100 shares of the underlying security, so in practice options on 10000 shares would amount to 100 options. To avoid confusion we will assume a one-to-one ratio of options to shares. 2 The Black-Scholes equation treats the number of shares as a continuous, not discrete, quantity. To reduce round-off error we will calculate all quantities in this example to at least six significant digits.

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

246

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

An Undergraduate Introduction to Financial Mathematics

112 110 108 106

S

104 102 100 0

2

4

6

8

10

12

week

Fig. 10.3 A realization of a random walk taken by the value of a security. The horizontal line indicates the strike price of a European call on the security.

The interest for the current week is added to the cumulative cost entry of the next week. Table 10.1 summarizes the weekly rebalancing up to expiry. The entries in this table are computed iteratively. The subscript i indicates a quantity evaluated at the beginning of the ith week. The interest cost initially is zero and the initial cumulative cost is Σ0 = 10000∆0S0 . When i ≥ 1, the interest cost is Σi−1 (er/52 − 1) and the new cumulative cost is Σi = (∆i − ∆i−1 )Si + Σi−1 er/52 . At the expiry date the investment firm has in its portfolio 10000 shares of the security for which the investor will pay a total of $1, 050, 000. Thus the net proceeds to the investment firm of selling the call option and hedging this position are 1050000 + 29615.50e0.04(13/52) − 1069458 = $10455.10. The reader should note that the value of the call options earns interest compounded continuously at the risk-free rate. In this case, the bank makes money by issuing the call option. If the bank had adopted a covered position from the moment the option contract was written, they would have made a profit of 1050000 + 29615.50e0.04(13/52) − 1000000e0.04(13/52) = $69862.50. 14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

247

Hedging Table 10.1 Week 0 1 2 3 4 5 6 7 8 9 10 11 12 13

BC8495/Chp. 10

Delta hedging using portfolio rebalancing at weekly intervals.

S 100.00 98.79 102.52 103.41 102.82 102.25 100.67 106.05 104.17 106.08 105.86 110.40 112.46 108.47

∆ 0.389813 0.339811 0.462922 0.490192 0.460541 0.428236 0.347145 0.589204 0.491348 0.595047 0.585915 0.878690 0.985811 1.000000

Shares Held 3898.13 3398.11 4629.23 4901.92 4605.42 4282.37 3471.46 5892.05 4913.49 5950.47 5859.15 8786.90 9858.11 10000.00

Share Cost 389813 335699 474588 506907 473528 437872 349471 624852 511838 631226 620250 970074 1108643 1084700

Interest Cost 0 299.972 262.190 359.517 381.494 358.327 333.184 270.620 468.369 390.286 475.238 468.164 717.253 810.509

Cumulative Cost 389813 340716 467192 495751 465646 432972 351671 608646 507177 617571 608379 932071 1053256 1069458

If the bank had held a naked position until expiry, they would have had a loss of 1050000 + 29615.50e0.04(13/52) − 1084700 = −$4787.36. Does this mean that adopting the covered position is preferable to hedging? It can only be sees as preferable in hindsight. The potential losses from the covered position are large. Had the random walk of the stock moved higher by expiry, the losses from the naked position could have been much larger. Alternatively, the value of the security may evolve in such as fashion that the call option is not exercised. Such a scenario is depicted in Fig. 10.4. Since the value of the security finishes below the strike price, the call option expires unused. Table 10.2 summarizes the weekly rebalancing activity of the investment firm as it practices Delta hedging. At the expiry date the investment firm owns no shares of the security. None are needed since the call option will not be exercised. The net proceeds to the investment firm of selling the call option and hedging its position are 29615.50e0.04(13/52) − 29667.7 = $245.44. In this case the firm earns a small amount of money on these transactions. Rebalancing of the portfolio can take place either more or less frequently than weekly as was done in the two previous extended examples. The quantity discussed in Sec. 9.3 known as Gamma (Γ) is the second partial deriva-

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

248

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

An Undergraduate Introduction to Financial Mathematics

105

100 S 95

90 0

2

4

6

8

10

12

week

Fig. 10.4 Another realization of a random walk taken by the value of a security. The horizontal line indicates the strike price of a European call on the security.

Table 10.2 Delta hedging using portfolio rebalancing at weekly intervals for an option which will expire unused. Week 0 1 2 3 4 5 6 7 8 9 10 11 12 13

S 100.00 101.71 100.43 100.91 103.37 97.69 91.95 91.12 92.81 95.45 97.75 96.58 95.40 95.10

∆ 0.389813 0.440643 0.386757 0.394649 0.482725 0.246176 0.071229 0.043022 0.050427 0.078574 0.110154 0.036209 0.001508 0.000000

Shares Held 3898.13 4406.44 3867.58 3946.49 4827.25 2461.76 712.290 430.230 504.272 785.741 1101.54 362.098 15.0811 0.00000

Share Cost 389813 448179 388421 398240 498993 240489 65495.1 39202.5 46801.4 74999.0 107675 34971.4 1438.74 0.00000

Interest Cost 0.00000 299.972 339.987 298.603 304.961 375.257 197.720 74.0827 54.3619 80.4120 104.228 49.3529 23.9154 0.00000

Cumulative Cost 389813 441813 388035 396297 487646 256937 96270.5 70643.2 77569.4 104495 135445 64134 31078.0 29667.7

tive of the portfolio with respect to the value of the underlying security. In other words Gamma is the rate of change of Delta with respect to S. If |Γ| is large then ∆ changes rapidly with relatively small changes in S. In this case frequent rebalancing of the investment firm’s position may be necessary. If |Γ| is small then ∆ is relatively insensitive to changes in S and hence rebalancing may be needed infrequently. Thus an investment

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Hedging

BC8495/Chp. 10

249

firm can monitor Gamma in order to determine the frequency at which the firm’s position should be rebalanced. 10.3

Delta Neutral Portfolios

In exercise (20) of Chapter 8 the reader was asked to verify that a stock or security itself satisfies the Black-Scholes PDE (7.15). Now suppose that a portfolio consists of a linear combination of options and shares of the underlying security. The portfolio contains a short position in a European call option and a long position in the security (hedged appropriately as described in the previous section). Thus the net value P of the portfolio is ∂C P = C − ∆S = C − S, (10.1) ∂S S0

where S0 is the price of the security at the time the hedge is created. The quantity P satisfies the Black-Scholes equation since C and S separately do, and the partial differential equation is linear. Thus it is reasonable to contemplate Delta for the portfolio. The partial derivative of the entire portfolio with respect to S represents the sensitivity of the value of the portfolio to changes in S. Differentiating both sides of Eq. (10.1) with respect to S produces ∂P ∂C ∂C = − ∂S ∂S ∂S S0 This quantity equals zero whenever S = S0 (for example at the moment the hedge is set up) and remains very close to zero so long as S is near S0 . For this reason a portfolio which is hedged using Delta hedging is sometimes called Delta neutral. Assuming that the risk-free interest rate remains constant and the volatility of the security does not change, then a Taylor series expansion of the value of the portfolio in terms of t and S yields ∂P ∂P ∂ 2 P (S − S0 )2 (t − t0 ) + (S − S0 ) + + ··· ∂t ∂S ∂S 2 2 1 δP = Θδt + ∆δS + Γ(δS)2 + · · · 2 P = P0 +

All omitted terms in the Taylor series involve powers of δt greater than 1. The Gamma term is retained since the stochastic random variable S follows

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

250

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

An Undergraduate Introduction to Financial Mathematics

√ a stochastic process dependent on δt, see Eq. (5.44). If the portfolio is hedged using Delta hedging, then ∆ for the portfolio is zero and thus 1 δP ≈ Θδt + Γ(δS)2 , 2

(10.2)

where the approximation omits terms involving powers of δt greater than 1. The term involving Θ is not stochastic and thus must be retained; however, the approximation can be further refined if the portfolio can be made Gamma neutral, i.e. if the makeup of the portfolio can be adjusted so that Γ = 0. 10.4

Gamma Neutral Portfolios

Gamma for a portfolio is the second partial derivative of the portfolio with respect to S. The portfolio cannot be made Gamma neutral using a linear combination of just an option and its underlying security, since the second derivative of S with respect to itself is zero. As explained in the previous section, the portfolio can be made Delta neutral with the appropriate combination of the option and the underlying security. The portfolio can be made Gamma neutral by manipulating a component of the portfolio which depends non-linearly on S. One such component is the option. However, as mentioned earlier, the portfolio cannot contain only the option and its underlying security. One way to achieve the goal of a Gamma neutral portfolio is to include in the portfolio two or more different types of option dependent on the same underlying security. For example, suppose the portfolio contains options with two different strike times written on the same stock. In concrete terms, an investment firm may sell a number of European call options with expiry three months hence and buy some other number of the European call options on the same stock with expiry arriving in six months. Let Γ1 be the Gamma of a sold call option (with the earlier strike time), Γ2 be the Gamma of a purchased call option (with the later strike time), x be the number of the early strike options sold, and y be the number of later strike options purchased. The Gamma of the portfolio would be ΓP = xΓ1 − yΓ2 . The numbers x and y can be chosen so that ΓP = 0. Once the portfolio is made Gamma neutral, the underlying security can be added to the portfolio in such a way so as to make the portfolio Delta neutral. The reader

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

251

Hedging

should keep in mind that the underlying security will have a Γ = 0 and thus Gamma neutrality for the portfolio will be maintained while Delta neutrality is achieved. Thus Eq. (10.2) for a Gamma neutral portfolio reduces to δP ≈ Θδt.

(10.3)

Equation (10.3) implies that the change in the value of a simultaneuously Gamma and Delta neutral portfolio is proportional to the size of the time step. Example 10.2 Suppose a stock’s current value is $100 while its volatility is σ = 0.22 and the risk-free interest rate is 2.5% per year. An investment firm sells European call options on this stock with a strike time of three months and a strike price of $102. The firm buys European call options on the same stock with the same strike price but with a strike time of six months. According to Eq. (9.10) the Gamma of the three-month option is Γ3 = 0.03618, while that of the six-month option is Γ6 = 0.02563. The portfolio becomes Gamma neutral at any point in the first quadrant of x3 x6 -space where the equation 0.03618x3 − 0.02563x6 = 0 is satisfied. Suppose x3 = 100000 of the three-month option were sold, then the portfolio is Gamma neutral if x6 = 141163 of the six-month options are purchased. Thus prior to the inclusion of the underlying stock in the portfolio, the Delta of the portfolio is x3 ∆3 − x6 ∆6 = (100000)(0.4728) − (141163)(0.5123) = −25038. Therefore the portfolio can be made Delta neutral if 25038 shares of the underlying stock are sold. Figure 10.5 shows that over a fairly wide range of values of the underlying stock the value P of the Gamma and Delta neutral portfolio remains nearly the same. This discussion of hedging is far from complete. The present chapter has focused mainly on making a portfolio’s value resistant to changes in the value of a security. In reality, the risk-free interest rate and the volatility of the stock also affect the value of a portfolio. The quantities Rho and Vega discussed in Chapter 9 can be used to set up portfolios hedged against changes in the interest rate and volatility. In the present chapter it was also assumed that the necessary options and securities could be bought so as to form the desired hedge. In practice this may not always be possible.

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

252

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

An Undergraduate Introduction to Financial Mathematics

106 P 2.5 2 1.5 1 0.5

80

90

100

110

120

130

S

Fig. 10.5 The aggregate value of a Gamma and Delta neutral portfolio is insensitive to a range of changes in the value of the underlying securities.

For example, an investment firm may not be able to purchase sufficient quantities of a stock to make their portfolio Delta or Gamma neutral. In this case they may have to substitute a different, but related security or other financial instrument in order to set up the hedge. This strategy will be discussed in the next chapter after the prerequisite statistical concepts are introduced. 10.5

Exercises

(1) A call option closes “in the money” if the market price of the underlying security exceeds the strike price for the call option at the time the option matures. Use Eq. (8.36) to show that ∆ → 1 as t → T − for an “in the money” European call option. (2) A call option closes “out of the money” if the strike price for the call option exceeds the market price of the underlying security at the time the option matures. Use Eq. (8.36) to show that ∆ → 0 as t → T − for an “out of the money” European call option. (3) A call option closes “at the money” if the strike price for the call option equals the market price of the underlying security at expiry. Find lim− ∆ for an “at the money” European call option. t→T

(4) Suppose the price of a stock is $54 per share with volatility 23% per

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

253

Hedging

annum and the risk-free interest rate is 4.5% per year. Calculate the price of a European call option with a strike price of $57 per share and a strike time of five months. Use Delta and a linear approximation to determine the price of a four-month call option on a stock whose current price is $54.75 per share (all other features of the stock and option are the same as before). Compare the linear approximation with the Black-Scholes value of the second option. (5) The linear approximation in exercise (4) over estimates the value of the second call option since it does not take into account the change in the value of the option due to the passage of time. Repeat the previous linear approximation including both Delta and Theta for the option. (6) The linear approximation calculated in exercise (5) is closer to the exact value of the second call option than the approximation found in exercise (4). Including the price correction associated with Gamma will produce an even more accurate approximation. Repeat the previous approximation including Delta, Gamma, and Theta for the option. (7) Suppose the price of a stock is $45 per share with volatility σ = 0.20 per annum and the risk-free interest rate is 4.5% per year. Create a table similar to Table 10.1 for European call options on 5000 shares of the stock with a strike price of $47 per share and a strike time of 15 weeks. Rebalance the portfolio weekly assuming the weekly prices of the stock are as follows. Week S Week S

0 45.00 8 47.79

1 44.58 9 48.33

2 46.55 10 48.81

3 47.23 11 51.36

4 47.62 12 52.06

5 47.28 13 51.98

6 49.60 14 54.22

7 50.07 15 54.31

(8) Repeat exercise (7) for the following scenario in which the call option finishes out of the money. Week S Week S

0 45.00 8 41.94

1 44.58 9 42.72

2 45.64 10 44.83

3 44.90 11 44.93

4 43.42 12 44.19

5 42.23 13 41.77

6 41.18 14 39.56

7 41.52 15 40.62

(9) Repeat exercise (7) assuming this time that the financial institution has issued a European put option for the stock. (10) Repeat exercise (8) assuming this time that the financial institution has issued a European put option for the stock.

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

254

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 10

An Undergraduate Introduction to Financial Mathematics

(11) Set up a Gamma neutral portfolio consisting of European call options on a security whose current value is $85. Suppose the risk-free interest rate is 5.5% and the volatility of the security is 17% per year. The portfolio should contain a short position in four-month options with a strike price of $88 and a long position in six-month options with the same strike price. (12) For the portfolio described in exercise (11) add a position in the underlying security so that the portfolio is also Delta neutral. (13) Set up a Gamma neutral portfolio consisting of European call options on a security whose current value is $95. Suppose the risk-free interest rate is 4.5% and the volatility of the security is 23% per year. The portfolio should contain a short position in three-month options with a strike price of $97 and a long position in five-month options with a strike price of $98. (14) For the portfolio described in exercise (13) add a position in the underlying security so that the portfolio is also Delta neutral. (15) Set up a Gamma neutral portfolio consisting of positions in threemonth European call options on a security whose current value is $60. Suppose the risk-free interest rate is 5.65% and the volatility of the security is 45% per year. The portfolio should contain a sold option with a strike price of $62 and a long position in x options with a strike price of $65. (16) For the portfolio described in exercise (15) add a position in the underlying security so that the portfolio is also Delta neutral. (17) Consider a sold call on a security whose current value of $60. The risk-free interest rate is 5.65% and the volatility of the security is 45% per year. The strike price of the call is $60 and the strike time is 90 days hence. If a Delta neutral portfolio is created determine the profit or loss after one day if the price of the security the next day is $60.50. (18) For the Delta neutral portfolio described in exercise (17) plot the profit or loss after one day if the price of the security is S. (19) Consider the Delta neutral portfolio described in exercise (17). Find the day 1 price of the security so that the profit/loss is zero. (20) Using the results of exercises (15) and (16) determine the profit or loss after one day if the price of the security the next day is $60.50. (21) For the Delta/Gamma neutral portfolio described in exercise (20) plot the profit or loss after one day if the price of the security is S.

14:30:15.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

Chapter 11

Extensions of the Black-Scholes Model

The work done to develop the Black-Scholes option pricing formulas in Chapters 7-8 can readily be extended to other types of options that generalize the European-style options on non-dividend paying stocks. In this chapter we will develop variations on the Black-Scholes model to price options on stocks which pay either discrete or continuous dividends and to options with payoffs other than the magnitude of the difference between the stock value and the strike price at expiry. Unless otherwise specified all options discussed in this chapter will still have the European style of exercise. 11.1

Options on Stocks Paying Continuous Dividends

The simplest extension to the Black-Scholes model is one which includes the effects of dividends paid continuously. While stock from an individual corporation may typically pay dividends 1–4 times per year, a portfolio made of stock from many different corporations each paying dividends on their own schedule may be approximated by a continuously paid stream of dividends. Assuming a stock worth S pays a dividend at rate δS then the dividend paid in a time interval of length dt will be δS dt. This changes the stochastic differential equation modeling the movements of the stock’s value from that given in Eq. (7.11) to one including the diminution of the value by the amount of dividend paid. dS = (µ − δ)S dt + σS dW (t)

(11.1)

If the value of the European option F depends on t and S then by use of Itˆ o’s Lemma (Lemma 5.4) the option follows a stochastic process described

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

256

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

An Undergraduate Introduction to Financial Mathematics

by dF =

  1 (µ − δ)SFS + σ 2 S 2 FSS + Ft dt + σSFS dW (t). 2

(11.2)

As in the earlier chapter if we assume a portfolio P consists of a long position in the option F and short position in ∆ shares of S then P = F − ∆S. An important difference is the effect of the continuous dividend in the value of the portfolio. Since the stock pays a dividend of δS dt in time dt then ∆ shares will pay a dividend of δ∆S dt in time dt which implies dP = d(F − ∆S) − δ∆S dt = dF − ∆ dS − δ∆S dt.

(11.3)

Letting ∆ = FS and substituting Eqs. (11.1) and (11.2) into Eq. (11.3) produces the differential equation   1 2 2 (11.4) dP = −δSFS + σ S FSS + Ft dt. 2 In the absence of arbitrage the change in value of the portfolio consisting of the option and stock should be equal to the change in value of an equivalent amount of cash invested at the risk-free rate r. This implies dP = r(F − ∆S) dt.

(11.5)

Equating Eq. (11.4) and Eq. (11.5) produces the following partial differential equation for a European option on a stock paying dividends at a continuous yield of δ. 1 rF = Ft + σ 2 S 2 FSS + (r − δ)SFS 2

(11.6)

Readers will note that Eq. (11.6) is nearly the same as Eq. (8.5), the only difference being the risk-free interest rate on the right-hand side of Eq. (11.6) is now discounted by the dividend yield rate. The similarities do not end there. If F (S, t) represents a call option, then at expiry the payoff the remains (K − S)+ and if the value of the stock vanishes then F (0, t) = 0. These are again the final condition and boundary condition of the European call option explored in Chapters 7 and 8. From the Put-Call Parity formula for stocks paying continuous dividends in Eq. (7.8) we see that   lim F (S, t) = lim Pe + Se−δ(T −t) − Ke−r(T −t) S→∞

S→∞

= Se−δ(T −t) − Ke−r(T −t) .

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Extensions of the Black-Scholes Model

BC8495/Chp. 11

257

As the value of the stock increases without bound the put option becomes worthless (since it is unlikely to the exercised) and the call option’s value approaches the difference between the stock without its dividend payments and the present value of the strike price. The boundary condition at infinity should be compared to Eq. (8.8). Defining the function G(S, t) = eδ(T −t) F (S, t) then the following final condition and boundary conditions for G(S, t) hold. G(S, T ) = F (S, T ) = (S − K)+ G(0, t) = e

δ(T −t)

G(S, t) → S − e

F (0, t) = 0

−(r−δ)(T −t)

as S → ∞.

(11.7) (11.8) (11.9)

Furthermore by differentiating G(S, t) and re-arranging terms FS = e−δ(T −t) GS FSS = e−δ(T −t) GSS Ft = e−δ(T −t) (δG + Gt ). Substituting these partial derivatives into Eq. (11.6) and simplifying produces the partial differential equation 1 (r − δ)G = Gt + σ 2 S 2 GSS + (r − δ)SGS . 2

(11.10)

Thus the function G(S, t) satisfies the Black-Scholes partial differential equation for a European call option on a non-dividend paying stock where the risk-free interest rate is interpreted as r − δ. The reader should compare Eq. (11.10) and the conditions in Eqs.(11.7)–(11.9) with their counterparts in Eqs.(8.6)–(8.8). There is no need to repeat the change of variables work of Chap. 8. Using the results found in Eqs.(8.36)–(8.38) we have ln(S/K) + (r − δ + σ 2 /2)(T − t) √ σ T −t   √ G(S, t) = SΦ (w) − Ke−(r−δ)(T −t) Φ w − σ T − t w=

(11.11) (11.12)

and therefore the value of European call option with continuously paid dividends is   √ C e,δ (S, t) = e−δ(T −t) SΦ (w) − Ke−r(T −t) Φ w − σ T − t . (11.13) 14:30:24.

June 25, 2012

13:21 WSPC/Book Trim Size for 9in x 6in Juliet

258

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

An Undergraduate Introduction to Financial Mathematics

C 20

15

10

5

90

100

110

120

SHTL

Fig. 11.1 A plot of C e,δ as a function of S. The lighter curve represents a stock paying no dividends and the heavier curve represents a stock paying continuous dividends.

The presence of the new decaying exponential factor on the right-hand side of Eq. (11.13) would suggest that, all other conditions and quantities remaining the same, the value of European call option on a stock paying continuous dividends should be less than one which paid no dividends. This is illustrated in Fig. 11.1 for a stock whose current value is S = 100, the risk-free interest rate is 3.25%, the volatility in the stock price is 25%, the strike price is K = 100, and expiry arrives in three months. The lighter curve is the value of a European call option assuming the stock pays no dividends, while the heavier curve assumes the stock pay dividends at a continuous rate of 5%. Example 11.1 Suppose the current price of a security is $62 per share. The continuously compounded interest rate is 10% per year. The volatility of the price of the security is σ = 20% per year. The stock pays dividends continuously at a rate of δ = 3% per year. The cost of a five-month European call option with a strike price of $60 per share can be found as below. 5 , t = 0, r = 0.10, σ = 0.20, T = 12 S = 62, K = 60, δ = 0.03 According to Eq. (11.11) w ≈ 0.544463 and therefore by Eq. (11.13) the price of the European call option is C e = $5.24. This value can be compared with the value of $5.80 calculated for a non-dividend paying stock in Example 8.3.

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Extensions of the Black-Scholes Model

BC8495/Chp. 11

259

In Exercise 3 the reader will be asked to show that the value of a European put option on a stock paying continuous dividends is  √  P e,δ (S, t) = Ke−r(T −t)Φ σ T − t − w − e−δ(T −t) SΦ (−w) . (11.14)

Example 11.2 Suppose the current price of a security is $97 per share. The stock pays a continuous dividend at a yield of 6.5% per year. The continuously compounded interest rate is 8% per year. The volatility of the price of the security is σ = 45% per year. The cost of a three-month European put option with a strike price of $95 per share can be found using Eqs. (11.11), and (11.14). T = 1/4, δ = 0.065,

t = 0, S = 97,

r = 0.08,

σ = 0.45,

K = 95.

Thus we have w ≈ 0.221763 and consequently the put option price is P = $7.34. The inclusion of the continuous dividend yield in the model introduces a new Greek, the partial derivative with respect to δ. Since the dividend yield rate functions like the risk-free interest rate, this Greek is often called “Rho” as well. To avoid confusion with the Rho (partial derivative with respect to r) defined earlier in Sec. 9.5, this new Greek will be denoted ρδ . For the European call option ∂C e,δ ∂δ = −S(T − t)e−δ(T −t) Φ (w)   ∂w √ ∂w + Se−δ(T −t) φ (w) − Ke−r(T −t) φ w − σ T − t ∂δ ∂δ = −S(T − t)e−δ(T −t) Φ (w) √ i √ 2 T − t h −δ(T −t) −w2 /2 − √ Se e − Ke−r(T −t) e−(w−σ T −t) /2 σ 2π = −S(T − t)e−δ(T −t) Φ (w) √ 2 i √ 2 e−w /2 T − t h −δ(T −t) √ − Se − Ke−r(T −t) ewσ T −t−σ (T −t)/2 σ 2π √ 2 i e−δ(T −t)−w /2 T − t h √ = −S(T − t)e−δ(T −t) Φ (w) − S − Keln(S/K) σ 2π −δ(T −t) = −S(T − t)e Φ (w) . (11.15)

ρδC =

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

260

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

An Undergraduate Introduction to Financial Mathematics

Thus the value of the European call option on a stock paying constant, continuous dividends at rate δ is a decreasing function of δ.

11.2

Options on Stocks Paying Discrete Dividends

Developing the pricing formula for an option on a stock paying discrete dividends is not as straight forward as pricing an option on a stock paying continuous dividends, yet it is also not difficult if we keep the absence of arbitrage assumption in mind. In this section the case of a stock which pays a single discrete dividend during the life of the option will be considered. The case of multiple, discrete dividends is readily seen as an extension of the approach described here. Recall that if a stock pays a dividend of dy S(t) at t = td then in the absence of arbitrage the price of the stock immediately before and after the dividend payment must satisfy Eq. (7.6), repeated below for convenience. − S(t+ d ) = S(td )(1 − dy )

(11.16)

The stock price follows a random walk, but any realization of the random variable must exhibit a jump discontinuity at t = td . A typical realization of a random walk is illustrated in Fig. 11.2. SHtL SHtd -L

SHtd +L

td

Fig. 11.2

t

The discontinuous jump in the price of a stock across a dividend date.

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

261

Extensions of the Black-Scholes Model

If S(t) is the continuous random variable describing the value of the stock on which the single, discrete proportional dividend dy S(t− d ) is paid then S(t) obeys the following stochastic differential equation. dS = (µ − dy D (t − td ))S dt + σS dW (t)

(11.17)

As before µ and σ are the drift and volatility associated with this stock. The expression dy D (t − td ) S is the dividend yield rate, the value of the dividend paid per unit time. The function D (t − td ) is the Dirac delta function. The Dirac delta function1 is frequently used in mathematical models to represent an instantaneous impulse or shock to a system. Properties of this function will be explored in the exercises of this chapter. As we work with Eq. (11.17) we will derive a relationship between dy and δ which was used in Eq. (11.16) to represent the fraction of the stock’s value paid as a discrete dividend. Defining Y = ln S and applying Itˆo’s Lemma (Lemma 5.4) we obtain the Wiener process   1 dY = µ − dy D (t − td ) − σ 2 dt + σ dW (t). (11.18) 2 Integrating yields the following piecewise-defined function.  Z t Z t 1 2 Y (t) − Y (0) = µ − dy D (s − td ) − σ ds + σ dW (s) 2 0 0   Z t 1 = µ − σ 2 t − dy D (s − td ) dt + σW (t) 2 0   1 2 if t < td ,  µ − 2 σ t + σW (t) Y (t) = Y (0)   µ − 12 σ 2 t − dy + σW (t) if t ≥ td .

Note that Y (t) is a normally distributed random variable with mean (µ − σ 2 /2)t for t < td , with mean (µ − σ 2 /2)t − dy for t ≥ td and with variance σ 2 t. Since Y (t) = ln S(t) then the stock price is a lognormal random variable with  (µ−σ2 /2)t+σW (t)  if t < td ,  S(0)e S(t) = (11.19)   S(0)e(µ−σ2 /2)t−dy +σW (t) if t ≥ t . d 1 Many

authors denote the Dirac delta function as δ(t). Since the symbol δ is already in use as the continuously compounded proportional dividend, the Dirac delta function is denoted D(t) to avoid confusion.

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

262

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

An Undergraduate Introduction to Financial Mathematics

Consider the one-sided limits at t = td . h i 2 lim S(t) − lim S(t) = lim S(0)e(µ−σ /2)t+σW (t) t→t− d

t→t+ d

t→t− d

h i 2 − lim+ S(0)e(µ−σ /2)t−dy +σW (t) t→td

Since the expression (µ − σ 2 /2)t + σW (t) is continuous the limits from the left and the right as t approaches td are the same and thus we may write + (µ−σ S(t− d ) − S(td ) = S(0)e

2

/2)td +σW (td )

− S(0)e(µ−σ   2 = S(0)e(µ−σ /2)td +σW (td ) 1 − e−dy .

(µ−σ Using the fact that S(t− d ) = S(0)e last equation to the following.

2

/2)td +σW (td )

2

/2)td −dy +σW (td )

we may simplify the

− −dy S(t+ d ) = S(td )e

(11.20)

Thus when a discrete dividend of size dy S(t− d )D(t − td ) is paid, the relative change in the value of the stock is 1 − e−dy . Using the fact that S(t) is a lognormal random variable, we may apply Lemma 3.1 to see that  S(0)eµt if t < td , E [S(t)] = S(0)eµt−dy if t ≥ td . Figure 11.3 illustrates the expected value of a stock for which µ = 1 and dy = 1/3. Naively one might expect the jump discontinuity in the price of the stock to induce a jump discontinuity in the value of a European option on the stock. In fact the value of an option must be a continuous function of time across the dividend date or arbitrage will result. Expressing this mathematically we see that lim C e (S(t), t) = lim+ C e (S(t), t)

t→t− d

− C (S(t− d ), td ) − C e (S(t− d ), td ) e

t→td

+ = C e (S(t+ d ), td ) −dy + = C e (S(t− , td ) d )e

(11.21)

by Eq. (11.20). The reader should carefully consider Eq. (11.21) and its implications. This equation states that the value of the call option will change discontinuously across the dividend date as a function of S, the price of the underlying stock. However, the price of the option is made continuous by equating the value of the option just before the dividend

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

263

Extensions of the Black-Scholes Model

E@SHtLD

SH0L

td

t

Fig. 11.3 The expected value of a stock with drift parameter µ = 1 for which a single discrete dividend is paid at t = td . The dividend is one third the value of the stock at t = td . − is paid C e (S(t− d ), td ) with the value of the option just after the dividend − −dy + e is paid C (S(td )e , td ) for which the value of the underlying has been − −dy modified to S(t+ ) = S(t . d d )e Now we can turn to the mechanics of pricing an option on a stock for which a single discrete dividend will be paid during the life of the option. This will require solving the Black-Scholes equation twice. Once for the interval (td , T ] from immediately after the dividend is paid until expiry and a second time for the interval [0, td ), from the moment of the option is written until immediately prior to the dividend date. To cross the dividend date the two solutions of the Black-Scholes equation will be connected using Eq. (11.21). On the interval (td , T ] no dividends are paid and thus the call option can be priced using the formula given in Eq. (8.38). At t = t+ d the value of the call option is   q + + + −r(T −t+ d )Φ T − t . C e (S(t+ ), t ) = S(t )Φ (w) − Ke w − σ d d d d

Immediately before the dividend date the value of the call option is − − −dy + e C e (S(t− , td ) d ), td ) = C (S(td )e −dy = S(t− Φ (w) d )e

  q + −r(T −t+ ) d − Ke Φ w − σ T − td .

14:30:24.

(11.22)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

264

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

An Undergraduate Introduction to Financial Mathematics

On the right-hand side of Eq. (11.22) the price of the underlying stock has been scaled by the factor e−dy . Consider the effect of scaling the value of the underlying stock by e−dy on the Black-Scholes initial, boundary value problem given in Eqs. (8.5)–(8.8). Making the change of variables Sˆ = Se−dy then the partial derivatives ∂ ∂ [F (S, t)] = e−dy [F (S, t)] and ∂S ∂ Sˆ ∂2 ∂2 [F (S, t)] = e−2dy [F (S, t)] . 2 ∂S ∂ Sˆ2 Therefore Eq. (8.5) is transformed to 1 ˆ ˆ. rF = Ft + σ 2 Sˆ2 FSˆSˆ + rSF S 2

(11.23)

Consequently scaling the stock price leaves the Black-Scholes partial differential equation unchanged. At expiry the payoff of the call option will be ˆ T ) = F (Se−dy , T ) F (S, = (Se−dy − K)+

= e−dy (S − Kedy )+ .

(11.24)

The payoff of the call option with underlying stock Sˆ is the same as the payoff of e−dy call options on the underlying stock S with strike price Kedy . The boundary condition at Sˆ = 0 remains the same as Eq. (8.7) and the boundary condition at infinity becomes ˆ t) = Sˆ − Ke−r(T −t) lim F (S, h i = e−dy S − Kedy −r(T −t) .

ˆ S→∞

(11.25)

Thus for times prior to the dividend date, the call option can be priced by equating it with the price of e−dy call options having a strike price of Kedy . For t < td , h  i √ (11.26) C(S, t) = e−dy SΦ (w) − Kedy −r(T −t) Φ w − σ T − t . The continuity of the value of the European call option across the dividend date can be verified in Fig. 11.4 which illustrates the value of an at-themoney call option over the interval [0, T ]. A more general picture of the

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Extensions of the Black-Scholes Model

BC8495/Chp. 11

265

behavior of the call option on a stock paying a discrete dividend can be seen in the surface plot of Fig. 11.5.

C

td

T

t

Fig. 11.4 The value of a European call option on a stock paying a single discrete, proportional dividend at t = td . The value of the stock is constant where S(t) = K, the strike price.

C

K

S

td t T

Fig. 11.5 The value of a European call option on a stock paying a single discrete, proportional dividend at t = td . The value of the call option is continuous across the dividend date for all values of S.

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

266

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

An Undergraduate Introduction to Financial Mathematics

Example 11.3 Suppose a stock pays a discrete dividend of 10% of its value at td = 1/2. The risk-free interest rate is 10% per year and the volatility of the stock price is 20% per year. Consider the value of a oneyear European call option on the stock with a strike price of $100. If the value of the stock immediately prior to the dividend date is $100, then immediately after the dividend is paid we have the following variables and parameters for the Black-Scholes European call option formula. td = 0.5 K = 100

T =1 S(t+ d)

= 90

r = 0.10 σ = 0.20

Using these values and Eq. (8.38) the value of the option immediately postdividend is + C e (S(t+ d ), td ) ≈ 3.05038.

Prior to the dividend payment we use the following values to price the option. td = 0.5 K = 100

T =1 S(t− d)

= 100

r = 0.10 dy = 0.105361 σ = 0.20

Substituting these values in Eq. (11.26) we see that − C e (S(t− d ), td ) ≈ 3.05038,

just as desired. Fig. 11.6 shows the value of the European call option immediately before (bold curve) and after the dividend date as a function of the underlying stock price. The before curve can be obtained by contracting the after curve in the horizontal direction by a factor of e−dy . To summarize, the value of a European call option written on a stock paying a single discrete, proportional dividend during the life of the option can be written as a piecewise-defined function. Using the formulas for w and C(S, t) defined in Eqs. (8.36) and (8.38), the option on the dividend-paying stock has value  √  −dy  e SΦ (w) − Kedy Φ w − σ T − t if t < td , e,δ √ C (S, t) = e−dy SΦ (w) − KΦ w − σ T − t if t ≥ td . To be very clear, in the pre-dividend portion of this formula Kedy is used for the strike price in Eqs. (8.36) and (8.38) while in the post-dividend formula Se−dy is used for the value of the underlying stock.

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

267

Extensions of the Black-Scholes Model

C 10

8

6

4

After Before

2

20

40

60

80

100

120

S

Fig. 11.6 The value of a European call option immediately before (bold) and after a discrete dividend payment. Note that while the value of the stock instantaneously decreases in value by 10% as the dividend is paid, the value of the call option is continuous.

Example 11.4 The value of a European put option on the underlying stock described in Example 11.3 can be determined from the value of the European call and the Put-Call Parity formula for options on stocks paying discrete dividends, Eq. (7.7). Adapting this formula to the situation of a single dividend payment gives −r(td −t) P e,δ (S, t) + S(t) − (1 − e−dy )S(t− d )e

= C e,δ (S, t) + Ke−r(T −t) .

(11.27)

Proceeding as in Example 11.3, C e,δ (100, 0) ≈ 6.94898. Assuming the stock price immediately prior to the dividend date is $100, then P e,δ (100, 0) ≈ 6.94502. Recall that immediately prior to and after the dividend payment C e,δ (S, t± d ) ≈ 3.05038. Computing the price of the European put just before and after the dividend payment gives P e,δ (S, t± d ) ≈ 8.17332. 11.3

Exercises

(1) Suppose a stock pays a dividend D at the end of the year. At the end of the following year the stock will pay a dividend of D(1 + g). The year after that the stock will pay a dividend of D(1 + g)2 . If this continues until k dividends have been paid, find a formula for the sum of the present values of the dividends assuming the interest rate is r

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

268

(2)

(3)

(4)

(5)

(6)

(7) (8)

(9)

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

An Undergraduate Introduction to Financial Mathematics

and is compounded annually. (Hint: find an expression for the sum of the dividends S and multiply it by 1 + r.) Find the sum of the present values of the dividends described in exercise (1) as the number of dividends increases to infinity assuming 0 < g < r. Use the Put-Call Parity formula Eq. (7.8), for a stock paying continuous dividends at yield δ to develop the pricing formula for a European put option on a stock paying continuous dividends. Find the value of a six-month European call option on a stock currently worth $85 per share which pays dividends at a continuous rate of 1.75% per year. The risk-free interest rate is 4.5% per annum. The volatility of the stock is 35% per year. The strike price for the option will be $90. Find the value of a four-month European put option on a stock currently worth $55 per share which pays dividends at a continuous rate of 2.5% per year. The risk-free interest rate is 5.5% per annum. The volatility of the stock is 28% per year. The strike price for the option will be $58. Find the value of a three-month European call option on a stock currently worth $110 per share which pays dividends at a continuous rate of 1% per year. The risk-free interest rate is 5.5% per annum. The volatility of the stock is 25% per year. The strike price for the option will be $120. Find the value of the European put option for the stock described in exercise (6). Using the Put-Call Parity formula Eq. (7.8) for options on a stock paying continuous, constant dividends at yield δ and the Rho for a European call option Eq. (11.15), find the Rho for a European put option. Define the piecewise constant function d (t) as follows.  1  if − < t < ,  2 d (t) =   0 if |t| ≥  where  > 0. Show that

Z



−∞

14:30:24.

d (t) dt = 1.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Extensions of the Black-Scholes Model

BC8495/Chp. 11

269

(10) Define the Dirac delta function D (t) to be D (t) = lim+ d (t) →0

where d (t) is defined in exercise (9). (a) Show that D (t) = 0 for all t 6= 0. (b) Show that Z ∞ D (t) dt = 1. −∞

(11) Show that if f is continuous then Z ∞ δ(t − t0 ) f (t) dt = f (t0 ). −∞

(12) (13) (14) (15)

(16) (17)

(18)

Use the definition of the Dirac delta function from exercise (10) and the Integral Mean Value Theorem. Starting with Eq. (11.17), derive Eq. (11.18). Show that dy = − ln(1 − δ). Show that if Eq. (11.21) does not hold, an arbitrage opportunity exists. Suppose the current price of a stock is $50 and the stock will pay a dividend of 3% of its value in two months. The risk-free interest rate is 2.5%, the strike price of a European call option is $50, the strike time is five months, and the volatility of the stock is 35%. Find the values of the call option at time t = 0 and at the dividend date. Assume the price of the stock immediately before the dividend date is $50. Find the values of a European put option at time t = 0 and at the dividend date for the stock described in exercise (15). Occasionally the dividend paid on a stock will be an absolute amount rather than an amount proportional to the value of the stock immediately before the dividend date. For example, under the conditions mentioned in exercise (15) the stock pays a dividend of $1.50. Find the value at t = 0 of a European call option on the stock described in exercise (15) using the Black-Scholes call option price formula given in Eq. (8.38) where the current value of the stock has been discounted by the present value of the dividend amount to be paid. How does this value compare to the value found in exercise (15)? For the stock described in exercise (15), suppose two dividends are paid. At two months a dividend of $0.75 is paid and at four months

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

270

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 11

An Undergraduate Introduction to Financial Mathematics

another dividend of $0.75 is paid. Use the discounting procedure of exercise (17) to approximate the value of the call option at time t = 0. (19) In this exercise we will approximate the value of a European put option on a stock which pays a single dividend of an absolute amount using the same approach as in exercise (17). Suppose the stock described in exercise (15) pays a dividend of $1.50. Find the value at t = 0 of a European put option on the stock using the Black-Scholes put option price formula given in Eq. (8.39) where the current value of the stock has been discounted by the present value of the dividend amount to be paid. How does this value compare to the value found in exercise (16)? (20) For the stock described in exercise (16), suppose two dividends are paid. At two months a dividend of $0.75 is paid and at four months another dividend of $0.75 is paid. Use the discounting procedure of exercise (19) to approximate the value of the put option at time t = 0.

14:30:24.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

Chapter 12

Optimizing Portfolios

There are several notions of the idea of optimizing a portfolio of securities, options, bonds, cash, etc. An optimal portfolio could be defined as one with a maximal rate of return or it could mean a portfolio for which the probability of a large fluctuation (usually in the downward direction) in the value of the portfolio is minimized. Alternatively an optimal portfolio for some investors could combine the two notions allowing the investor to specify for example an acceptable level of return and then designing the portfolio with the minimum chance of deviating from that return. The probability of deviating from a desired rate of return will serve as our definition of risk associated with a portfolio. Thus in this chapter we will explore portfolio optimality in the sense of maximizing the rate of return while minimizing the variance in the rate of return, and hence minimizing the risk to the investor. We will also introduce the Capital Assets Pricing Model which attempts to relate the rate of return of a specific investment to the rate of return for the entire market of investments. We will see in this chapter that the difference in the expected rates of return for a specific security and the risk-free interest rate is proportional to the difference in the expected rates of return for the market and the risk-free interest rate. In order to explain this proportionality relationship we will introduce two additional statistical measures, covariance and correlation. We will also make use of these concepts in developing additional hedging strategies. 12.1

Covariance and Correlation

The concept of covariance is related to the degree to which two random variables tend to change in the same or opposite direction relative to one another. If X and Y are the random variables then mathematically the

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

272

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics Table 12.1 Child 1 2 3 4 5 6 7 8 9 10

A sample of heights and arm spans for children.

Ht. (cm) 142 148 152 150 141 142 149 151 147 152

Span (cm) 138 144 148 145 136 139 144 145 144 148

Child 11 12 13 14 15 16 17 18 19 20

Ht. (cm) 150 152 148 152 144 148 150 138 145 142

Span (cm) 147 141 144 148 140 143 146 134 142 138

covariance, denoted Cov (X, Y ), is defined as Cov (X, Y ) = E [(X − E [X])(Y − E [Y ])] .

(12.1)

Making use of the definition and properties of expected value one can see that Cov (X, Y ) = E [XY − Y E [X] − XE [Y ] + E [X] E [Y ]]

= E [XY ] − E [Y ] E [X] − E [X] E [Y ] + E [X] E [Y ]

= E [XY ] − E [X] E [Y ] ,

(12.2)

where Eq. (12.2) is generally more convenient to use than the expression in the right-hand side of Eq. (12.1). Example 12.1 Table 12.1 lists the heights and arm spans of a sample of 20 children [Shodor (2007)]. This data will be used to illustrate the concept of covariance. Let X represent height and Y represent arm span for each child. The reader can readily calculate that E [X] = 147.15 cm and E [Y ] = 142.70 cm. The expected value of the pairwise product of height and arm span is E [XY ] = 21013.8. Thus the covariance of height and arm span for this sample is Cov (X, Y ) = 15.445. Note that the covariance is positive, indicating that, in general, as height increases so does arm span. From the definition of covariance several properties of this concept follow almost immediately. If X and Y are independent random variables then E [XY ] = E [X] E [Y ] by Theorem 2.5 and thus the covariance of independent random variables is zero. The following set of relationships can also be established. Theorem 12.1 Suppose X, Y , and Z are random variables, then the following statements are true:

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

273

Optimizing Portfolios

(1) Cov (X, X) = V (X), (2) Cov (X, Y ) = Cov (Y, X), (3) Cov (X + Y, Z) = Cov (X, Z) + Cov (Y, Z). The proofs of the first two statements are left for the reader as exercises. Justification for statement 3 is given below. Proof.

Let X, Y , and Z be random variables, then

Cov (X + Y, Z) = E [(X + Y )Z] − E [X + Y ] E [Z]

= E [XZ] + E [Y Z] − E [X] E [Z] − E [Y ] E [Z]

= E [XZ] − E [X] E [Z] + E [Y Z] − E [Y ] E [Z]

= Cov (X, Z) + Cov (Y, Z) .

 The third statement of Theorem 12.1 can be generalized as in the following corollary. Corollary 12.1 Suppose {X1 , X2 , . . . , Xn } and {Y1 , Y2 , . . . , Ym } are sets of random variables where n, m ≥ 1, then Cov

n X

Xi ,

i=1

m X

Yi

i=1

!

=

n X m X

Cov (Xi , Yj ) .

(12.3)

i=1 j=1

Proof. We begin by demonstrating the result when m = 1. The corollary holds trivially when n = 1 and follows from the third statement of Theorem 12.1 when n = 2. Suppose the claim holds when n ≤ k where k ∈ N. Let {X1 , X2 , . . . , Xk , Xk+1 } be random variables. Cov

k+1 X i=1

Xi , Y1

!

= Cov

k X

Xi , Y1

i=1

=

k X

!

+ Cov (Xk+1 , Y1 )

Cov (Xi , Y1 ) + Cov (Xk+1 , Y1 )

i=1

=

k+1 X

Cov (Xi , Y1 )

i=1

Therefore by induction we may show that the result is true for any finite, integer value of n (at least when m = 1). When m is an integer larger than

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

274

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

1 we can argue that     n m n m X X X X Cov  Xi , Yj  = Cov Xi , Yj  i=1

j=1

i=1

=

n X i=1

= =

(shown above)

j=1



Cov 

n X m X

i=1 j=1 n X m X

m X j=1



Yj , Xi 

(by Theorem 12.1 (2))

Cov (Yj , Xi ) Cov (Xi , Yj ) .

i=1 j=1

Therefore Eq. (12.3) holds for all finite, positive integer values of m and n.  Yet another corollary follows from Corollary 12.1. This corollary generalizes statement 1 of Theorem 12.1. Corollary 12.2

If {X1 , X2 , . . . , Xn } are random variables then ! n n n X n X X X V Xi = V (Xi ) + Cov (Xi , Xj ) . i=1

Proof. Let Y = orem 12.1,

i=1

Pn

i=1

Xi , then according to the first statement of The-

V (Y ) = Cov (Y, Y )   ! n n n X X X V Xi = Cov  Xi , Xj  i=1

i=1

= =

n X n X

j=1

Cov (Xi , Xj )

i=1 j=1 n X

Cov (Xi , Xi ) +

i=1

=

(12.4)

i=1 j6=i

n X i=1

14:30:40.

V (Xi ) +

(by Corollary 12.1)

n X n X

Cov (Xi , Xj )

i=1 j6=i

n n X X i=1 j6=i

Cov (Xi , Xj ) .

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Optimizing Portfolios

BC8495/Chp. 12

275

Once more the first statement of Theorem 12.1 was used to reintroduce the variance in the last line of the derivation.  Often a quantity related to covariance is used as a measure of the degree to which large values of a random variable X are associated with large values of another random variable Y . This quantity is known as correlation and is denoted ρ (X, Y ). The correlation of two random variables is defined as Cov (X, Y ) ρ (X, Y ) = p . V (X) V (Y )

(12.5)

The correlation of two random variables can be interpreted as a measure of the degree to which monotonic changes (increases or decreases) in one of the variables are reflected in similar changes (increases with increases and decreases with decreases) in the other variable. The correlation is more than just a simple re-scaling of the covariance. While the covariance of X and Y may numerically be positive, negative, or zero, the correlation always lies in the interval [−1, 1]. Once again we see that independent random variables have a correlation of zero and hence are described as uncorrelated. Theorem 12.2 Suppose X and Y are random variables such that Y = aX + b where a, b ∈ R with a 6= 0. If a > 0 then ρ (X, Y ) = 1, while if a < 0 then ρ (X, Y ) = −1. Proof.

We start by calculating the covariance of X and Y . Cov (X, Y ) = Cov (X, aX + b) = E [X(aX + b)] − E [X] E [aX + b]   = E aX 2 + bX) − E [X] (aE [X] + b)   = aE X 2 + bE [X] − aE [X] E [X] − bE [X]     2 = a E X 2 − E [X]

= aV (X)

The reader should make note of the use of Theorem 2.3. Therefore using the result of exercise (22) from Chapter 2 aV (X) a ρ (X, Y ) = p = 2 |a| V (X) · a V (X)

which is −1 when a < 0 and 1 when a > 0. 14:30:40.



May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

276

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

The converse of Theorem 12.2 is false. A correlation close to unity does not indicate a linear relationship between the random variables. See exercise (6). Before we can bound the correlation in the interval [−1, 1] we will need to prove the following lemma. Lemma 12.1 (Schwarz If X and Y are random variables  2  Inequality)  2 2 then (E [XY ]) ≤ E X E Y .

 2  2  2 Proof. The cases in which E X = 0, E X = ∞, E Y = 0, or  2 E Y = ∞ are left as exercises.  Suppose for the purposes of this proof  that 0 < E X 2 < ∞ and 0 < E Y 2 < ∞. If a and b are real numbers then the following two inequalities hold:       0 ≤ E (aX + bY )2 = a2 E X 2 + 2abE [XY ] + b2 E Y 2       0 ≤ E (aX − bY )2 = a2 E X 2 − 2abE [XY ] + b2 E Y 2

    If we let a2 = E Y 2 and b2 = E X 2 then the first inequality above yields

p         0 ≤ E Y 2 E X 2 + 2 E [Y 2 ] E [X 2 ]E [XY ] + E X 2 E Y 2 p     −2E X 2 E Y 2 ≤ 2 E [Y 2 ] E [X 2 ]E [XY ] p − E [X 2 ] E [Y 2 ] ≤ E [XY ] . By a similar set of steps the second inequality produces E [XY ] ≤ Therefore, since

p

E [X 2 ] E [Y 2 ].

p p − E [X 2 ] E [Y 2 ] ≤ E [XY ] ≤ E [X 2 ] E [Y 2 ]     2 (E [XY ]) ≤ E X 2 E Y 2 . 2

2

Occasionally (E [XY ]) is written as E [XY ] or even as E2 [XY ]. A careful reading of the expression will avoid any possible confusion.  Now we are in a position to prove the following theorem. Theorem 12.3 If X and Y are random variables then −1 ≤ ρ (X, Y ) ≤ 1.

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

277

Optimizing Portfolios

Proof.

BC8495/Chp. 12

Consider the covariance of X and Y . 2

2

(Cov (X, Y )) = (E [(X − E [X])(Y − E [Y ])])     ≤ E (X − E [X])2 E (Y − E [Y ])2

(by Lemma 12.1)

= V (X) V (Y ) p Thus |Cov (X, Y ) | ≤ V (X) V (Y ), which is equivalent to the inequality, Cov (X, Y ) ≤1 −1 ≤ p V (X) V (Y ) −1 ≤ ρ (X, Y ) ≤ 1,

which follows from the definition of correlation.



Example 12.2 Referring to the data in Table 12.1 the correlation between height and arm span is calculated as 15.445 Cov (X, Y ) ρ (X, Y ) = p =p ≈ 0.870948. V (X) V (Y ) (18.6605)(16.8526)

For this data set there seems to be an especially strong linear relationship between height and weight. Figure 12.1 illustrates this as well. Arm span HcmL 148 146 144 142 140 138 136 140

Fig. 12.1

142

144

146

148

150

152

Height HcmL

A scatter plot of height versus arm span for a sample of twenty children.

We end this section with a more financial application of the concepts of covariance and correlation. A portfolio may consist of n different investments. Suppose that an investor may invest an amount wi for one time period and at the end of the time period, the investment will return wealth

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

278

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

in the amount of wi Xi . In Sec. 1.6 the concept of rate of return was defined to be the interest rate equivalent to this growth in value. For the ith investment the rate of return will be determined by wi (1 + Ri ) = wi Xi

which implies

Ri = Xi − 1.

Thus the total accumulated wealth after one time period from all n investPn ments is W = i=1 wi (1 + Ri ). If the rate of return for each of the n investments is treated as a random variable, then " n # n n X X X E [W ] = E wi (1 + Ri ) = wi + wi E [Ri ] . i=1

i=1

i=1

The variance in the accumulated wealth is, according to Eq. (12.4), ! n X V (W ) = V wi (1 + Ri ) i=1

=V

n X i=1

=

n X

wi Ri

!

V (wi Ri ) +

i=1

=

n X

wi2 V (Ri ) +

i=1

n X X

i=1 j6=i n X X

Cov (wi Ri , wj Rj ) wi wj Cov (Ri , Rj ) .

i=1 j6=i

Later sections will expand on this idea and develop a method of selecting a portfolio of investments which minimizes the variance in the wealth generated. Minimizing the variance in the rate of return on a portfolio of investments makes the return more “predictable”. Remember that the smaller the variance of a random variable, the smaller the spread about the expected value of that variable. Before concluding this section we will extend our understanding of covariance and lognormal random variables to derive a technical result which will be needed in Sec. 12.6. Lemma 12.2 If X is a lognormal random variable with drift parameter µ and volatility σ and K > 0 is a constant then      Cov X, (X − K)+ = E X(X − K)+ − E [X] E (X − K)+ , 14:30:40.

(12.6)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Optimizing Portfolios

where   2 2 E X(X − K)+ = e2(µ+σ ) Φ (w + 2σ) − Keµ+σ /2 Φ (w + σ) ,

BC8495/Chp. 12

279

(12.7)

with w = (µ − ln K)/σ.

Proof. Equation (12.6) follows from the definition of covariance. The technical portion of the lemma is the derivation of Eq. (12.7). Z ∞   2 2 1 1 E X(X − K)+ = √ x(x − K)+ e−(ln x−µ) /2σ dx x 2πσ 0 Z ∞ 2 2 1 = √ (x − K)e−(ln x−µ) /2σ dx 2πσ K Z ∞ 2 1 = √ (eσz+µ − K)eσz+µ e−z /2 dz 2π (ln K−µ)/σ The last equation is derived by making the substitution σz = ln x − µ. Therefore 2 Z   e2(µ+σ ) ∞ 2 E X(X − K)+ = √ e−(z−2σ) /2 dz 2π (ln K−µ)/σ µ+σ2 /2 Z ∞ 2 Ke e−(z−σ) /2 dz − √ 2π (ln K−µ)/σ   2 µ − ln K = e2(µ+σ ) Φ + 2σ σ   2 µ − ln K − Keµ+σ /2 Φ +σ . σ The proof is complete if we let w = (µ − ln K)/σ. 12.2



Optimal Portfolios

In the previous chapter on hedging, all of the discussion and examples assumed that a hedged position with respect to a particular option could be set up by taking a position in the stock or security underlying the option. However, it may not always be possible to purchase (or sell) sufficient shares of the underlying security (or the option itself) to create the hedged position. In these cases the manager of a portfolio may have to find a surrogate for the security. Upon reflection, a particular financial instrument will be a better or worse surrogate for a security depending on how the values of the security and the other instrument change. If the changes in the two values

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

280

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

are highly correlated then the portfolio manager will have more confidence in using the surrogate. Suppose an investment firm sells a European call option for stock A and wishes to create a Delta neutral portfolio. If sufficient shares of stock A cannot be purchased to create the hedge then the firm can investigate hedging their short position in the option for stock A with a long position in n shares of stock B. If the hedge could be created with stock A in place of stock B then the Delta neutral hedge is achieved when n = ∆ as we saw in Sec. 10.2. The issue confronting the investment firm now is, what should n be when the hedge must be created with stock B? An optimal portfolio is one for which the variance in the value of the portfolio is minimized. The value of the portfolio consisting of a short position on a call option for stock A and a long position of n shares of stock B will be denoted P = CA − nB. Note that CA denotes a European call option on the underlying security A (earlier the notation C a was used to denote an American call option). The variance in the value of the portfolio is   2 V (P) = E (CA − nB)2 − E [CA − nB] = n2 V (B) − 2nCov (CA , B) + V (CA ) .

Thus it is seen that the variance in the value of the portfolio is a quadratic function of the hedging parameter n. Thus the minimum variance is achieved when s Cov (CA , B) V (CA ) n= = ρ (CA , B) . (12.8) V (B) V (B) Thus if the correlation between the values of the option on stock A and stock B were unity then the hedging parameter would be the ratio of the standard deviations in the value of the option on stock A and the value of stock B. A small correlation between CA and B would indicate that stock B is a poor surrogate for stock A. Equation (12.8) also implies that n will decrease as V (B) increases. The variance in the portfolio will be minimized by decreasing the amount of large variance component in the portfolio. The analysis leading up to Eq. (12.8) overlooks several important assumptions about the hedging situation. First the expiry date of the call option did not come into play. If expiry is some finite (as opposed to instantaneously short) time from the moment the hedge is created, then rightfully

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Optimizing Portfolios

BC8495/Chp. 12

281

we should consider the stochastic behavior of B over the interval to expiry. The concept of hedging as described in Chapter 10 is a continuous dynamic process (even though in practice it may be carried out discretely). To illustrate the importance of considering the time horizon of the calculation consider a hedging problem for which we already know the correct answer. Namely consider the calculation of the value of n for the case where the long position in the European call option CA is hedged with a short position in n shares of security A (rather than security B). This time the call option is hedged using the underlying security. We know from the derivation of the Black-Scholes PDE that the n which minimizes the variance of P is the Delta (∆ = ∂CA /∂A) of the call option. Consider the value of n given in Eq. (12.8) with B replaced by A and considering only the change in the value of the call option and the security over an instantaneously short time interval ∆T . According to Eq. (12.8) n=

Cov (∆C, ∆A) . V (∆A)

We have dropped the subscript of the call option since there is only one security relevant to the portfolio. Making use of differentials  Cov ∂C Cov (∆C, ∆A) ∂A ∆A, ∆A = V (∆A) V (∆A) ∂C Cov (∆A, ∆A) = ∂A V (∆A) ∂C n= . ∂A Thus the minimizing n is the familiar hedging ratio ∆ for the call option. The quantity ∂C/∂A can be pulled out of the covariance since it is an instantaneously determined quantity and thus constant. The last sequence of equations rightly should have been a sequence of approximations which become equations by passing to the limit as ∆T → 0. The reader requiring more rigor may review the stochastic calculus introduced in Chapter 5 and then consult exercise (8). The idea of selecting a portfolio by minimizing the variance of its value will be extended in later sections of this chapter. Before returning to minimum variance analysis, we must introduce the concept known as utility and describe its properties.

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

282

12.3

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

Utility Functions

This section will focus on defining and understanding a class of functions which can be used as the basis of rational decision making. Suppose that an investor is faced with the choice of two different investment products. Suppose further that the set of outcomes resulting from investing in either of the products is {C1 , C2 , . . . , Cn }. The reader should think of this set of outcomes as the union of the possible outcomes of investing in the first product and the possible outcomes resulting from investing in the second product. The probability of outcome Ci coming about as the result of investing in the first product will be denoted pi for i = 1, 2, . . . , n. If outcome Ci cannot result from investing in the first product then, of course, pi = 0. For the second product the values of the probabilities will be denoted qi . The investor can rank the outcomes in order of desirability. Without loss of generality suppose the outcomes have been ranked from least to most desirable as C1 ≤ C2 ≤ · · · ≤ Cn . To each of the possible outcomes a utility can be assigned. The utility function u(Ci ) is defined as follows. To start, u(C1 ) = C1 and u(Cn ) = Cn . The values of u(Ci ) for 1 < i < n will be defined by referring to C1 and Cn . Suppose that for each i the investor is given the following choice: participate in a random experiment in which they receive outcome Ci with certainty, or participate in a random experiment where they will receive C1 with probability φi or receive Cn with probability 1 − φi . The expected value of the outcome of the first experiment is Ci and the expected value of the outcome of the second experiment is Ei = φi C1 + (1 − φi )Cn . If φi = 1 then the investor will surely participate in the random experiment with the certain outcome of Ci , since they rank this as a more desirable outcome than C1 . If φi = 0 then the investor will participate in the second random experiment since its expected outcome is Cn . They would not better their result by taking the sure outcome Ci . At some value of φi ∈ [0, 1] the investor will be indifferent to the choice. The utility of Ci , in other words u(Ci ) is defined to be the Ei for which the investor is indifferent to the choice of experiment. Utility functions are specific to individual investors, like personality traits. In general a utility function is an increasing function. Most rational investors assign greater utility to more preferable outcomes. It is assumed

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

283

Optimizing Portfolios

that the value of φi at which the investor is indifferent to the choice of experiment is unique. The rational investor, once having decided to participate in the second random experiment, will not at a higher expected value decide to once again accept the certain result. Thus the utility function is well-defined. The utility function provides a means by which two outcomes of unequal desirability may be compared. The utility of outcome Ci is equivalent to receiving C1 with probability φi or receiving Cn with probability 1−φi . The reader should beware that there are multiple notions of probability at work here. The utility function was defined in terms of a person’s preference for receiving outcome Ci with certainty or willingness to participate in a specific type of random experiment. However, the probability of the occurrence of outcome Ci is not u(Ci ). Recall we have assumed that P (Ci ) = pi or qi depending on which of two investment products the hypothetical investor chooses. Returning to the original task of deciding between two different investment products, suppose the investor decides to invest in the first product. For the first investment product the expected value of the utility function is then E [u(p)] =

n X

pi u(Ci )

i=1

= C1

n X

pi φi + Cn

i=1

n X i=1

pi (1 − φi ).

Thus the expected utility of the first investment is equivalent to the expected value of a simple random experiment in which the investor will P receive the least desirable outcome C1 with probability ni=1 pi φi and the Pn most desirable outcome Cn with probability i=1 pi (1 − φi ). Similarly the expected value of the utility function for the second investment product is found to be E [u(q)] =

n X

qi u(Ci ).

i=1

Therefore the investor will choose the first product whenever n X i=1

pi u(Ci ) >

n X

qi u(Ci ),

i=1

and otherwise will choose the second product.

14:30:40.

(12.9)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

284

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

In this section we have spoken of outcomes or consequences Ci of making an investment decision. More concretely these outcomes can be thought of as receiving differing amounts of money for an investment (some of which may be negative). Thus in general the utility function denoted u(x) is the investor’s utility of receiving an amount x. In the remainder of this section we will explore categories and properties of utility functions. A general property of utility functions is that the amount of extra utility that an investor experiences when x is increased to x+∆x is non-increasing. In other words u(x + ∆x) − u(x) is a non-increasing function of x. This property is illustrated in Fig. 12.2. We will describe a function f (t) as concave on an open interval (a, b) if for every x, y ∈ (a, b) and every λ ∈ [0, 1] we have λf (x) + (1 − λ)f (y) ≤ f (λx + (1 − λ)y).

(12.10)

uHtL

uHx+DxL

uHxL x Fig. 12.2

x+Dx

t

The extra utility received, u(x + ∆x) − u(x), is a decreasing function of x.

Graphically this may be interpreted as meaning that all the secant lines lie below the graph of f (t). See Fig. 12.3. A utility function u(t) obeying inequality (12.10) will also be called concave. An often repeated statement about investing is that an investor may receive greater rewards by taking greater risks. However, for most rational investors there is a level of reward beyond which, in order to reap greater reward, they are unwilling to accept the higher level of risk. While a casino gambler may be willing to roll the

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

285

Optimizing Portfolios

uHtL

uHyL

uHxL t x

y

Fig. 12.3 For concave functions the secant line parametrized by λ ∈ [0, 1] will lie below the corresponding point on the graph of the function.

dice for a chance to double an investment of $20, it would be the rare gambler who would be willing to roll the dice in order to double $20, 000, 000. The utility of that extra reward is not as high as the initial reward. These are gamblers and investors who will avoid games or investments for which the risk is (in their belief) too high even if the potential rewards are commensurately high. An investor whose utility function is concave is said to be risk-averse. An investor with a linear utility function of the form u(x) = ax + b with a > 0 is said to be risk-neutral. An investor whose utility function increases more rapidly as the reward increases is said to be risk-loving. The following result can be demonstrated regarding concave functions. Theorem 12.4 If f ∈ C 2 (a, b) then f is concave on (a, b) if and only if f 00 (t) ≤ 0 for a < t < b. Proof. If f is concave on (a, b) then by definition f satisfies inequality (12.10). Let x, y ∈ (a, b). Without loss of generality we may assume x < y. If w = λx + (1 − λ)y and if 0 < λ < 1 then a < x < w < y < b. Inequality (12.10) is then equivalent to (1 − λ) [f (y) − f (w)] ≤ λ [f (w) − f (x)] . 14:30:40.

(12.11)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

286

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

By the definition of w, w−x and y−x y−w . λ= y−x

1−λ =

(12.12) (12.13)

Substituting Eqs. (12.12) and (12.13) into inequality (12.11) and rearranging terms produces another inequality which is equivalent to the definition of a concave function stated in inequality (12.10). f (y) − f (w) f (w) − f (x) − ≤0 y−w w−x

(12.14)

Applying the Mean Value Theorem to each of the difference quotients of inequality (12.14) implies that for some α and β satisfying with x < α < w < β < y, f 0 (β) − f 0 (α) ≤ 0. Using the Mean Value Theorem once more proves that for some t with α E [u(Y )] . Example 12.4 “investments”:

An investor must choose between the following two

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

289

Optimizing Portfolios

A: Flip a fair coin, if the coin lands heads up the investor receives $10, otherwise they receive nothing. B: Receive an amount $M with certainty. The investor is risk-averse with a utility function u(x) = x − x2 /25. The rational investor will select the investment with the greater expected utility. The expected utility for investment A is   1 102 1 1 u(10) + u(0) = 10 − = 3. 2 2 2 25 The expected utility for B is u(M ) = M − M 2 /25. Thus the investor will choose the coin flip whenever 3>M−

M2 25

M 2 − 25M + 75 > 0 √ 25 − 5 13 > M. 2 Thus investment A is preferable to B whenever M < $3.49. The astute reader will note that the quadratic inequality solved above is satisfied for M in the sets defined by the relationships √ √ 25 − 5 13 25 + 5 13 M< ≈ 3.49 or M > ≈ 21.51. 2 2 Mathematically this implies the investor would prefer investment A even if the certain amount of investment B is greater than $21.51. A wise investor would never accept a chance at a maximum payoff of $10 if they can receive a guaranteed amount of $21.51 or more. This type of situation leads us to adopt the following logical convention. If M < 3.49 the investor of the previous example will choose investment A, while for M ≥ $3.49 the investor will choose the investment B. This example illustrates the concept known as the certainty equivalent which is defined as the minimum value C of a random variable X at which u(C) = E [u(X)]. Example 12.5 An investor wishes to find the certainty equivalent C for the following investment choice: A: Flip a fair coin, if the coin lands heads up the investor receives 0 < X ≤ 10, otherwise they receive 0 < Y < X. 14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

290

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

B: Receive an amount C with certainty. Assuming the investor’s utility function is the same as in the previous example, the certainty equivalent and payoffs of investment A must satisfy the following equation.   C2 1 X2 Y2 C− = X− +Y − 25 2 25 25 "  2 2  2 # 25 1 25 25 C− = X− + Y − 2 2 2 2 s 2  2 1 25 25 25 −√ + Y − C= X− 2 2 2 2 Again note that we choose the certainty equivalent to be the smallest value of C satisfying the equation. The design of investment A specifies that 0 < Y < X < 10. Thus the certainty equivalent can be thought of as a surface plotted over the triangular region bounded by 0 < X < 10 with 0 < Y < X. Figure 12.4 illustrates the certainty equivalent as a function of X and Y .

12.5

Portfolio Selection

In the previous sections we have treated the investor’s choice between two different investments as a binary, all-or-nothing decision. In reality an investor may choose to split investment capital between two (or more) investments. In this section we will explore the portfolio selection problem, the task of allocating funds in an optimal fashion among several investment options. We begin with a simple example. Suppose an investor has a total of x amount of capital to invest. Assuming that the investor may use any proportion of this capital, let α ∈ [0, 1] be the proportion they invest. The investment is structured such that an allocation of αx will earn αx with probability p and lose αx with probability 1 − p. Thus for an investment of αx the random variable representing the investor’s financial position after the conclusion of the investment is  x(1 + α) with probability p, X= x(1 − α) with probability 1 − p. 14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

291

Optimizing Portfolios

X 2.5

BC8495/Chp. 12

0

5 7.5 10 10

7.5

C

5

2.5

0 0

2.5

5 Y

7.5 10

Fig. 12.4 The certainty equivalent in this example is a section of an ellipsoid. The domain of interest in the XY -plane is drawn as a shadow below the surface.

The allocation proportion α will be optimal when the expected value of the utility is maximized. E [u(X)] = pu(x(1 + α)) + (1 − p)u(x(1 − α)) d E [u(X)] = pxu0 (x(1 + α)) − x(1 − p)u0 (x(1 − α)) dα 0 = pu0 (x(1 + α)) − (1 − p)u0 (x(1 − α)) The critical value of α which solves the last equation above will correspond to a maximum for the expected value of the utility as long as the utility function is concave. Example 12.6 If u(x) = ln x, then the expected value of the investor’s utility is maximized when 1−p p − = 0, 1+α 1−α 14:30:40.

which implies

α = 2p − 1.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

292

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

Notice that by the choice of utility function, the total amount of capital to invest, x, becomes irrelevant when maximizing utility. For the sake of concreteness we may assume x = 1 and then multiply by the appropriate scaling factor for other total amounts. If p > 1/2 the investor should allocate 100(2p − 1)% of their capital. If 0 ≤ p ≤ 1/2, then E [u(X)] is maximized when α = 0, i.e. when no investment is made. A curve depicting the expected value of the investor’s utility as a function of p is presented in Fig. 12.5.

E@uHXLD 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.2

0.4

0.6

0.8

1

p

Fig. 12.5 The expected value of utility for a risk-averse investor allocating 100(2p − 1)% of their capital.

With this simple example understood, we are now ready to consider a more general situation. Suppose that an infinitely divisible unit amount of capital may be allocated among n different investments. A proportion xi will be allocated to security i for i = 1, 2, . . . , n. The vector notation hx1 , x2 , . . . , xn i will be used to represent the portfolio of proportions of investments. The return from investment i will be denoted Wi for i = 1, 2, . . . , n. The quantity Wi = 1 + Ri where Ri is the rate of return for investment i. The portfolio selection problem is then defined to be that of determining xi for i = 1, 2, . . . , n such that (1) 0 ≤ xi ≤ 1 for i = 1, 2, . . . , n, and Pn (2) i=1 xi = 1, 14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Optimizing Portfolios

BC8495/Chp. 12

293

and which maximizes the investor’s total expected utility E [u(W )], where W =

n X

xi Wi .

i=1

Throughout this section it will be assumed that n is large and that the Wi ’s are not too highly correlated. Under these assumptions the Central Limit Theorem of Sec. 3.5 implies that W is a normally distributed random variable. The allocation of investment funds can be driven by the objective of maximizing the expected value of the investor’s utility function. Suppose an investor’s utility function is given by u(x) = 1 − e−bx where b > 0. The reader is asked in exercise (15) to show that this utility function is concave and monotonically increasing. Assuming that W is a normally distributed random variable then −bW is also normally distributed with E [−bW ] = −bE [W ]

and V (−bW ) = b2 V (W ) .

Consequently,     E [u(W )] = E 1 − e−bW = 1 − E e−bW .

Making the assignment Y = e−bW , then Y is a lognormal random variable (see Sec. 3.6). According to Lemma 3.1,   2 E [Y ] = E e−bW = e−bE[W ]+b V(W )/2 which implies that

E [u(W )] = 1 − e−b(E[W ]−bV(W )/2) . Thus the expected utility is maximized when E [W ] − bV (W ) /2 is maximized. If two different portfolios, hx1 , x2 , . . . , xn i and hy1 , y2 , . . . , yn i, give rise respectively to returns X and Y then the portfolio represented by hx1 , x2 , . . . , xn i will be preferable if E [X] ≥ E [Y ]

and V (X) ≤ V (Y ) .

To extend this discussion and make it more concrete numerically, suppose b = 0.005 and that an investor wishes to choose among an infinite number of different portfolios which can be created by investing in two securities denoted A and B. The investor has $100 to invest and will allocate

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

294

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

y dollars to security A and 100−y dollars to B. The following table summarizes what is known about the rates of return on the hypothetical securities A and B. Security E [rate of return] p V (rate of return)

A 0.16 0.20

B 0.18 0.24

Assume that the correlation between the rates of return is ρ = −0.35. Once again W will represent the wealth returned. E [W ] = 100 + 0.16y + 0.18(100 − y) = 118 − 0.02y

V (W ) = y 2 (0.20)2 + (100 − y)2 (0.24)2 + 2y(100 − y)(0.20)(0.24)(−0.35) = 0.04y 2 + 0.0576(100 − y)2 − 0.0336y(100 − y)

Since the optimal portfolio is the one which maximizes b E [W ] − V (W ) 2 = 118 − 0.02y − 0.0025(0.04y 2 + 0.0576(100 − y)2 − 0.0336y(100 − y)) = −0.000328y 2 + 0.0172y + 116.56.

This occurs when y ≈ 26.2195. For this choice of y E [W ] ≈ 117.476

V (W ) ≈ 276.049

E [u(W )] ≈ 0.442296. In the next section we will generalize and extend these ideas to situations in which investment capital is partitioned in n ways. 12.6

Minimum Variance Analysis

Building on the ideas contained in the previous section we consider the case of an investor allocating a portion of their capital between two potential investments. Without loss of generality it is assumed that the investor has an infinitely divisible unit of capital to invest and will invest a fraction α ∈ [0, 1] in one security and 1 − α in a second security. The rates of return of the two securities are respectively the random variables R1 and R2 . The variances of these rates of return will be denoted σ12 and σ22 . The covariance

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

295

Optimizing Portfolios

in the rates of return is assumed to be c. Thus the variance in the wealth W returned from the investments is V (W ) = α2 σ12 + (1 − α)2 σ22 + 2cα(1 − α).

(12.17)

One measure of an optimal portfolio of investments would be the one with the least variance in the returned wealth. Differentiating V (W ) with respect to α and solving for the critical value of α produces  0 = 2 ασ12 − (1 − α)σ22 + c(1 − 2α) α∗ =

σ12

σ22 − c . + σ22 − 2c

The denominator of α∗ is the covariance of R1 − R2 with itself and thus is non-negative. In order for the critical value of α to be defined, it is assumed that σ12 + σ22 > 2c. This condition is also sufficient to guarantee that the variance of the wealth in Eq. (12.17) will have a global minimum. The parameter α∗ must also fall within the closed interval [0, 1]. This occurs whenever c ≤ min{σ12 , σ22 }. If c ≤ 0 then this condition is always satisfied. According to the Extreme Value Theorem [Stewart (1999)], the minimum variance will occur when α = 0, α = 1, or α = α∗ . V (W )|α=0 = σ22 V (W )|α=1 = σ12 V (W )|α=α∗ =

σ12 σ22 − c2 + σ22 − 2c

σ12

Consider the special case when the rates of return of the two investments are uncorrelated. Under this assumption c = 0 and the critical value of α at which the variance in the returned wealth is minimized reduces to σ2 α = 2 2 2 = σ1 + σ2 ∗

1 σ12 1 σ12

+

1 σ22

.

There is an appealing simplicity and symmetry to this critical value. This is not a coincidence, it is in fact a pattern seen in the more general case of allocating a unit of investment among n different potential investments.

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

296

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

Theorem 12.7 Suppose that 0 ≤ αi ≤ 1 will be invested in security i for i = 1, 2, . . . , n subject to the constraint that α1 + α2 + · · · + αn = 1. Suppose the rate of return of security i is a random variable Ri and that all the rates of return are mutually uncorrelated. The optimal, minimum variance portfolio described by the allocation vector hα∗1 , α∗2 , . . . , α∗n i, is the one for which α∗i

=

where σi2 = V (Ri ).

1 σi2 Pn 1 j=1 σj2

for i = 1, 2, . . . , n,

Proof. Since the rates of return are uncorrelated then the variance in the returned wealth W is V (W ) =

n X

α2i σi2 ,

i=1

P and is subject to the constraint that 1 = ni=1 αi . Minimizing the variance subject to the constraint is accomplished by use of Lagrange Multipliers [Stewart (1999)]. This technique states that V (W ) will be optimized at one of the solutions (α1 , α2 , . . . , αn , λ) of the set of simultaneous equations: ! ! n n X X 2 2 ∇ αi σi = λ∇ αi i=1

i=1

n X

αi = 1.

i=1

The symbol ∇ denotes the gradient operator. These equations are equivalent to respectively: 2αi σi2 = λ for i = 1, 2, . . . , n, and n X αi = 1. i=1

Solving for αi in the first equation and substituting into the second equation determines that λ = Pn

2

1 j=1 σj2

14:30:40.

.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

297

Optimizing Portfolios

Substituting this expression for λ into the first equation yields αi =

1 σi2 Pn 1 j=1 σj2

for i = 1, 2, . . . , n.

The reader can confirm that for each i, αi ∈ [0, 1].



The previous discussion can be generalized yet again to the situation in which the portfolio of securities is financed with borrowed capital. Let w = hw1 , w2 , . . . , wn i represent the portfolio of investments. As before, Ri will represent the rate of return on investing wi in the ith security. For the sake of simplicity we will assume this is a one-period model in which investments are purchased by borrowing money which must be paid back at simple interest rate r. The net wealth generated by the portfolio financed with borrowed capital after one time period is R(w) =

n X i=1

wi (1 + Ri ) − (1 + r)

n X

wi =

i=1

n X i=1

wi (Ri − r).

(12.18)

The expected value of the net wealth generated by the portfolio and the variance in the net wealth are functions of the vector w and are defined to be respectively, r(w) = E [R(w)]

(12.19)

2

(12.20)

σ (w) = V (R(w)) .

In this situation, the borrowed amounts w1 , w2 , . . . , wn do not have to sum to unity, or to any other prescribed value. Lemma 12.3 Assuming the rates of return on the securities are uncorrelated, the optimal portfolio generating an expected unit amount of net wealth with the minimum variance in the net wealth is * + r1 −r r2 −r rn −r w∗ =

σ12 (rj −r)2 j=1 σj2

Pn

σ22 (rj −r)2 j=1 σj2

, Pn

2 σn (rj −r)2 j=1 σj2

, . . . , Pn

(12.21)

where ri = E [Ri ] and σi2 = V (Ri ) for i = 1, 2, . . . , n. The proof of this lemma is similar to the proof of Theorem 12.7 and is left as an exercise. The existence of the optimal portfolio w∗ provided by Lemma 12.3 will be used in the following result known as the Portfolio Separation Theorem. The name is suggestive of the result which indicates to an

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

298

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

investor how portions of a portfolio should be invested so as to minimize the variance in the wealth generated. Theorem 12.8 (Portfolio Separation Theorem) If b is any positive scalar, the variance of all portfolios with expected wealth generated equal to b is minimized by portfolio bw∗ where w∗ is described in Eq. (12.21). Proof.

Suppose x is a portfolio for which r(x) = b, then   1 1 r(x) = r x (by exercise (18)) b b = 1.

Thus we see that the portfolio w∗ ,

1 bx

is a portfolio with unit expected rate of return. For

σ 2 (bw∗ ) = b2 σ 2 (w∗ )   1 2 2 ≤b σ x b = σ 2 (x)

(by Lemma 12.3)

(by exercise (18)).



As a consequence of the Portfolio Separation Theorem all expected wealths generated by portfolios can be normalized to unity when determining the optimal portfolio. As a special case of the preceding discussion the investor may wish to divide their portfolio between a risk-free investment such as a savings account and a “risky” investment such as a security. The risk-free interest rate will be symbolized by rf . By assumption the rate of return on the risk-free investment is guaranteed, i.e. has a variance of zero. If the rate of return of the security is RS and x ∈ [0, 1] is invested in the security while 1−x is placed in the risk-free investment, the rate of return on the portfolio is R = (1 − x)rf + xRS , which implies the expected rate of return for this portfolio is E [R] = (1 − x)E [rf ] + xE [RS ] = (1 − x)rf + xrS

r = rf + (rS − rf )x.

(12.22)

To achieve a more compact notation we have set r = E [R] and rS = E [RS ]. Since the risk-free return is a constant, its expected value is this same

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Optimizing Portfolios

BC8495/Chp. 12

299

constant value. The expected rate of return on the portfolio is a linear function of x. The line has its r-intercept at rf and slope rS − rf . Thus if the expected rate of return of the security exceeds the risk-free rate of return, the slope of the line is positive and the expected rate of return of the portfolio increases with x, otherwise it decreases. The variance in the rate of return on the portfolio is V (R) = (1 − x)2 V (rf ) + x2 V (RS ) + 2x(1 − x)Cov (rf , RS ) σ 2 = x2 σS2

(12.23)

where V (R) = σ 2 and V (RS ) = σS2 . The reader will note that since the rate of return on the risk-free investment is constant, its variance is zero and the covariance with the rate of return on the security also vanishes. The standard deviation of the return on the portfolio is also an increasing linear function of x. Equation (12.23) can be used to eliminate the parameter x from Eq. (12.22) and to derive a formula relating the expected rate of return and the standard deviation of the rate of return. r = rf +

rS − rf σ σS

Once again, if the expected rate of return on the security exceeds the rate of return of the risk-free investment, the expected rate of return of the portfolio increases with the standard deviation in the return on the portfolio. Thus an investor willing to accept a wider variation in the return on funds invested can expect a higher rate of return. If the idealized single security of the previous discussion is replaced with an investment spread evenly throughout the securities market then the same form of linear relationship is determined, but the symbols rM and σM will be used in place of rS and σS respectively. This is known as the equation of the capital market line. r = rf +

rM − rf σ σM

(12.24)

The capital market line is illustrated in Fig. 12.6. Equation (12.24) can also be written as r − rf rM − rf = σ σM which can be interpreted as saying that the excess rate of return (above the risk-free rate) normalized by the standard deviation in the rate of return

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

300

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

for an investment spread evenly throughout the market is (naturally) the same as the normalized excess rate of return on the entire market.

r

rM rf

Σ ΣM Fig. 12.6 The capital market line illustrates the linear relationship between the standard deviation in the return on a portfolio and the return on the portfolio.

Before moving on, the reader should reflect on the σr-plane. For a fixed level of expected return, risk-adverse investors will want to minimize the variance (or standard deviation) in the rate of return. A fixed level of return corresponds to a horizontal line in the σr-plane. Thus a risk-adverse investor would prefer a portfolio for which the standard deviation in the rate of return is as far to the left on the horizontal line as is feasible. The phrase “as is feasible” is used because it may not be possible to achieve a given rate of return with zero standard deviation. Referring to the illustration of the capital market line in Fig. 12.6, for a rate of return above rf it is not feasible to have zero variance in the rate of return. Similarly, for a fixed standard deviation in the rate of return, an investor should prefer the highest expected return. Thus along a vertical line in the σr-plane an investor should prefer a portfolio as high on the line as is feasible. Again, for a fixed σ not all levels of r are achievable. These ideas could be extended a great deal and the interested reader is encouraged to consult [Luenberger (1998)]. Finally we arrive at a discussion of the main topic of this chapter, the Capital Asset Pricing Model (CAPM). This mathematical model will relate the return on the investment in an individual security to the return

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

301

Optimizing Portfolios

on the entire market. In this way an investor can weigh the return on the investment against the risk involved in the investment. The expression Ri will be the return on investment i and RM will denote the return on the entire market. An investor will invest a portion x of his wealth in investment i and the remainder 1−x in the market. The following calculations will seem familiar to the reader. They are similar to those calculations done when determining the optimal allocation of a portfolio between two securities. This time the second security is the entire market. The rate of return on the portfolio is given by R = xRi + (1 − x)RM , which implies the expected rate of return on the portfolio and the variance in the expected rate of return are respectively, E [R] = xE [Ri ] + (1 − x)E [RM ]

V (R) = x2 V (Ri ) + (1 − x)2 V (RM ) + 2x(1 − x)Cov (Ri , RM ) . Without further information we cannot assume that the rate of return for investment i is uncorrelated from the rate of return for the market. To simplify the notation we will make the following assignments: r = E [R], 2 ri = E [Ri ], rM = E [RM ], σ 2 = V (R), σi2 = V (Ri ), and σM = V (RM ). Furthermore we will replace Cov (Ri , RM ) by its slightly more compact equivalent ρi,M σi σM .1 Thus the equations above become r = xri + (1 − x)rM 2

σ =

x2 σi2

+ (1 −

2 x)2 σM

(12.25) + 2x(1 − x)ρi,M σi σM .

(12.26)

Equations (12.25) and (12.26) can be thought of as the parametric form of a curve in the σr-plane. When x = 0, corresponding to the situation in which the entire portfolio is invested in the market, the parametric curve intersects the capital market line. In fact the parametric curve must be tangent to the capital market line at x = 0. If the intersection were transverse rather than tangential, then for some value of x near zero the parametric curve would lie above the capital market line and hence represent an infeasible expected return. See Fig. 12.7. Thus we can derive an equation relating ri , rM , and rf . Using Eq. (12.24) to obtain the slope of the capital market 1 This is a small abuse of notation since we normally use ρ (R , R ) to denote the i M correlation of Ri and RM . Here there is little chance of confusion.

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

302

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

r

Σ Fig. 12.7 The solid line shows once again the capital market line. The solid curve tangent to the capital market line represents feasible allocations of a portfolio. The dashed curve is infeasible since a portion of it lies above the capital market line, and hence would have a greater return with lower risk than the entire market.

line we have rM − rf dr = σM dσ x=0 dr dx x=0 = dσ dx x=0

=

ri − rM ρi,M σi σM − σM

σM

.

Rewriting the last equation in the form ri − rf rM − rf = ρi,M σi σM allows us to interpret the equation as saying that the normalized excess rate or return on investment i is proportional to the normalized excess rate of return on the market, where the proportionality constant is the correlation in the rates of return on investment i and the market. Solving the last equation for ri − rf yields ri − rf = 14:30:40.

ρi,M σi (rM − rf ). σM

(12.27)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Optimizing Portfolios

BC8495/Chp. 12

303

The expression ρi,M σi /σM is commonly denoted βi and is referred to as the beta of security i. The Capital Asset Pricing Model, or CAPM for short, is the name given to Eq. (12.27). The CAPM can be interpreted as stating that the excess rate of return of security i above the risk-free interest rate is proportional to the excess rate of return of the market above the risk-free interest rate. The proportionality constant is βi . For the case in which the rate of return on security i is uncorrelated with the rate of return on the market, the excess expected return will be zero. When the correlation is positive, excess positive expected return for the market implies excess positive expected return for security i. When the correlation is negative, excess positive expected return for the market will imply excess negative return for security i. Example 12.7 Suppose the risk-free interest rate is 5% per year, the expected rate of return on the market is 9% per year, and the standard deviation in the return on the market is 15% per year. If the covariance in the expected returns on a particular stock and the market is 10% then β=

0.10 = 4.44444. (0.15)2

Therefore the expected rate of return on the stock will be r = 0.05 + 4.44444(0.09 − 0.05) = 0.22777, or expressed as a percentage return, 22.78%. The variance in the rate of return of a portfolio can be thought of as a measure of the risk in the portfolio. For a given level of variance in the rate of return, the investor will expect the highest expected value of the rate of return. Thus once again we turn to Lagrange Multipliers to solve this constrained optimization problem. The maximum and minimum value of expected rate of return will occur when d(σ 2 ) dr =λ dx dx

(12.28)

subject to the constraint expressed by Eq. (12.26). The reader will be asked

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

304

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

to verify in the exercises that the solution to this set of equations is ri − rM λ=± q 2 2 2 2 − 2ρ 2 2 (ρi,M − 1)σi σM + (σi2 + σM i,M σi σM )σ x=

(12.29)

2 σM − ρi,M σi σM 2 − 2ρ σi2 + σM i,M σi σM q 2 + (σ 2 + σ 2 − 2ρ 2 (ρ2i,M − 1)σi2 σM i,M σi σM )σ i M ± . (12.30) 2 − 2ρ σi2 + σM i,M σi σM

Under these conditions the maximum expected rate of return is  2 (ri − rM ) ρi,M σi σM − σM r = rM − (12.31) 2 − 2ρ σi2 + σM i,M σi σM q 2 + (σ 2 + σ 2 − 2ρ 2 (ri − rM ) (ρ2i,M − 1)σi2 σM i,M σi σM )σ i M ∓ . 2 − 2ρ σi2 + σM i,M σi σM In most scenarios the variance in the rate of return for investment i will exceed the variance for the market. Thus we will assume that σi > σM . As two special cases, we will explore the return on the portfolio when the rates of return on the market and investment i are perfectly correlated, i.e. when ρi,M = 1. In this case r = rM −

(ri − rM )(σM ± σ) . σi − σM

If the rate of return on investment i is anti-correlated with the rate of return on the market, then r = rM −

(ri − rM )(−σM ± σ) . σi + σM

Example 12.8 Suppose the expected rate of return for investment i is ri = 0.07 with standard deviation σi = 0.25 and the expected rate of return for the market is rM = 0.05 with standard deviation σM = 0.20. Suppose the rates of return have a correlation of ρi,M = 0.57. If an investor is willing to hold a portfolio with a standard deviation in its rate of return at σ = 0.22 then the maximum rate of return can be found via the method outlined above. According to Eq. (12.31), the maximum rate of return on the portfolio is 0.0650248. This occurs when x = 0.751242 or when slightly more than 75% of investment capital is placed in investment i. The curve depicting r and x as parametric functions of σ is shown in Fig. 12.8.

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

305

Optimizing Portfolios

r 0.1 0.08 0.06 0.04 0.02

0.2

0.4

0.6

0.8

1

x

Fig. 12.8 This curve shows the relationship between the rate of return r, and the proportion of capital invested x.

12.7

Mean-Variance Analysis

This chapter concludes with a discussion of the stochastic process approach to determining the expected rate of return of a portfolio. If necessary, the reader may wish to review Chapter 5 before proceeding. In this section the portfolio will consist of positions in a security and a European call option on the security. The reader should not be confused by the portfolio. The investor will purchase a units of the security and b units of the call option, where either a or b could be positive or negative. For example, the investor might take a short position in the security and hedge this position with a long position in the call option. Likewise, a short position in the call option can be offset with a long position in the security. We will assume the present value of the security is S(0), the value of the option is C, the strike price is K, and the expiry date is T . The risk-free interest rate is r. Suppose the value of the security obeys a Brownian-type motion, stochastic process with drift parameter µ and volatility σ. In other words, dS(t) = µS(t) dt + σS(t) dW (t). The reader will recall that Itˆ o’s Lemma (Lemma 5.4) implies the logarithm

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

306

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

of the security follows a Wiener process of the form: d(ln S(t)) =

  1 µ − σ 2 dt + σ dW (t). 2

At time t = 0 the investor devotes resources in the amount of aS(0) + bC to create the portfolio. At the expiry date T the positions in the portfolio are zeroed out. The present value of any gain or loss is then  R = e−rT aS(T ) + b(S(T ) − K)+ − (aS(0) + bC).

Using the linearity property of the expected value, the expected return is   E [R] = e−rT aE [S(T )] + bE (S(T ) − K)+ − aS(0) − bC.

According to Eqs. (5.28) and (5.29) the expected value and variance of ln S(T )/S(0) are respectively   1 2 E [ln(S(T )/S(0))] = µ − σ T 2

(12.32)

V (ln(S(T )/S(0))) = σ 2 T.

(12.33)

and

Now from the expressions in Eqs. (12.32) and (12.33) and by making use of Theorems 3.1 and 3.4 we have E [ln S(T )] =

  1 µ − σ 2 T + ln S(0) 2

V (ln S(T )) = V (ln S(T )) = σ 2 T.

Thus by Lemma 3.1, E [S(T )] = S(0)eµT . Provided that r < µ the expression e−rT E [S(T )] = S(0)e(−r+µ)T > S(0). In order to determine a closed form expression for E [(S(T ) − K)+ ] we will make use of Corollary 3.2. S(T ) is a lognormal random variable with

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

307

Optimizing Portfolios

√ parameters ln S(0)+(µ−σ 2 /2)T (the drift parameter) and σ T (the volatility parameter). If we let w=

ln(S(0)/K) + (µ − σ 2 /2)T √ σ T

then according to Corollary 3.2,  √    E (S(T ) − K)+ = S(0)eµT Φ w + σ T − KΦ (w) .

Thus the expected value of the rate of return is r = E [R]

(12.34)      √  (µ−r)T −rT µT = aS(0) e − 1 + be S(0)e Φ w + σ T − KΦ (w) − C .

The level sets of the expected return are lines in the ab-plane. The variance in the rate of return can be calculated as  V (R) = V e−rT aS(T ) + b(S(T ) − K)+  = e−2rT V aS(T ) + b(S(T ) − K)+   = e−2rT a2 V (S(T )) + b2 V (S(T ) − K)+  + 2abCov S(T ), (S(T ) − K)+ .

(12.35)

Referring to the second part of Lemma 3.1 we have  2  V (S(T )) = S 2 (0)e2µT eσ T − 1 . According to Corollary 3.4,

 √   2 V (S(T ) − K)+ = S 2 (0)e(2µ+σ )T Φ w + 2σ T  √  − 2KS(0)eµT Φ w + σ T + K 2 Φ (w)   2 √  − S(0)eµT Φ w + σ T − KΦ (w) .

Lemma 12.2 implies that

 Cov S(T ), (S(T ) − K)+   √  = KS(0)eµT Φ (w) − Φ w + σ T  2   √  √  + S 2 (0)e2µT eσ T Φ w + 2σ T − Φ w + σ T . 14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

308

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

Putting these last three results together yields, at long last, the complicated, but complete, expression for V (R). h  2  V (R) = e−2rT a2 S 2 (0)e2µT eσ T − 1 (12.36)     √ √  2 + b2 S 2 (0)e(2µ+σ )T Φ w + 2σ T − 2KS(0)eµT Φ w + σ T  2   √  2 µT + K Φ (w) − S(0)e Φ w + σ T − KΦ (w)  h 2   √  √ i + 2ab S 2 (0)e2µT eσ T Φ w + 2σ T − Φ w + σ T h  √ ii + KS(0)eµT Φ (w) − Φ w + σ T The level sets of V (R) are parabolas in the ab-plane.

Example 12.9 Suppose a share of a security has a current price of $100 while the drift parameter and volatility of the price of the security are respectively 0.08 and 0.21 per annum. The risk free interest rate is 3.5% annually. The value of a European call option for a strike price of $102 with an exercise time of 12 months on the security is $9.075. An investor can use the preceding analysis to determine that the expected rate of return on a portfolio consisting of a position x in the security and a position y in the call option is given by Eq. (12.34) as E [R] = 4.60279x + 3.12945y. The line of zero return is given by the equation y = −1.4708x. Thus at a point in the second or fourth quadrant above this line the expected return will be positive. We look for a point in the second or fourth quadrant since typically x and y must be of opposite sign due to the fact that the investor will take opposite positions in the security and the option. A density plot in Fig. 12.9 shows the return in various subsets of the plane. If the investor decides that a return of E [R] = 2 is desirable they will then select x and y so that the variance in the return is minimized over all portfolios for which E [R] = 2. This is yet another constrained optimization exercise. According to Eq. (12.36) the variance in the return is V (R) = 493.329x2 + 681.025xy + 271.757y 2. Rather than use the method of Lagrange multipliers to find the minimum of V (R) we will take the more elementary approach of solving the equation 2 = 4.60279x + 3.12945y for y and substituting into the expression for

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

309

Optimizing Portfolios

1.0

0.5

y

0.0

- 0.5

- 1.0 - 1.0

- 0.5

0.0

0.5

1.0

x

Fig. 12.9 The expected return of the portfolio is positive above the dashed diagonal line shown in the density plot. The solid line illustrates the locus of points at which the expected return equals 2.

variance. In this case we get the quadratic expression V (R) = 110.995 − 75.6514x + 79.5567x2 which is minimized when x = 0.475456. Under these conditions y = −0.0602088 and the minimum variance is V (R) = 93.0107. It is instructive to understand that constructing a portfolio of minimum variance is different than constructing a hedged portfolio. Recall that a portfolio is Delta neutral if for every option in the portfolio (either a long or short position) ∆ units of the underlying security are held (in the opposite position). Considering a portfolio of the form P = C − aS (with a > 0 for the sake of simplicity) then we may rewrite the variance in the rate of return on this portfolio using Eq. (12.35) as   V (R) = e−2rT a2 V (S(T )) + V (S(T ) − K)+  (12.37) − 2aCov S(T ), (S(T ) − K)+ . 14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

310

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

In the final example of this chapter we will compare the values of a which minimize V (R) and ∆ which produce a Delta neutral portfolio. Example 12.10 we may calculate

Using the same values as those given in Example 12.9 V (R) = 493.329a2 − 681.025a + 271.757

which is minimized when a = 0.690234. In fact the minimum variance is min(V (R)) = 36.7241. According to Eqs. (8.36) and (9.8) the portfolio is Delta neutral when a = ∆ = 0.570391. For this value of a the variance in the rate of return on the portfolio is V (R) = 43.8095 and is, as expected, larger than the minimum variance. 12.8

Exercises

(1) Using the definition of covariance and variance prove the first two statements of Theorem 12.1. (2) Show that for any two random variables X and Y V (X + Y ) = V (X) + V (Y ) + 2Cov (X, Y ) . If X and Y are independent random variables, show this formula reduces to the result mentioned in Theorem 2.7. (3) Fill in the remaining details    of the proof  of Lemma 12.1  for the cases in which E X 2 = 0, E X 2 = ∞, E Y 2 = 0, or E Y 2 = ∞ (4) The data shown in the table below was originally published in the New York Times, Section 3, pg. 1, 05/31/1998. It lists the name of a corporate CEO, the CEO’s corporation, the CEO’s golf handicap, and a rating of the stock of the CEO’s corporation. Determine the covariance and correlation between the CEO’s golf handicap, and the corporation’s stock rating. Name Terrence Murray William T. Esrey Hugh L. McColl Jr. James E. Cayne John R. Stafford John B. McCoy Frank C. Herringer

14:30:40.

Corp. Fleet Financial Sprint Nationsbank Bear Stearns Amer. Home Prod. Banc One Transamerica

Handicap 10.1 10.1 11 12.6 10.9 7.6 10.6

Rating 67 66 64 64 58 58 55

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

311

Optimizing Portfolios

Ralph S. Larsen Paul Hazen Lawrence A. Bossidy Charles R. Shoemate James E. Perrella William P. Stiritz

BC8495/Chp. 12

Johnson&Johnson Wells Fargo Allied Signal Bestfoods Ingersoll-Rand Ralston Purina

16.1 10.9 12.6 17.6 12.8 13

54 54 51 49 49 48

(5) The table below lists the names, heights, and weights of a sample of Playboy centerfolds [Dean et al. (2001)]. Let X represent height and Y represent weight for each woman. Name Allias, Henriette Anderson, Pamela Broady, Eloise Butler, Cher Clark, Julie Sloan, Tiffany Stewart, Liz Taylor, Tiffany Witter, Cherie York, Brittany

Height (in.) 68.5 67 68 67 65 66 67 67 69 66

Weight (lb.) 125 105 125 123 110 120 116 115 117 120

Determine the covariance and correlation between the heights and weights listed. (6) Consider the match paired data in the table below. X Y

0 y

1 1

1 2

1 3

1 4

1 5

1 6

1 7

1 8

1 9

Show that while the functional relationship between X and Y is not linear, as y → −∞, the correlation between X and Y approaches 9/10. (7) Suppose that a company sells a European call option on a security. The standard deviation on the price of the call option is σO = 0.045. The company hedges its position by buying a different security for which the standard deviation in value is σD = 0.037. The correlation between the values of the two instruments is ρ (O, D) = 0.86. What is the optimal ratio of the number of shares of security purchased to the number of options sold so that the variance in the value of the portfolio is minimized?

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

312

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

(8) Show using the concepts of the stochastic calculus that the variance in the change in value of the portfolio P = C − nS, where C is a European call option on the underlying security S, is minimized when n = ∆, the Delta of the call option. (9) Which of the following utility functions are concave on their domain? (a) u(x) = ln x (b) u(x) = (ln x)2 (c) u(x) = tan−1 x (10) The mean of a discrete random variable as defined in Eq. (2.4) is sometimes called the arithmetic mean. The harmonic mean is defined as H= P

1

.

(12.38)

The geometric mean is defined as Y G= X P(X) .

(12.39)

X

P(X) X

X

(11) (12)

(13)

(14)

(15)

Use Jensen’s inequality (12.15) to show that H ≤ G ≤ E [X] and that equality holds only when X1 = X2 = · · · = Xn . You may assume that all the random variables are positive. Verify inequality (12.16) for f (x) = tanh x and g(t) = t. Find the certainty equivalent for the choice between (a) flipping a fair coin and either winning $10 or losing $2, or (b) receiving an amount C with certainty. Assume that your utility function is f (x) = x − x2 /50. Suppose you must choose between receiving an amount C with certainty or playing a game in which a fair die is rolled. If a prime number results you win $15, but if a non-prime number results you lose that amount of money. Assume that your utility function is f (x) = x − x2 /2. Suppose a risk-averse investor with utility function u(x) = ln x will invest a proportion α of their total capital x in an investment which will pay them either 2αx with probability p or nothing with probability 1 − p. The amount of capital not invested will earn interest for one time period at the simple rate r = 11%. What proportion of their capital should the investor allocate if they wish to achieve the maximum expected utility? Consider the function u(x) = 1 − e−bx where b > 0. 14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

313

Optimizing Portfolios

(a) Show that u(x) is concave. (b) Show that 0 ≤ x1 < x2 implies u(x1 ) < u(x2 ).

(16) Suppose an investor whose utility function is u(x) = 1 − e−x/100 has a total of $1000 to invest in two securities. The expected rate of return for the first investment is r1 = 0.08 with a standard deviation in the rate of return of σ1 = 0.03. The expected rate of return for the second investment is r2 = 0.13 with a standard deviation in the rate of return of σ2 = 0.09. The correlation in the rates of return is ρ = −0.26. What is the optimal amount of money to place into each investment? (17) Suppose that an investor will split a unit of wealth between the following securities possessing variances in their rates of return as listed in the table below. Assuming that the rates of return are uncorrelated, determine the proportion of the portfolio which will be allocated to each security so that the variance in the returned wealth is minimized. Security A B C D E

σ2 0.24 0.41 0.27 0.16 0.33

(18) For any real scalar c show that for r(w) and σ 2 (w) as defined in Eqs. (12.19) and (12.20) the following results hold. r(cw) = cr(w) σ 2 (cw) = c2 σ 2 (w) (19) Prove Lemma 12.3. (20) In this exercise we will extend the work that was done in exercise (17) by determining the optimal portfolio which minimizes the variance in the return under the condition that the portfolio is financed with money borrowed at the interest rate of 11%. Security A B C D E

14:30:40.

r 0.13 0.12 0.15 0.11 0.17

σ2 0.24 0.41 0.27 0.16 0.33

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

314

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 12

An Undergraduate Introduction to Financial Mathematics

(21) Verify that the solutions to the set of Eqs. (12.26) and (12.28) are given by the values of λ and x in (12.29) and (12.30). (22) Suppose the risk-free interest rate is 4.75% per year, the expected rate of return on the stock market is 7.65% per year, and the standard deviation in the return on the market is 22% per year. If the covariance in the expected returns on a particular stock and the market is 15%, determine the expected rate of return on the stock. (23) Suppose a share of a security has a current price of $83 while the drift parameter and volatility of the price of the security are respectively 0.13 and 0.25 per annum. The risk free interest rate is 5.5% annually. Find the value of 3-month European call option on the security with a strike price of $86. Use the mean-variance analysis of Sec. 12.7 to determine the positions of the optimal portfolio with an expected return of $10.

14:30:40.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

Chapter 13

American Options

For the most part the preceding chapters have been concerned with European-style options, their properties, and their uses in hedging. For the reader who has come this far, the final chapter will explore the pricing and properties of American-style options. Recall that American-style options may be exercised at any time up to expiry. In this chapter we will see there are important differences between the European- and American-style options as well as nearly paradoxical similarities. We will develop a means of approximating the values of American puts and calls using the binomial approach outlined in Section 8.5. Before tackling the pricing procedure for American-style options, readers should understand the binomial model presented in the earlier section. 13.1

Parity and American Options

In this chapter we will use the superscripts “a” and “e” to distinguish between American and European options respectively. Unless otherwise stated the underlying security is assumed to pay no dividends. Earlier an arbitrage argument was used to show that American and European options, on the same underlying security and with the same strike price and expiry date, obey the following inequalities (see Section 7.1 and exercise (1) in that same chapter). Ce ≤ Ca

and P e ≤ P a

Intuitively these inequalities hold because the American options give the holder all the rights of the European options with the addition of the possibility of early exercise. There are market scenarios under which an American option may be exercised to yield a profit while the equivalent Euro-

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

316

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

SHtL

K

T

t

Fig. 13.1 The shaded region indicates the intervals before expiry in which the price of the underlying security S(t) exceeds the strike price K. An American call option could be exercised profitably in these intervals, but a European call option could not, and would ultimately expire unused.

pean option would expire out of the money. See Figure 13.1. Consider an American call versus a European call on the same underlying security with the same strike price K and expiry time T . If the value of the security S(t) > K + C a for some time 0 ≤ t < T , the American option could be exercised and generate a positive profit. Since t < T the European option could not be exercised. The European option would only generate a positive profit if S(T ) > K + C e and there is no guarantee that will occur. The European put and call prices also obey the Put-Call Parity formula expressed in Eq. (7.1). However, the American options do not obey a PutCall Parity formula. In this section we will develop some bounds on the prices of the American options and explore a relationship between C a and Ce. Assume that the American options are created at time t = 0 when the price of the underlying security is S. The risk-free interest rate is r compounded continuously. If the strike price is K then the values of the American put and call options with identical strike price and expiry T > 0 obey the following inequality. Ca + K ≥ S + P a

(13.1)

To prove this, assume to the contrary that C a + K < S + P a . In this case an investor could short the security, sell the put, and buy the call.

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

American Options

BC8495/Chp. 13

317

This produces a cash flow of S + P a − C a . If this amount is positive it can be invested at the risk-free rate, otherwise it is borrowed at the riskfree rate. If the holder of the American put chooses to exercise the put at time 0 ≤ t ≤ T , the investor can exercise the call option and purchase the security for K. At that time the investor’s balance is (S + P a − C a )ert − K > Kert − K ≥ 0. If the American put expires out of the money, the investor will close their short position in the security at time T by exercising the call option. In this case (S + P a − C a )erT − K > KerT − K > 0. Thus the investor receives a non-negative profit in either case, violating the principle of no arbitrage. Consequently the inequality in (13.1) holds. Now it is also the case that S + P a ≥ C a + Ke−rT .

(13.2)

A proof by contradiction and a no arbitrage argument will establish this inequality as well. Suppose S + P a < C a + Ke−rT . An investor could sell an American call and buy the security and the American put. This generates a cash flow of C a − S − P a at t = 0. If necessary the investor will borrow funds at the risk-free rate r compounded continuously. If the holder of the call decides to exercise it at any time 0 ≤ t ≤ T , the investor may sell the security for the strike price K. Thus at time t the investor’s asset balance is (C a − S − P a )ert + K = (C a + Ke−rt − S − P a )ert

≥ (C a + Ke−rT − S − P a )ert

since r > 0. By assumption S + P a < C a + Ke−rT , so the last expression above is positive. The investor has earned a risk-less positive profit, but this contradicts the no arbitrage assumption. Therefore the inequality in (13.2) is true. By rearranging terms in (13.1) and (13.2) and combining the two inequalities we have proved the following theorem. Theorem 13.1 If the risk-free interest rate is r compounded continuously, if C a and P a are the values of American call and put options re-

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

318

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

spectively both with strike price K and expiry T written on a non-dividendpaying security, and if the value of the underlying security is S, then S − K ≤ C a − P a ≤ S − Ke−rT .

(13.3)

Perhaps one of the most unexpected results governing the value of American-style options is the equality holding between C e and C a . We have seen that C a ≥ C e for American and European call options with the same strike price, expiry time, and underlying security. We have also seen a simulation of an example for which the American call is in the money for a period before expiry and could be exercised to generate a positive profit while the European option would be out of the money at expiry (Figure 13.1). Since the American call, through possible early exercise, may generate a greater profit than its European counterpart, it is surprising that C a = C e . However, the usual type of no-arbitrage proof will establish the equality. Suppose that C a > C e . Since the American call is (supposedly) worth more, an investor could sell the American call and buy a European call with the same strike price K, expiry date T , and underlying security. The net cash flow C a − C e > 0 would be invested at the risk-free rate r. Assuming the interest is compounded continuously, at any time 0 ≤ t ≤ T , the amount due is (C a − C e )ert . If the holder of the American call chooses to exercise the option at some time t ≤ T , the investor may sell short a share of the security for amount K and add the proceeds to the amount invested at the risk-free rate. At time T the investor must close out the short position in the security and may use the European option to do so. Upon settlement of the short position the holdings of the investor are (C a − C e )erT + K(er(T −t) − 1) > 0. If the American option is not exercised, the investor has holdings of (C a − C e )erT > 0 at expiry. A rational investor will only exercise the European option if profit is increased, else the European option will be allowed to expire unused. In either case the investor earns a risk-free positive profit. This completes the proof of the next theorem. Theorem 13.2 If C a and C e are the values of American and European call options respectively on the same underlying security with identical strike prices and expiry times, then Ca = Ce.

14:31:02.

(13.4)

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

American Options

BC8495/Chp. 13

319

Now that we have a means of determining the value of an American call on a non-dividend paying security we may use it to establish a bound on the value of an American call on the same underlying security with the same expiry and strike price. Example 13.1 The price of a security is currently $36, the risk-free interest rate is 5.5% compounded continuously, and the strike price of a sixmonth American call option worth $2.03 is $37. The range of no arbitrage values of a six-month American put on the same security with the same strike price can be found by making use of the inequality in (13.3). 36 − 37 ≤ 2.03 − P a ≤ 36 − 37e−0.055(6/12) 2.03 ≤ P a ≤ 3.03

The American put may be worth more than the European put (all parameters of the option being equal) due to the possibility of early exercise. The preceding discussion assumes the underlying stock pays no dividends. In practice options are frequently written on stocks that do pay dividends. Typically individual stocks pay dividends one to four times per year. Stock indices formed by bundling a large number of stocks may be assumed to pay dividends continuously. Options are sometimes written on foreign currencies which pay a continuously compounded “dividend” in the form of the risk-free interest rate on the currency. The payment of a dividend may also trigger the early exercise of an American call option. Fortunately it is not difficult to generalize Eq. (13.3) to include dividends. Theorem 13.3 If the risk-free interest rate is r compounded continuously, if C a and P a are the values of American call and put options respectively both with strike price K and expiry T written on a security whose value is S, and if PV(div) is the present value of all dividends paid by the security before expiry, then S − K − PV(div) ≤ C a − P a ≤ S − Ke−rT .

(13.5)

Proof. The proof will proceed as before by contradiction. Assuming S − K − PV(div) > C a − P a is equivalent to the inequality C a + K + PV(div) < P a + S. An investor can short the security and sell an American put on the security while purchasing an American call. The investor also invests K +PV(div) in savings at the risk-free interest rate. Since the security has been borrowed

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

320

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

in order to be shorted, it is the responsibility of the investor to pay the dividends to the buyer of the security, hence the need to invest not only the strike price but all the present value of the dividends. During the life of the American put, the investor pays the buyer of the security the dividends due out of this investment of PV(div). The initial cash flow to the investor upon creating this portfolio is P a + S − (C a + K + PV(div)) > 0. If the owner of the American put chooses to exercise at time t ∈ [0, T ], the investor pays them K for the security and receives the security which they use to close out the short position. The investor’s portfolio now is worth C a + Kert − K > 0. If the owner of the American put allows the option to expire, then the investor may exercise the American call at expiry to purchase the security for K and close out their short position. At expiry the investor’s portfolio is worth (S(T ) − K)+ + KerT − S(T ) = KerT − K > 0. In either case the investor receives a positive profit. Thus the inequality on the left of Eq. (13.5) has been established. The inequality on the right is the same as that in Eq. (13.3) and thus the same proof as in Theorem (13.1) holds.  Comparing Eqs. (13.3) and (13.5) we can see that the payment of dividends may widen the separation in the prices of the American call and put options. When no dividend is paid the width of the interval potentially separating the two options is K(1 − e−rT ). Including the payment of dividends the width is increased to K(1 − e−rT ) + PV(div). This increased separation between the prices of the American call and put can be attributed to the increased uncertainty associated with whether the call option will be exercised prior to the dividend date, so as to receive the dividend. In the next section we will develop a binomial model approximation to the value of an American put.

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

American Options

13.2

BC8495/Chp. 13

321

American Puts Valued by a Binomial Model

As earlier in Section 8.5 we will assume for the purposes of pricing an American put on a non-dividend paying security that the • • • • •

strike price of the American put is K, expiry date of the American put is T > 0, price of the security at time t with 0 ≤ t ≤ T is S(t), continuously compounded risk-free interest rate is r, and price of the security follows a geometric Brownian motion with variance σ 2 .

If the interval [0, T ] is divided into n subintervals of length ∆t = T /n where n ∈ N, we will assume that at time k∆t with k = 0, 1, . . . , n − 1 the security price is S(k∆t). At time (k + 1)∆t the price from the previous time step may have increased by a factor u > 1 or decreased by a factor 0 < d < 1. The probability of an increase is p and naturally that of a decrease is 1 − p. The discrete dynamics of the movement of the price of the security should approximate geometric Brownian motion for which the volatility in the security price is σ 2 . Following the same line of reasoning as in Section 8.5 we obtain the following relationships.   q  1 −r∆t 2 )∆t 2 (r+σ2 )∆t −r∆t (r+σ e +e + u= e +e −4 (13.6) 2   q 2 2 1 −r∆t d= e + e(r+σ )∆t − e−r∆t + e(r+σ2 )∆t − 4 (13.7) 2 p=

er∆t − d u−d

(13.8)

These values of the parameters of the discrete model insure that the discrete model is arbitrage-free and that the discrete version of the security has the same expectation and variance as the continuous version. An American put is always worth at least as much as the payoff generated by immediate exercise. Thus P a (t) ≥ (K − S(t))+ for any 0 ≤ t ≤ T . The quantity (K − S(t))+ is called the intrinsic value of the American put. Since we will make frequent reference to the intrinsic value of the option, we will define the function Q(t) = (K − S(t))+ and use it where the intrinsic value is needed. The reader should bear in mind that the intrinsic value represents the payoff of an in the money put option. To develop a pricing formula for the American put, consider a coarse time discretization of the interval [0, T ]. For the moment we will assume the American put can

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

322

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

only be exercised at t = 0 or at t = T . The value of the underlying security would evolve along one of the two branches shown in Fig. 13.2. The corre-

S(T)=u S(0)

p

S(0)

1– p

S(T)=d S(0) Fig. 13.2

A single-step discrete approximation to the evolution of the value of a security.

sponding intrinsic values of the American put are shown in Fig. 13.3. At expiry the value of an American put is its intrinsic value. Thus we always have the relationship P a (T ) = Q(T ), assuming t = T is the expiration date of the option. In this single step model, the American put will be worth the greater of its intrinsic value at t = 0 or the present value of the expected value of the American put at t = T . This can be expressed as    P a (0) = max Q(0), e−rT p(K − uS(0))+ + (1 − p)(K − dS(0))+    = max Q(0), e−rT E (K − S(T ))+ . Consequently if the intrinsic value of the put at time t = 0 is greater than the present value of the expected value of the option at time t = T , the put

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

323

American Options

(K – u S(0))+

p

(K – S(0))+

1–p

(K – d S(0))+ Fig. 13.3 The intrinsic values of an American put option at the discrete times t = 0 and t = T .

could be exercised early. This approach to pricing the American put can be generalized to a multi-step evolution of the security price. Suppose the interval [0, T ] has been partitioned into n subintervals of length ∆t = T /n. The partition consists of the points t0 < t1 < · · · < tn where ti = i∆t. For convenience we will assign Q(∞) = 0, indicating that an expired option has zero value. If i = n then P a (n∆t) = P a (T ) = Q(T ),

(13.9)

i.e. the value of the option is the intrinsic value. If i = n − 1 then the value of the put is defined to be the greater of the intrinsic value at t = (n − 1)∆t and the present value at t = (n − 1)∆t of the expected value of the put at t = n∆t. In other words P a ((n − 1)∆t) = max{Q((n − 1)∆t), e−r∆t E [Q(n∆t)]} =

14:31:02.

max

{e−r(j−n+1)∆t Q(j∆t)}.

j∈{n−1,n,∞}

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

324

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

This relationship may be generalized to define the value of the American put at any time value in the partition of [0, T ]. For 0 ≤ i < n we define P a (i∆t) =

max

{e−r(j−i)∆t E [Q(j∆t)]}.

j∈{i,i+1,...,n,∞}

(13.10)

Therefore, prior to expiry, the value of the American put is the greater of its intrinsic value and the discounted value of its expected value at the next time step. Equations (13.9) and (13.10) constitute a recursive algorithm for pricing an American put option. A binomial lattice of values of the underlying security determines the intrinsic values of the American put. At expiry the value of the put is given by Eq. (13.9). At every time step prior to the expiration of the option, the value of the American put is determined by Eq. (13.10). Before exploring some of the properties of the binomial lattice formula for the price of an American put, we will work through the details of a two-step binomial model. Example 13.2 Suppose the current price of a security is $32, the riskfree interest rate is 10% compounded continuously, and the volatility of Brownian motion for the security is 20%. The price of a two-month American put with a strike price of $34 on the security can be found as outlined below. The length of a time step will be ∆t = 1/12. The parameters u and d given in Eqs. (13.6) and (13.7) and governing the proportional increase and decrease in the price of the security are u ≈ 1.0603 and d ≈ 0.9431. The probability of an increase in the price of the security occurring between time steps is given by the formula in Eq. (13.8). p ≈ 0.5567 The time until expiry will be divided into two single-month time steps and a binomial lattice of security prices will be created. The binomial lattice of security values is shown in Fig. 13.4. Now we may create a corresponding binomial lattice of intrinsic values for the American put. Each node in this lattice is merely the positive part of the payoff generated by the exercise of the put. These intrinsic values are shown in Fig. 13.5. Now we can begin to recursively calculate the values of the American put at the nodes of the binomial lattice. At the leaf (right-most) nodes, the price of the option is its intrinsic value according to Eq. (13.9). Thus

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

325

American Options

35.9798

33.9316

32.

32.

30.1783

28.4604

Fig. 13.4

A two-step discrete approximation to the evolution of the value of a security.

a partial lattice of put values showing only those at expiry is shown in Fig. 13.6. At t = 1/12, for the upper of the two nodes in the binomial lattice at this time step, the value of the American put according to Eq. (13.10) is P a (1/12) = max{(34 − 33.9316)+, e−0.10/12 (p(0) + (1 − p)2)} = max{0.0683805, 0.879248} = 0.879248.

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

326

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

0

0.0683805

2.

2.

3.82166

5.53962

Fig. 13.5 The intrinsic values of an American put with a strike price of $34 on the security shown in Fig. 13.4.

In a similar fashion we may calculate the value of the American put at the lower of the two nodes in the lattice at t = 1/2. The partial binomial lattice of values of the American put for t = 1/12 and t = 2/12 is shown in Fig. 13.7. Finally the value of the American put at t = 0 is P a (0) = max{(34 − 32)+ , e−0.10/12 [p(0.879248) + (1 − p)(3.82166)]} = 2.16551.

The binomial lattice of values of the American put is shown in Fig. 13.8.

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

327

American Options

0

2.

5.53962

Fig. 13.6 The values of an American put with a strike price of $34 at expiry on the security shown in Fig. 13.4.

The next section contains some results describing the properties of the American put option. One simple property is evident from the formulas for the put detailed in Eqs. (13.9) and (13.10), namely the value of the American put is at least as great as the intrinsic value of the option. P a (t) ≥ Q(t) A binomial lattice providing a side-by-side comparison of the value of the American put and the intrinsic value of the option for the previous example is shown in Fig. 13.9.

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

328

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

0

0.879248

2.

3.82166

5.53962

Fig. 13.7 More values of an American put with a strike price of $34 at expiry on the security shown in Fig. 13.4.

13.3

Properties of the Binomial Pricing Formula

The pricing algorithm for an American-style put option given in Eqs. (13.9) and (13.10) is simple to implement in a spreadsheet or computer program. Its simplicity may hide some of the mathematical properties of the value of the option. In this section, brief descriptions and justifications of some of the properties of the put option will be presented. These results will be helpful in deciding the optimal time to exercise an American put option. Lemma 13.1

Suppose the value of the security underlying an American

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

329

American Options

0

0.879248

2.16551

2.

3.82166

5.53962

Fig. 13.8 The values of the American-style puts corresponding to the values of the securities in Fig. 13.4.

put option follows a path through the n-step binomial lattice such that for some i ∈ {0, 1, . . . , n − 1} we have (K − dS(ti ))+ = 0, then Q(ti ) = 0. Proof.

Since d < 1 < u we have dS(ti ) < S(ti ) < uS(ti ) which implies (K − uS(ti )) < (K − S(ti )) < (K − dS(ti )),

from which it follows that (K − uS(ti ))+ ≤ (K − S(ti ))+ ≤ (K − dS(ti ))+ . We have assumed (K − dS(ti ))+ = 0, which implies (K − S(ti ))+ = Q(ti ) = 0. 

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

330

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

0 0

0.879248 0.0683805

2.16551 2.

2. 2.

3.82166 3.82166

5.53962 5.53962

Fig. 13.9 The top number in each box represents the price of the American put option for the previous example while the bottom number is the intrinsic value of the option.

This lemma can be thought of as stating that if the intrinsic value of an American put at the next downward movement of the price of the underlying security would be zero, then the current intrinsic value of the option is zero as well. Likewise the intrinsic value of the American put at higher values of the security price would be zero. The following lemma reveals the effect this has on the value of the option. Lemma 13.2 Suppose the value of the security underlying an American put option follows a path through the n-step binomial lattice such that for some i ∈ {0, 1, . . . , n − 1} we have (K − dS(ti ))+ = 0, then P a (ti ) = 14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

American Options

BC8495/Chp. 13

331

e−r∆t P a (ti+1 ). Proof.

By Lemma 13.1 Q(ti ) = 0 and by Eq. (13.10) P a (ti ) = =

max

{e−r(j−i)∆t E [Q(tj )]

j∈{i,i+1,...,n,∞}

{e−r(j−i)∆t E [Q(tj )]

max

j∈{i+1,i+2,...,n,∞}

= e−r∆t

max

{e−r(j−[i+1])∆t E [Q(tj )]

j∈{i+1,i+2,...,n,∞}

= e−r∆t P a (ti+1 ).



Now we may explore a property of the intrinsic value which holds when the American put will expire in the money. Lemma 13.3 Suppose the value of the security underlying an American put option follows a path through the n-step binomial lattice such that for some i ∈ {0, 1, . . . , n − 1} we have (K − uS(ti ))+ > 0, then Q(ti ) > e−r∆tE [Q(ti+1 )]. Proof. 13.4

See exercise (7).



Optimal Exercise Time

Since the owner of the American put may exercise it at any time step between t = 0 and t = T (or not at all), naturally the owner will be interested in the optimal time for exercising the option. In this section we will determine optimal time to exercise an American put option on a stock that pays no dividends. The optimal exercise time is another example of a stopping time. The option holder may make the decision to exercise only on the information represented in the binomial lattice model. The optimal time for exercise depends on many factors including whether we are thinking of “optimal” from the perspective of the option owner or optimal from the perspective of the option writer. From the perspective of the option holder, the optimal time for exercise may be the worst time for exercise for the option writer. In other words the optimal exercise for the option owner may yield the smallest profit for the option writer. If the option seller has determined the price of the put correctly, then even in the worst case scenario, there should be no arbitrage opportunity for the option buyer. For the option seller the optimal outcome may be for the option to expire

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

332

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

unused. In Chapter 12 the concept of the utility function was introduced. The holder of an American option may act so as to maximize the expected utility of the option, which could lead to early exercise of both puts and calls. To keep things simple we will consider for the moment only the value of the American option and not the investor’s utility function. As long as the value of the American put exceeds the intrinsic value, the owner should not exercise the option. Mathematically this is stated as the inequality P a (t) > Q(t)

=⇒

option owner should not exercise.

To act otherwise is for the option holder to accept a smaller payoff than would be generated if the option were sold to another party. Care must be taken in the determination of the optimal time to exercise the put, since we have defined P a (T ) = Q(T ), i.e. the value of the put is the intrinsic value at expiry (see Eq. (13.9)). If the option has not been used prior to expiry, it would only be used at expiry if P a (T ) > 0. Ideally the owner would like to exercise the option at a time for which the payoff is as large as possible. In Fig. 13.9 this may appear to be at the second time step since a payoff of approximately 5.53962 appears there. However, this particular payoff can only be achieved if the security follows a path of two consecutive downward movements in price. This behavior is not certain to occur. The American put should be exercised the first time the value of the option is equal to the intrinsic value, provided that value is positive. This allows for the case of an option expiring unused (in which case its value would always have been higher than the intrinsic value prior to expiry and its value at expiry was zero). With these conditions in mind we will define τ ∗ to be the optimal stopping time having value  a a   ∞ if P (t) > Q(t) for t ∈ [0, T ) and P (T ) = 0, a a τ ∗ = T if P (t) > Q(t) for t ∈ [0, T ) and P (T ) > 0, a   min {t | P (t) = Q(t)} otherwise. t∈[0,T )

In the discrete setting it is convenient to define i∗ ∈ {0, 1, . . . , n, ∞} as  a a   ∞ if P (i∆t) > Q(i∆t) for i ∈ {0, 1, . . . , n − 1} and P (n∆t) = 0, a a i∗ = n if P (i∆t) > Q(i∆t) for i ∈ {0, 1, . . . , n − 1} and P (n∆t) > 0,   min {i | P a (i∆t) = Q(i∆t)} otherwise. i∈{0,1,...,n−1}

(13.11)

Then τ ∗ = i∗ ∆t.

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

American Options

BC8495/Chp. 13

333

We will give an informal proof that τ ∗ is the optimal stopping time below. The reader interested in a more complete proof should consult [Shreve (2004a)]. Theorem 13.4 Let i∗ be defined as in Eq. (13.11) then τ ∗ = i∗ ∆t is the optimal exercise time for the owner of the American option and furthermore ∗

P a (0) = e−rτ E [P a (τ ∗ )] . Proof.

For notational convenience we will define Pˆa (t) as Pˆa (t) = P a (min{t, τ ∗ }).

(13.12)

Suppose we are at the ith time step of the binomial lattice with i ∈ {0, 1, . . . , n − 1} and i∗ > i. According to the definition of i∗ given in Eq. (13.11), the value of the put exceeds its intrinsic value, or symbolically P a (i∆t) > Q(i∆t). According to Eq. (13.10) Pˆa (i∆t) = P a (i∆t) (since i∗ > i) =

max

{e−r(j−i)∆t E [Q(j∆t)]}

j∈{i,i+1,...,n,∞}

= e−r∆t

max

{e−r(j−[i+1])∆t E [Q(j∆t)]}

j∈{i+1,i+2,...,n,∞}

(since P a (i∆t) > Q(i∆t)) = e−r∆t P a ((i + 1)∆t) i h = e−r∆t E Pˆa ((i + 1)∆t) .

(13.13)

On the other hand if i∗ ≤ i then the put has been exercised and its value is set. Pˆa (i∆t) = P a (i∗ ∆t) (since i∗ ≤ i)

= pP a (i∗ ∆t) + (1 − p)P a (i∗ ∆t) = pPˆa ((i + 1)∆t) + (1 − p)Pˆa ((i + 1)∆t) h i = E Pˆa ((i + 1)∆t) .

(13.14)

The American put must be exercised on or before the nth time step or expire unused. Thus if we set i = 0, then if i∗ = 0 we have P a (0) = Q(0) = E [Q(0)] = e0 E [Q(0)] = e0 E [P a (0)] ,

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

334

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

which satisfies Eq. (13.12). Note that E [Q(0)] = Q(0) since at t = 0 the option’s payoff is known with certainty. If i∗ > 0 then by Eq. (13.13) h i P a (0) = Pˆa (0) = e−r∆t E Pˆa (∆t) . Now if i∗ = 1 then Pˆa (∆t) = P a (∆t) and

P a (0) = e−r∆t E [P a (∆t)] , which once again satisfies Eq. (13.12). The other possibility is that i∗ > 1. Another application of Eq. (13.13) gives us h h ii h i P a (0) = e−r∆t E e−r∆t E Pˆa (2∆t) = e−r(2∆t)E Pˆa (2∆t) .

We are assuming that the risk-free interest rate is constant which allows it to be brought out of the expected value. To finish the proof we must consider two cases: (1) i∗ is finite, and (2) ∗ i = ∞. For the case that i∗ is finite we may apply the reasoning above for i = 0, 1, . . . , i∗ − 1 to obtain ∗

P a (0) = e−ri

∆t

E [P a (i∗ ∆t)] .

∗ ∗ ˆ Now h for i = ii , i + 1, . . . , n we have from Eq. (13.14) Pa (i∆t) = E Pˆa ((i + 1)∆t) . Thus

i h E Pˆa (i∆t) = Pˆa (i∗ ∆t) = P a (i∗ ∆t)

for all i = i∗ , i∗ + 1, . . . , n. Finally if i∗ = ∞ we know the payoff of the American put is zero and Eq. (13.12) is satisfied.  Some readers may recognize the concept of the martingale used in the proof of Theorem 13.4. This chapter concludes with an example illustrating the result of Theorem 13.4. Example 13.3 Suppose the current price of a non-dividend paying security is $36, the volatility in the security price is 20% annually, and the risk-free interest rate is 6% annually. If an American put option with maturity occurring in 4 months is written its price may be approximated using the binomial model with ∆ = 1/12 (or one month). The binomial lattice is shown in Fig. 13.10. The upper number in each box is the value of the put

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

335

American Options

0 0 0 0 0.880188 0 2.33928 1.84511 4.06039 4.

0 0 1.87637 1.84511

4. 4. 6.03319 6.03319

4. 4. 6.03319 6.03319

7.95155 7.95155

7.95155 7.95155 9.76156 9.76156 11.4694 11.4694

Fig. 13.10 The top number in each box represents the price of the American put option while the bottom number is the intrinsic value of the option.

at that node in the lattice. The lower number is the corresponding intrinsic value (payoff from immediate exercise). The first time these values are equal is at t∗ = 1/12. Thus we may verify that ∗

P a (0) = e−rt E [P a (t∗ )] = e−0.06/12 [(0.528557)(2.33928) + (1 − 0.528557)(6.03319)]

= 4.06039.

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

336

13.5

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. 13

An Undergraduate Introduction to Financial Mathematics

Exercises

(1) The price of a security is currently $50, the risk-free interest rate is 6% compounded continuously, and the strike price of a three-month American put option worth $9.75 is $51. Find the range of no arbitrage values of a three-month American call on the same security with the same strike price. (2) The price of a security is currently $93, the risk-free interest rate is 5.74% compounded continuously, and the strike price of a two-month American call option worth $11.77 is $90. Find the range of no arbitrage values of a two-month American put on the same security with the same strike price. (3) The price of a security is currently $115, the risk-free interest rate is 3.75% compounded continuously, and the strike price of otherwise identical six-month American call and put options is $110. Find the maximum difference in the prices of the American call and put options if the security will pay a single dividend of 10% in three months time. Assume the price of the security remains constant during the lives of the options. (4) The price of a security is currently $98, the risk-free interest rate is 2.95% compounded continuously, and the strike price of otherwise identical four-month American call and put options is $100. Find the maximum difference in the prices of the American call and put options if the security pays a continuous stream of dividends at a rate of $1 per month. (5) The price of a security is $56, the risk-free interest rate is 12% compounded continuously, and the volatility of the security is 25%. Find the price of a two-month American put on the security with a strike price of $58. (6) Verify that, in the absence of arbitrage, the expression for p in Eq. (13.8) lies in the interval [0, 1] and thus is a valid value for a probability. (7) Complete the proof of Lemma 13.3. (8) Verify the conclusion of Theorem 13.4 for an American put option maturing in 3 months on a stock whose current price is $80. The strike price of the option in $78, the risk-free interest rate is 5% per annum, and the volatility of the stock price is 25%. (9) Verify the conclusion of Theorem 13.4 for an American put option maturing in 4 months on a stock whose current price is $100. The

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

American Options

BC8495/Chp. 13

337

strike price of the option in $100, the risk-free interest rate is 3.25% per annum, and the volatility of the stock price is 29%. (10) The Put-Call Parity formula for American-style options on nondividend-paying stocks has been expressed in other ways. For example in [Guo and Su (2006)] it is written as C a + Ke−rT ≤ P a + S ≤ C a + K. Show that the inequality in Eq. (13.3) is equivalent to this form. (11) In Chap. 7 it was demonstrated that in the absence of arbitrage, the value of a European option on a non-dividend paying stock satisfies the inequality C e ≥ S − Ke−rT , where K is the strike price, r is the risk-free interest rate, and T is the expiry date. Consider another type of option we will describe as a perpetual call. The perpetual call can be exercised at any time and never expires. Show that the value of the perpetual call is C = S. (12) Use Eq. (7.7) to show that for a European call option on a dividendpaying stock the following inequality holds. C e ≥ S(0) − Ke−rT − PV(div) (13) Consider an American call option on a stock which will pay a dividend of D at time 0 < td < T where T is the expiry of the option. Show that if h i D ≤ K 1 − e−r(T −td )

the option will not be exercised at time td . (Hint : consider the value of the option just before and just after td using the result of exercise (12).) (14) Consider American options which are identical except for the value of the underlying stock S1 < S2 . Show that the following two inequalities hold. C a (S1 ) ≤ C a (S2 ) P a (S1 ) ≥ P a (S2 )

(15) Consider American options on a non-dividend-paying stock which are identical except for the value of the underlying stock S1 < S2 . Show that the following two inequalities hold. C a (S2 ) − C a (S1 ) ≤ S2 − S1 P a (S1 ) − P a (S2 ) ≤ S2 − S1 14:31:02.

May 25, 2012

14:36

338

WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp.

An Undergraduate Introduction to Financial Mathematics

(16) Consider American options which are identical except for the value of the underlying stock S1 < S2 . If 0 < γ < 1 show that the following two inequalities hold. C a (γS1 + (1 − γ)S2 ) ≤ γC a (S1 ) + (1 − γ)C a (S2 ) P a (γS1 + (1 − γ)S2 ) ≤ γP a (S1 ) + (1 − γ)P a (S2 )

(17) In exercise (13) of Chap. 7 readers were asked to prove the following two inequalities. C(K2 ) − C(K1 ) C(K3 ) − C(K2 ) ≤ K2 − K1 K3 − K2 P (K3 ) − P (K2 ) P (K2 ) − P (K1 ) ≤ K2 − K1 K3 − K2

for options with strike prices K1 < K2 < K3 . Show that there exists a number 0 < γ < 1 such that the following two inequalities hold. C(K2 ) ≤ γC(K1 ) + (1 − γ)C(K3 ) P (K2 ) ≤ γP (K1 ) + (1 − γ)P (K3 )

(18) Suppose you observe call options with the characteristics listed in the table below. C

K

4.57 3.80 2.13

100 105 115

Does the possibility of arbitrage exist? (19) Consider American options which are identical except for strike times T1 < T2 . Show that the following two inequalities hold. C a (T1 ) ≤ C a (T2 ) P a (T1 ) ≤ P a (T2 )

(20) Consider American options written on an underlying stock whose value is S. The strike prices of the options are both K. Show that C a (S(t)) − (S(t) − K)+ and P a (S(t)) − (K − S(t))+ are maximized when S(t) = K.

14:31:02.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. A

Appendix A

Sample Stock Market Data

In this appendix are end-of-day closing stock market prices for Sony Corporation stock. The data were collected from the website www.siliconinvestor.com and cover the one year period of August 13, 2001 until August 12, 2002.

Date Aug-12-2002 Aug-9-2002 Aug-8-2002 Aug-7-2002 Aug-6-2002 Aug-5-2002 Aug-2-2002 Aug-1-2002 Jul-31-2002 Jul-30-2002 Jul-29-2002 Jul-26-2002 Jul-25-2002 Jul-24-2002 Jul-23-2002 Jul-22-2002 Jul-19-2002 Jul-18-2002 Jul-17-2002 Jul-16-2002

14:31:20.

Close 42.86 44.44 44.0 43.85 42.5 42.01 42.71 44.55 45.33 46.65 46.14 44.7 45.49 47.1 45.21 44.97 46.3 48.95 47.82 49.59

Date Jul-15-2002 Jul-12-2002 Jul-11-2002 Jul-10-2002 Jul-9-2002 Jul-8-2002 Jul-5-2002 Jul-3-2002 Jul-2-2002 Jul-1-2002 Jun-28-2002 Jun-27-2002 Jun-26-2002 Jun-25-2002 Jun-24-2002 Jun-21-2002 Jun-20-2002 Jun-19-2002 Jun-18-2002 Jun-17-2002

Close 50.3 50.9 51.2 50.47 52.75 51.98 53.17 51.75 50.0 51.55 53.1 50.3 48.95 49.41 50.1 48.63 50.23 50.06 51.8 52.78

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

340

Juliet

Undergrad Introd to... 3rd edn

An Undergraduate Introduction to Financial Mathematics

Date Jun-14-2002 Jun-13-2002 Jun-12-2002 Jun-11-2002 Jun-10-2002 Jun-7-2002 Jun-6-2002 Jun-5-2002 Jun-4-2002 Jun-3-2002 May-31-2002 May-30-2002 May-29-2002 May-28-2002 May-24-2002 May-23-2002 May-22-2002 May-21-2002 May-20-2002 May-17-2002 May-16-2002 May-15-2002 May-14-2002 May-13-2002 May-10-2002 May-9-2002 May-8-2002 May-7-2002 May-6-2002 May-3-2002 May-2-2002 May-1-2002 Apr-30-2002 Apr-29-2002

14:31:20.

Close 52.45 53.27 54.31 54.01 54.36 55.5 55.5 56.7 55.91 56.3 58.11 58.69 57.8 58.23 59.4 59.7 59.56 58.05 57.65 58.81 56.48 56.15 55.6 55.0 54.75 54.6 55.14 52.65 54.0 54.02 53.45 54.8 54.2 55.48

Date Apr-26-2002 Apr-25-2002 Apr-24-2002 Apr-23-2002 Apr-22-2002 Apr-19-2002 Apr-18-2002 Apr-17-2002 Apr-16-2002 Apr-15-2002 Apr-12-2002 Apr-11-2002 Apr-10-2002 Apr-9-2002 Apr-8-2002 Apr-5-2002 Apr-4-2002 Apr-3-2002 Apr-2-2002 Apr-1-2002 Mar-28-2002 Mar-27-2002 Mar-26-2002 Mar-25-2002 Mar-22-2002 Mar-21-2002 Mar-20-2002 Mar-19-2002 Mar-18-2002 Mar-15-2002 Mar-14-2002 Mar-13-2002 Mar-12-2002 Mar-11-2002

Close 55.5 56.0 53.98 53.92 53.93 53.25 53.87 53.8 53.7 52.17 50.86 51.7 52.72 51.31 52.35 51.5 52.71 52.46 51.15 51.1 51.7 51.45 50.85 50.85 52.3 53.16 52.01 55.21 53.66 55.25 55.09 52.4 53.9 56.2

BC8495/Chp. A

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

341

Sample Stock Market Data

Date Mar-8-2002 Mar-7-2002 Mar-6-2002 Mar-5-2002 Mar-4-2002 Mar-1-2002 Feb-28-2002 Feb-27-2002 Feb-26-2002 Feb-25-2002 Feb-22-2002 Feb-21-2002 Feb-20-2002 Feb-19-2002 Feb-15-2002 Feb-14-2002 Feb-13-2002 Feb-12-2002 Feb-11-2002 Feb-8-2002 Feb-7-2002 Feb-6-2002 Feb-5-2002 Feb-4-2002 Feb-1-2002 Jan-31-2002 Jan-30-2002 Jan-29-2002 Jan-28-2002 Jan-25-2002 Jan-24-2002 Jan-23-2002 Jan-22-2002 Jan-18-2002 Nov-28-2001

14:31:20.

Close 56.59 56.05 53.4 50.3 50.7 47.8 46.2 47.22 45.75 47.25 46.71 45.3 45.4 44.87 44.85 45.55 44.85 43.77 43.45 42.4 42.5 41.0 41.75 43.25 43.5 44.76 44.35 43.7 45.2 46.4 45.86 44.91 43.56 44.7 46.11

Date Jan-17-2002 Jan-16-2002 Jan-15-2002 Jan-14-2002 Jan-11-2002 Jan-10-2002 Jan-9-2002 Jan-8-2002 Jan-7-2002 Jan-4-2002 Jan-3-2002 Jan-2-2002 Dec-31-2001 Dec-28-2001 Dec-27-2001 Dec-26-2001 Dec-24-2001 Dec-21-2001 Dec-20-2001 Dec-19-2001 Dec-18-2001 Dec-17-2001 Dec-14-2001 Dec-13-2001 Dec-12-2001 Dec-11-2001 Dec-10-2001 Dec-7-2001 Dec-6-2001 Dec-5-2001 Dec-4-2001 Dec-3-2001 Nov-30-2001 Nov-29-2001 Oct-10-2001

BC8495/Chp. A

Close 45.51 44.5 45.91 47.02 47.55 48.92 48.0 47.82 49.31 49.57 47.25 45.95 45.1 46.25 44.0 44.11 44.36 44.36 45.5 46.75 47.0 45.55 46.2 45.28 47.3 46.95 46.61 48.5 49.86 49.7 46.8 45.99 47.7 47.0 37.0

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

342

Juliet

Undergrad Introd to... 3rd edn

An Undergraduate Introduction to Financial Mathematics

Date Nov-27-2001 Nov-26-2001 Nov-23-2001 Nov-21-2001 Nov-20-2001 Nov-19-2001 Nov-16-2001 Nov-15-2001 Nov-14-2001 Nov-13-2001 Nov-12-2001 Nov-9-2001 Nov-8-2001 Nov-7-2001 Nov-6-2001 Nov-5-2001 Nov-2-2001 Nov-1-2001 Oct-31-2001 Oct-30-2001 Oct-29-2001 Oct-26-2001 Oct-25-2001 Oct-24-2001 Oct-23-2001 Oct-22-2001 Oct-19-2001 Oct-18-2001 Oct-17-2001 Oct-16-2001 Oct-15-2001 Oct-12-2001 Oct-11-2001 Aug-16-2001 Aug-15-2001

14:31:20.

Close 48.25 48.55 46.55 46.49 45.6 48.14 46.2 45.3 40.83 40.5 39.95 39.8 39.79 39.3 41.8 41.2 39.72 39.14 38.2 38.12 38.6 40.33 40.43 40.84 41.7 41.74 41.0 40.5 40.1 40.75 39.58 40.01 39.96 50.48 51.65

Date Oct-9-2001 Oct-8-2001 Oct-5-2001 Oct-4-2001 Oct-3-2001 Oct-2-2001 Oct-1-2001 Sep-28-2001 Sep-27-2001 Sep-26-2001 Sep-25-2001 Sep-24-2001 Sep-21-2001 Sep-20-2001 Sep-19-2001 Sep-18-2001 Sep-17-2001 Sep-10-2001 Sep-7-2001 Sep-6-2001 Sep-5-2001 Sep-4-2001 Aug-31-2001 Aug-30-2001 Aug-29-2001 Aug-28-2001 Aug-27-2001 Aug-24-2001 Aug-23-2001 Aug-22-2001 Aug-21-2001 Aug-20-2001 Aug-17-2001 Aug-14-2001 Aug-13-2001

Close 34.31 35.7 35.25 35.3 34.4 33.72 33.74 33.2 37.19 36.25 37.05 38.7 36.75 37.56 37.25 36.35 37.0 40.4 41.48 42.2 42.7 44.0 44.9 45.92 47.33 47.75 47.72 47.91 46.49 48.55 47.1 48.25 47.9 50.1 49.16

BC8495/Chp. A

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

Appendix B

Solutions to Chapter Exercises

B.1

The Theory of Interest

(1) The principal amount is P = $3659, the annual interest rate is r = 0.065, the period is t = 5 years. Thus the final compound balance is P (1 + rt) = 3659(1 + (5)(0.065)) = $4848.18, where the final balance has been rounded to the nearest cent. (2) The principal amount is P = $3993, the annual interest rate is r = 0.043, the period is t = 2 years. If there are n compounding periods per year, the account balance is  r nt . A=P 1+ n (a) If the interest is compounded monthly n = 12 and A = 3993(1 +

0.043 24 ) = $4350.93. 12

(b) If the interest is compounded weekly n = 52 and A = 3993(1 +

0.043 104 ) = $4351.44. 52

(c) If the interest is compounded daily n = 365 and A = 3993(1 +

0.043 730 ) = $4351.57. 365

(d) If the interest is compounded continuously A = P ert = 3993e(0.043)(2) = $4351.60. (3) If the principal is P = $3750 and t = 8, then

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

344

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(a) for 1.5% simple annual interest, A = P (1 + rt) = 3750(1 + 0.015(8)) = 4200.00. (b) for 1.5% effective annual compound interest, A = P (1 + r)t = 3750(1 + 0.015)8 = 4224.35. (c) for 0.75% six-month interest compounded every six months, A = P (1 + r)t = 3750(1 + 0.0075)16 = 4226.22. (d) for 0.375% three-month interest compounded every three months. A = P (1 + r)t = 3750(1 + 0.00375)32 = 4227.16. (4) The period is t = 1, the interest rate is r = 0.08 compounded n = 4 times per period. The effective simple annual interest rate R is R = −1 + (1 +

0.08 4 ) ≈ 0.08243. 4

(5) Since the competitor pays interest at rate r = 0.0525 compounded daily we will compute his effective simple annual interest rate and then determine the monthly compounded rate which matches it. The effective annual interest rate re is re = −1 + (1 +

0.0525 365 ) ≈ 0.05390. 365

Thus we wish to solve the following equation for i. 0.05390 = −1 + (1 +

√ i 1.05390 = 1 + 12 0.05261 = i

i 12 ) 12

12

Thus if you pay an interest rate compounded monthly of at least 5.261122% you will compete favorably. (6) The continuously compounded interest-bearing account will increase in value more rapidly, thus we wish to solve the following equation. 1000e0.0475t − 1000(1 +

0.0475 365t ) =1 365

This equation cannot be solved algebraically for t, so we use Newton’s method to approximate the solution, t ≈ 42.66 years. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

345

Solutions to Chapter Exercises

(7) We focus on finding the limit lim (1 + h)1/h . If we let y = (1 + h)1/h , h→0

then

  ln y = ln (1 + h)1/h 1 ln(1 + h) h 1 = e h ln(1+h)

=

eln y

y=e

ln(1+h) h

.

Therefore lim (1 + h)1/h = lim e

h→0

ln(1+h) h

h→0

= elimh→0

ln(1+h) h

since the exponential function is continuous everywhere. The limit in the exponent is the definition of the derivative of ln x at x = 1. Thus the exponent limit is 1 and lim (1 + h)1/h = e1 = e. h→0

(8) To determine the preferable investment we must determine the present value of the payouts. Since the interest rate is r = 0.0275 compounded continuously we have PA = 200e−0.0275 + 211e−0.055 + 198e−0.0825 + 205e−0.11 = 760.25 PB = 198e−0.0275 + 205e−0.055 + 211e−0.0825 + 200e−0.11 = 760.12. Thus investment A is preferable. (9) If the price of the house is $200000 and your down payment is 20% then you will borrow P = $160000. If the monthly payment on a t = 30 year, fixed rate mortgage should not exceed x = $1500 then we can substitute these quantities into Eq. (1.9) and use Newton’s Method to approximate r.   h n r i−nt P =x 1− 1+ r n   h 12 r i−(12)(30) 160000 = 1500 1− 1+ r 12 r ≈ 0.1080 Thus the interest rate must not exceed 10.80% annually.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

346

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(10) The effective annual real rate of interest is ri =

0.0505 − 0.0202 = 0.0297. 1 + 0.0202

The equivalent nominal annual rate compounded quarterly is then  r 4 1+ = 1 + 0.0297 4 r ≈ 0.0294. (11) Using the product rule for differentiation,  h R  i R d − t r(s) ds i d h − R t r(s) ds − 0t r(s) ds 0 0 0 e A(t) = e A (t) + e A(t) dt dt = e− + = e−

Rt 0



Rt 0

r(s) ds

A0 (t)  Z t  R t d − r(s) ds e− 0 r(s) ds A(t) dt 0 r(s) ds

A0 (t) − r(t)e−

Rt 0

r(s) ds

A(t),

by the Fundamental Theorem of Calculus. (12) Differentiating the function f (r) given in Eq. (1.15) yields 0

f (r) =

n X

−i−1

Ai (−i)(1 + r)

i=1

=−

n X

Ai (i)(1 + r)−(i+1) < 0

i=1

for −1 < r < ∞. Thus if there exist two distinct rates of return r1 and r2 then by the Mean Value Theorem f 0 (r) = 0 for some r between r1 and r2 , contradicting the inequality above. (13) Using Eq. (1.15) we setup the equation 10000 =

2000 3000 4000 3000 + + + 1+r (1 + r)2 (1 + r)3 (1 + r)4

whose solution is approximated using Newton’s Method. The rate of return r ≈ 0.0718 or equivalently 7.18% per year. (14) (a) The present value of initial investment and payout amounts for investment A is (assuming an annual interest rate of r = 0.0433), −$10.91. The present value of the initial investment and payout amounts for investment B is $18.63. Thus investment B is preferable if present value is the decision criterion. (b) The rate of return of investment A is rA ≈ 0.0394 while the rate of return of investment B is rB ≈ 0.0499. Thus investment B would 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

347

Solutions to Chapter Exercises

be the preferable investment if rate of return were the deciding criterion. (15) The present value of the income stream is P =

Z

3

(50000)e−0.01te−0.0425t dt

0

=

Z

3

(50000)e−0.0525t dt

0

3 50000 −0.0525t = e −0.0525 0 ≈ 138, 783.99. (16) Let the continuous rate of investment be S, then since the cumulative future value of the investment must be $1, 000, 000 then Z 6 1, 000, 000 = Se−0.0249t dt 0

6 S −0.0249t e = −0.0249 0 = 5.573311S

S ≈ $179, 426.56/yr.

(17) At the end of the 7th year Alice’s account balance is 12000(1 + 0.03)7. At the end of the 8th year her account balance is 12000(1.03)8 − (1.05)T. At the end of the 9th year her account balance is 12000(1.03)9−(1.05)(1.03)T −(1.05)T = 12000(1.03)9−(1.05)T [1 + 1.03] . At the end of the 10th year her account balance is 12000(1.03)10 − (1.05)(1.03)2 T − (1.05)(1.03)T − (1.05)T   = 12000(1.03)10 − (1.05)T 1 + (1.03) + (1.03)2 .

At the end of the 11th year Alice’s account balance is 12000(1.03)11 − T − (1.05)T

14:58:13.

3 X i=1

(1.03)i .

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

348

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Finally at the end of the 15th year Alice’s account balance is " # 3 X 15 4 i 12000 = 12000(1.03) − (1.03) T 1 + (1.05) (1.03) i=1

T = 1369.84.

(18) The homeowner has no preference between the two payment plans when the present values of the payments on August 31st are equal. 0.98(4500) = 1500 +

1500

1+ r ≈ 0.122666

 r 2 12

+

1500 1+

 r 4 12

(19) The present value of the first option is 1600 1 + 1+r

r 2



,

while the present value of the second option is 1600 2 + 50. 1 + r2

If the present values are equal then  1600 1 + 2r 1600 = 2 + 50 1+r 1 + r2 r ≈ 0.0645794. (20) Using Eq. (1.14), (a) the average annual compound rate is Z 1 5 r(5) = r(t) dt 5 0

5 1 (0.045t − 0.005 ln(1 + t)) 5 0 = 0.0432082.

=

(b) The amount due at t = 3 for a unit deposit made at t = 2 is A=e

R3 2

r(t) dt 3

= e (0.055t−0.005 ln(1+t))|2 = 1.04452.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

349

Solutions to Chapter Exercises

Therefore the effective annual interest rate is 1 + re = e

R3 2

r(t) dt

= 1.0445243 re ≈ 0.0445243. (c) The present value at time t = 1 is P (1) = Ae−

R5 1

r(t) dt 5

= 1750e− (0.045t−0.005 ln(1+t))|0 = 1750e−0.174507 = 1469.77.

B.2

Discrete Probability

(1) Assuming the regular tetrahedron is fair so that it equally likely to land on any of its four faces, the probability of it landing on 3 is p = 1/4. (2) Let the ordered pair (m, n) represent the numeric outcomes of the rolling of the dice. The outcome of the first die is m and the outcome of the second die is n. If you are concerned about the problem of keeping track of which is the first die and which is the second, assume that the first die is painted green while the second is painted red. According to the Multiplication Rule the probability of the event (m, n) is P (m, n) = P (first die shows m) P (second die shows n) since the dice are independent of each other. Since each of the die has six equally likely simple outcomes, then P (m, n) = 1/36. The table of 36 outcomes is shown below. (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1)

(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2)

(1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3)

(1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4)

(1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5)

(1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)

Since we are interested in the sums of the numbers shown on the upward faces of the dice we will tabulate them.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

350

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

Thus we can see the sample space of sums for the fair dice is the set {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Each of the entries in the previous table occurs with probability 1/36 and therefore the probabilities of the outcomes of rolling a pair of fair dice are as follows. Outcome

Prob.

Outcome

Prob.

2

1 36 1 18 1 12 1 9 5 36

7

1 6 5 36 1 9 1 12 1 18 1 36

3 4 5 6

8 9 10 11 12

(3) Let the probability that the batter strikes out in the first inning be denoted P (1) = 1/3. Let the probability that the batter strikes out in the fifth inning be denoted P (5) = 1/4. Let the probability that the batter strikes out in both innings be denoted P (1 ∧ 5) = 1/10. According to the Addition Rule the probability that the batter strikes out in either the first or the fifth inning is P (1 ∨ 5) = P (1) + P (5) − P (1 ∧ 5) =

1 1 1 29 + − = . 3 4 10 60

(4) The eight possible outcomes of the experiment are listed in the table below. Person

Hat

14:58:13.

1 Red Red Red Blue Red Blue Blue Blue

2 Red Red Blue Red Blue Blue Red Blue

3 Red Blue Red Red Blue Red Blue Blue

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

351

In 6 of the outcomes two of the three people will see their companions wearing mis-matched hats and pass while the third will see their two companions wearing matching hats opposite in color to their own hat’s color. Thus one of the three will guess the correct color in these six cases. In the remaining two cases all three hats are of the same color, so the strategy of guessing the opposite color would fail. Thus the probability of winning under the strategy is 6/8 = 3/4 = 0.75. (5) Since the cards are drawn without replacement the outcome of the second draw is dependent on the outcome of the first draw. Using the Multiplication Rule and conditional probability we have P (2 aces) = P (2nd ace|1st ace) P (1st ace)    3 4 = 51 52 1 = . 221 (6) Since the cards are drawn without replacement the outcome of the second draw is dependent on the outcome of the first draw. Using the Multiplication Rule and conditional probability we have P (2nd ace ∧ 1st not ace) = P (2nd ace|1st not ace) P (1st not ace)    4 48 = 51 52 16 = . 221 (7) Since the cards are drawn without replacement the outcome of the fourth draw is dependent on the outcome of the third draw which is dependent on the outcome of the second draw which is dependent on the outcome of the first draw. Using the Multiplication Rule and conditional probability we have P (4 aces) = P (4th ace|3rd ace) · P (3rd ace|2nd ace) · P (2nd ace|1st ace) · P (1st ace)     1 2 3 4 = 49 50 51 52 1 . = 270725 

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

352

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(8) If n ∈ {2, . . . , 49} is the position of the first ace drawn then P (n) =

48 47 4 50 − n · ··· · . 52 51 54 − n 53 − n

When n = 1, P (1) = 1/13. By calculating P (2) , P (3) , . . . , P (49) we obtain the following probabilities for the first ace appearing on the nth draw. n

P (n)

2

16 221 376 5525 17296 270725 3243 54145 3036 54145 2838 54145 1892 38675 1763 38675 328 7735 164 4165 152 4165 703 20825

3 4 5 6 7 8 9 10 11 12 13

n

P (n)

14

8436 270725 222 7735 12 455 11 455 352 15925 5456 270725 992 54145 899 54145 116 7735 522 38675 36 2975 9 833

15 16 17 18 19 20 21 22 23 24 25

n

P (n)

n

P (n)

26

8 833 92 10829 2024 270725 253 38675 44 7735 38 7735 228 54145 57 15925 48 15925 8 3185 16 7735 1 595

38

4 2975 22 20825 44 54145 33 54145 24 54145 12 38675 8 38675 1 7735 4 54145 2 54145 4 270725 1 270725

27 28 29 30 31 32 33 34 35 36 37

39 40 41 42 43 44 45 46 47 48 49

The draw with the highest probability of being the first ace is the first draw. (9) Let H represent an occurrence of heads (with probability p) and T represent an occurrence of tails (with probability 1 − p). (a) S = {H, T H, T T H, . . . , T T · · · T H, . . .} (b) P (T T T T H) = (1 − p)4 p

(10) List the 4! = 24 permutations of the four DVDs and count how many leave at least one DVD in its correct case. n 1 2 3 4

1 1 1 1 1

14:58:13.

2 2 2 3 3

3 3 4 2 4

4 4 3 4 2

Any Correct? Y Y Y Y

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

n 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4

2 4 4 1 1 3 3 4 4 1 1 2 2 4 4 1 1 2 2 3 3

3 2 3 3 4 1 4 1 3 2 4 1 4 1 2 2 3 1 3 1 2

4 3 2 4 3 4 1 3 1 4 2 4 1 2 1 3 2 3 1 2 1

BC8495/Chp. B

353

Any Correct? Y Y Y N Y N N Y Y N Y Y N N N Y Y Y N N

We see that 15 of the 24 possible permutations leave at least one DVD in its correct case, thus the desired probability is p=

5 15 = = 0.625. 24 8

(11) The outcomes of different spins of a roulette wheel are assumed to be independent, thus the probability that any spin of the wheel has an outcome of black is P (black) = 9/19. (12) The outcomes of different spins of a roulette wheel are assumed to be independent, thus the probability that any spin of the wheel has an outcome of 00 is P (00) = 1/38. (13) To be a probability distribution function 10 X

10 10 X X c 1 7381c f (x) = =c = = 1, x x 2520 x=1 x=1 x=1

thus c = 2520/7381.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

354

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(14) 5 20 15 P (1 black) = 20 15 P (2 black) = 20 15 P (3 black) = 20

4 19 5 · 19 14 · 19 14 · 19

P (0 black) =

·

3 18 4 · 18 5 · 18 13 · 18 ·

1 114 5 + · 20 15 + · 20 91 = 228 =

15 19 5 19

4 5 4 15 5 + · · = 18 20 19 18 38 14 5 15 14 35 · + · · = 18 20 19 18 76 ·

(15) This situation can be thought of as a binomial experiment with 25 trials and the probability of success on a single trial of 0.0016. Thus the probability that a manufacturing run will be rejected is P (X ≥ 2) = =

25 X

P (X = n)

n=2 25   X 25 (0.0016)n (1 − 0.0016)25−n n n=2

≈ 0.00075.

(16) The probability of a child being born female is p = 100/205 = 20/41. Thus the expected number of female children in a family of 6 total children is 6 X

x=0

(x · P (x)) = =

6 X

x=0

6−x !    x  20 6 20 1− x· 41 41 x

120 ≈ 2.93. 41

(17) Analysis of the table game “craps”. (a) Win on first roll. 1 6 1 = 18

p7 = p11 Thus P (7 ∨ 11) = 2/9. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

355

Solutions to Chapter Exercises

(b) Since 1 36 1 p3 = 18 1 p12 = 36 p2 =

then P (2 ∨ 3 ∨ 12) = 1/9. (c) The only two outcomes of importance in this situation are 4 and 7. Thus we need the probability the outcome is 4 given that the outcome is 4 or 7. P (4 | (4 ∨ 7)) =

P (4) P (4 ∧ (4 ∨ 7)) = = P (4 ∨ 7) P (4 ∨ 7)

1 12 1 12

+

1 6

=

1 3

(d) Since the rolls of the dice are independent, p4 = P (4) P (4 | (4 ∨ 7)) =



1 12

  1 1 = . 3 36

(e) In the same way we calculated the probability of the last event, p5 = p6 = p8 = p9 = p10 =

   1 2 2 P (5) P (5 | (5 ∨ 7)) = = 9 5 45    5 5 25 P (6) P (6 | (6 ∨ 7)) = = 36 11 396    5 5 25 P (8) P (8 | (8 ∨ 7)) = = 36 11 396    1 2 2 P (9) P (9 | (9 ∨ 7)) = = 9 5 45    1 1 1 P (10) P (10 | (10 ∨ 7)) = = . 12 3 36

(f) The player wins with probability p = p7 + p11 + p4 + p5 + p6 + p8 + p9 + p10   2 1 2 25 244 = +2 + + = ≈ 0.492929. 9 36 45 396 495

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

356

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(18) Using the probabilities calculated in exercise (8) and the definition of expected value we have the expected position of the first ace, µ=

49 X

n=1

(n · P (n)) =

53 = 10.6. 5

(19) By the definition of expected value X E [aX + b] = ((aX + b)P (X)) X

=a

X

(XP (X)) + b

X

X

P (X)

X

= aE [X] + b.

(20) Assume that X, Y and Z are jointly distributed discrete random variables and c is a constant. (a) E [X + Y |Z = z] = =

XX X

Y

XX X

+

Y

(XP (X, Y |Z = z))

XX X

=

((X + Y )P (X, Y |Z = z))

X

Y

(Y P (X, Y |Z = z))

(XP (X|Z = z)) +

X

X Y

(Y P (Y |Z = z))

= E [X|Z = z] + E [Y |Z = z] (b) E [cX|Y = y] =

X

(cXP (X|Y = y))

X

=c

X

(XP (X|Y = y))

X

= cE [X|Y = y] (21) Using the probabilities calculated in exercise (8) and the expected value calculated in exercise (18), the variance in the position of the first ace drawn is 49 X

n=1

 1696 901 2809 − = = 67.84. n2 · P (n) − µ2 = 5 25 25 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

357

This implies the standard deviation in the appearance of the first ace is approximately 8.2365. (22) By the definition of variance   2 V (aX + b) = E (aX + b)2 − (E [aX + b])   2 = E a2 X 2 + 2abX + b2 − (aE [X] + b)   2 = a2 E X 2 + 2abE [X] + b2 − a2 (E [X]) − 2abE [X] − b2   2 = a2 E X 2 − a2 (E [X]) = a2 V (X) .

(23) P (X = x|X + Y = n) = P ((X = x) ∧ (Y = n − x) | X + Y = n) P ((X = x) ∧ (Y = n − x)) = P (X + Y = n) P (X = x) P (Y = n − x) = P (X + Y = n) The last equality holds using the independence of X and Y . Expressions for the two probabilities in the numerator can be found using the binomial probability density formula in Eq. (2.3). Since the probabilities of success on single trials are the same for the two binomial experiments the random variable X + Y can be thought of as a binomial random variable with M + N trials. Thus the binomial formula can also be used to express the denominator.  x  n−x N p (1 − p)M−x n−x p (1 − p)N −(n−x) P (X = x|X + Y = n) =  M+N n p (1 − p)M+N −n n   M N n N +M−n n−x p (1 − p)  = xM+N pn (1 − p)M+N −n n   M N M x

=

x

n−x  M+N n

This probability distribution is known as the hypergeometric distribution. (24) The probability of at least two people sharing a birthday is one minus the probability of no one sharing a birthday. If there are n people

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

358

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

then the probability they all have different birthdays is       364 363 362 365 − n + 1 q= ··· 365 365 365 365 =

=

n−1 Y

k=1 n−1 Y k=0

=

365 − k 365 365 − k 365

365! . (365 − n)! 365n

Therefore the probability of at least two people having the same birthday is p=1−q =1−

365! . (365 − n)! 365n

Note that p30 = 1 −

365! ≈ 0.706316. (365 − 30)! 36530

(25) For the three random variables described: (a) SY = {0, 1}, (b) SZ = {0}, (c) the events X = x and Y = y are dependent, thus P ((X = x) ∧ (Y = y)) = P ((Y = y) | (X = x)) P (X = x) . Therefore we have P ((X = 0) ∧ (Y = 0)) = P ((Y = 0) | (X = 0)) P (X = 0) = (0)(1 − p) = 0

P ((X = 1) ∧ (Y = 0)) = P ((Y = 0) | (X = 1)) P (X = 1) = (1)(p) = p

P ((X = 0) ∧ (Y = 1)) = P ((Y = 1) | (X = 0)) P (X = 0) = (1)(1 − p) = 1 − p

P ((X = 1) ∧ (Y = 1)) = P ((Y = 1) | (X = 1)) P (X = 1) = (0)(p) = 0

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

359

Solutions to Chapter Exercises

(d) The events X = x and Z = z are dependent, thus P ((X = x) ∧ (Z = z)) = P ((Z = z) | (X = x)) P (X = x) . Therefore we have P ((X = 0) ∧ (Z = 0)) = P ((Z = 0) | (X = 0)) P (X = 0) = (1)(1 − p) = 1 − p

P ((X = 1) ∧ (Z = 0)) = P ((Z = 0) | (X = 1)) P (X = 1) = (1)(p) = p

B.3

Normal Random Variables and Probability

(1) The probability distribution function for X is given by the piecewise defined function 1 if −4 ≤ x ≤ 1 f (x) = 5 0 otherwise. Therefore P (X ≥ 0) =

Z



f (x) dx =

0

Z

1

0

1 1 dx = . 5 5

(2) A probability distribution must satisfy Eq. (3.2). (a) Constant C satisfies the equation below. 1=

Z



f (x) dx

−∞ Z ∞

C dx 1 + x2 0 Z M 1 = lim 2C dx M→∞ 1 + x2 0 M  = 2C lim tan−1 x 0 =2

M→∞

= 2C lim tan−1 M = Cπ M→∞

Thus C = 1/π.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

360

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(b) Using Eq. (3.1) P (X > 1) =

Z



f (x) dx

1

Z



1 dx π(1 + x2 ) 1 Z M 1 dx = lim M→∞ 1 π(1 + x2 ) M 1 −1 = lim tan x M→∞ π =

1

1 lim (tan−1 M − tan−1 1) = π M→∞ π 1 1 π = − = . π 2 4 4 (3) We know that for any randomR variable X with probability distribution ∞ function f (x) we must have −∞ f (x) dx = 1, thus Z



−∞

Z



c dx 3 x 1 Z M 1 = c lim dx M→∞ 1 x3 M ! −1 = c lim M→∞ 2x2 1   −1 1 = c lim + M→∞ 2M 2 2 c 1= 2 c = 2.

f (x) dx =

(4) 1= =

Z



−∞ Z a

f (x) dx f (x) dx +

−∞

Z



f (x) dx

a

= P (X < a) + P (X ≥ a)

1 − P (X < a) = P (X ≥ a) 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

361

Solutions to Chapter Exercises

(5) By definition, the mean of a continuously distributed random variable is Z ∞ µ= xf (x) dx. −∞

Thus for the given probability distribution function Z ∞ x µ= dx π(1 + x2 ) −∞ Z ∞ Z 0 x x dx + dx = 2 π(1 + x2 ) 0 −∞ π(1 + x ) Z 0 Z N x x = lim dx + lim dx M→−∞ M π(1 + x2 ) N →∞ 0 π(1 + x2 ) 0 ! N ! 1 1 2 2 = lim ln(1 + x ) + lim ln(1 + x ) N →∞ M→−∞ 2π 2π M 0     1 1 = lim − ln(1 + M 2 ) + lim ln(1 + N 2 ) . M→−∞ N →∞ 2π 2π The improper integral above does not converge and hence the mean of the random variable does not exist. (6) E [aX + b] =

Z

=a



(ax + b)f (x) dx Z xf (x) dx + b

−∞ Z ∞

−∞

= aE [X] + b



f (x) dx

−∞

(7) Proof. Let X1 , X2 , . . . , Xk be continuous random variables with probability distribution function f (x1 , x2 , . . . , xk ). By the definition of expected value E [X1 + X2 + · · · Xk ] Z ∞Z ∞ Z ∞ = ··· (x1 + x2 + · · · + xk )f (x1 , x2 , . . . , xk ) dx1 dx2 . . . dxk −∞ −∞ −∞ Z ∞Z ∞ Z ∞ = ··· x1 f (x1 , x2 , . . . , xk ) dx1 dx2 . . . dxk −∞ −∞ −∞ Z ∞Z ∞ Z ∞ + ··· x2 f (x1 , x2 , . . . , xk ) dx1 dx2 . . . dxk −∞

−∞

−∞

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

362

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Z



Z

Z



+ ···+ ··· −∞ −∞ Z ∞ Z = x1 f1 (x1 ) dx1 + −∞



xk f (x1 , x2 , . . . , xk ) dx1 dx2 . . . dxk Z ∞ x2 f2 (x2 ) dx2 + · · · + xk fk (xk ) dxk

−∞ ∞

−∞

−∞

= E [X1 ] + E [X2 ] + · · · + E [Xk ] .

To keep the notation compact we have written the marginal probability density of Xi as fi (xi ) for i = 1, 2, . . . , k, where fi (xi ) =

Z



−∞

Z



−∞

···

Z



f (x1 , x2 , . . . , xk ) dx1 . . . dxi−1 dxi+1 . . . dxk .

−∞



(8) Proof. Let X1 , X2 , . . . , Xk be pairwise independent, continuous random variables with probability distribution function f (x1 , x2 , . . . , xk ). Since the component random variables are assumed to be pairwise independent the joint probability distribution can be rewritten as f (x1 , x2 , . . . , xk ) = f1 (x1 )f2 (x2 ) · · · fk (xk ). Thus by the definition of expected value E [X1 X2 · · · Xk ] Z ∞Z ∞ Z ∞ = ··· x1 x2 · · · xk f (x1 , x2 , . . . , xk ) dx1 dx2 . . . dxk −∞ −∞ −∞ Z ∞Z ∞ Z ∞ = ··· x1 x2 · · · xk f1 (x1 )f2 (x2 ) · · · fk (xk ) dx1 dx2 . . . dxk −∞ −∞ −∞ Z ∞  Z ∞  Z ∞  = x1 f1 (x1 ) dx1 x2 f2 (x2 ) dx2 · · · xk fk (xk ) dxk −∞

= E [X1 ] E [X2 ] · · · E [Xk ] .

−∞

−∞



(9) The conditional probability distribution f (y|x) is calculated as f (x, y) −3 f (x, y) dy

f (y|x) = R x =

14:58:13.

5xy 2 R x 162 5xy 2 −3 162

dy

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

363

Solutions to Chapter Exercises

y2 2 −3 y dy

= Rx =

3y 2 . + 27

x3

Consequently the conditional expected value is x  Z x  4 3y 2 3y 4 = 3(x − 81) , E [Y |X = x] = y dy = 3 3 x + 27 4(x + 27) −3 4(x3 + 27 −3

for −3 < x < 3. (10) Assume that X, Y and Z are jointly distributed continuous random variables and c is a constant. (a) E [X + Y |Z = z] = =

Z



−∞ ∞

Z

−∞

Z

Z



(x + y)f (x, y|Z = z) dx dy

−∞ ∞

Z

xf (x, y|Z = z) dy dx

−∞ ∞ Z ∞

+ yf (x, y|Z = z) dx dy −∞ −∞ Z ∞ Z ∞ = xfX (x|Z = z) dx + yfY (y|Z = z) dy −∞

−∞

= E [X|Z = z] + E [Y |Z = z] (b) E [c X|Y = y] =

Z

=c



cxf (x|y) dx

−∞ Z ∞

xf (x|y) dx

−∞

= c E [X|Y = y] (11) Using the joint probability distribution we find the following. (a) E [XY ] =

Z

0

=

Z

0

14:58:13.

2

Z

x

(xy)f (x, y) dy dx   x 3 3 (xy) xy dy dx 8 0

0

2Z

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

364

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

=

3 8

Z

2

2

Z

0

Z

x

x2 y 4 dy dx

0

Z 2 3 x7 dx 40 0 12 3 8 2 = x 0= = 2.4 320 5 =

(b) 

2



x

(x2 y)f (x, y) dy dx   Z 2Z x 3 3 2 = (x y) xy dy dx 8 0 0 Z Z 3 2 x 3 4 = x y dy dx 8 0 0 Z 2 3 = x8 dx 40 0 3 9 2 64 = x 0= ≈ 4.26667 360 15

E X Y =

(c)

Z

0

h √ i Z E X Y =

2

0

Z

x

√ (x y)f (x, y) dy dx 0 0   Z 2Z x 3 3 xy = (xy 1/2 ) dy dx 8 0 0 Z Z 3 2 x 2 7/2 = x y dy dx 8 0 0 Z 2 1 = x13/2 dx 12 0 √ 1 15/2 2 64 2 = x ≈ 2.01133 = 90 45 0

(12) Proof. Let X be a continuous random variable with probability distribution function f (x). Suppose that E [X] = µ. By the definition of variance Z ∞ V (X) = (x − µ)2 f (x) dx −∞ Z ∞ Z ∞ Z ∞ 2 2 = x f (x) dx − 2µ xf (x) dx + µ f (x) dx −∞

14:58:13.

−∞

−∞

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

365

Solutions to Chapter Exercises

  = E X 2 − 2µ2 + µ2   = E X 2 − µ2 .



(13) Proof. Let X be a continuous random variable with probability distribution function f (x). Suppose a, b ∈ R,   2 V (aX + b) = E (aX + b)2 − (E [aX + b]) Z ∞ 2 = (ax + b)2 f (x) dx − (aE [X] + b) −∞ Z ∞ Z ∞ Z ∞ 2 2 2 =a x f (x) dx + 2ab xf (x) dx + b f (x) dx −∞ 2

2

−∞

2

−∞

− a (E [X]) − 2abE [X] − b   2 = a2 E X 2 + 2abE [X] + b2 − a2 (E [X]) − 2abE [X] − b2     2 = a2 E X 2 − (E [X]) = a2 V (X) .



(14) Proof. Let X1 , X2 , . . . , Xk be pairwise independent, continuous random variables with probability distribution function f (x1 , x2 , . . . , xk ). Let the mean of Xi be µi for i = 1, 2, . . . , k. By the definition of variance V (X1 + X2 + · · · + Xk ) h i 2 2 = E (X1 + X2 + · · · + Xk ) − (E [X1 + X2 + · · · + Xk ]) Z ∞ Z ∞ = ··· (x1 + · · · + xk )2 f (x1 , . . . , xk ) dx1 . . . dxk −∞

−∞

−∞

−∞

− (µ1 + · · · + µk )2 Z ∞ Z ∞ = ··· (x1 + · · · + xk )2 f1 (x1 ) · · · fk (xk ) dx1 . . . dxk − (µ1 + · · · + µk )2 Z ∞Z ∞ Z ∞ = ··· (x21 + x22 + · · · + x2k + 2x1 x2 −∞

−∞

−∞

+ · · · + 2xk−1 xk )f1 (x1 )f2 (x2 ) · · · fk (xk ) dx1 dx2 . . . dxk

− (µ21 + µ22 + · · · + µ2k + 2µ1 µ2 + · · · + 2µk−1 µk )       = E X12 + E X22 + · · · + E Xk2 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

366

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

+ 2E [X1 X2 ] + · · · + 2E [Xk−1 Xk ]

− (µ21 + µ22 + · · · + µ2k + 2µ1 µ2 + · · · + 2µk−1 µk )       = E X12 + E X22 + · · · + E Xk2 + 2E [X1 ] E [X2 ] + · · · + 2E [Xk−1 ] E [Xk ]

− (µ21 + µ22 + · · · + µ2k + 2µ1 µ2 + · · · + 2µk−1 µk )       = E X12 + E X22 + · · · + E Xk2 + 2µ1 µ2 + · · · + 2µk−1 µk

− (µ21 + µ22 + · · · + µ2k + 2µ1 µ2 + · · · + 2µk−1 µk )       = E X12 − µ21 + E X22 − µ22 + · · · + E Xk2 − µ2k

= V (X1 ) + V (X2 ) + · · · + V (Xk ) .



(15) The expected value of X is calculated as E [X] =

Z



xf (x) dx

−∞ Z 2

2 x|x| dx 5 Z 0  Z 2 2 2 2 = (−x ) dx + x dx 5 −1 0 14 µ= . 15 =

−1

The variance is calculated as Z ∞ V (X) = x2 f (x) dx − µ2 −∞ 2

Z

2 2 196 x |x| dx − 225 −1 5 Z 0  Z 2 196 2 = (−x3 ) dx + x3 dx − 5 225 −1 0 373 = . 450 =

(16) Proof.

Suppose m, n ∈ Z, then n−m

is even ⇐⇒ n − m + 2m = n + m

14:58:13.

is even. 

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

367

Solutions to Chapter Exercises

(17) If f (x) is three times continuously differentiable at x = x0 then we can write f (x) as f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) + f 00 (x0 )

(x − x0 )2 (x − x0 )3 + f 000 (z) 2 6

where z lies between x and x0 . Hence f (x0 + h) = f (x0 ) + f 0 (x0 )h + f 00 (x0 )

h2 h3 + f 000 (z1 ) 2 6

f (x0 − h) = f (x0 ) − f 0 (x0 )h + f 00 (x0 )

h3 h2 − f 000 (z2 ) . 2 6

Subtracting the second equation from the first and solving for f 0 (x0 ) yields f 0 (x0 ) =

h2 f (x0 + h) − f (x0 − h) − (f 000 (z1 ) + f 000 (z2 )) . 2h 12

Thus when h is small, f 0 (x0 ) ≈

f (x0 + h) − f (x0 − h) . 2h

(18) The limit as stated is indeterminate of the form 0 · ∞, thus with some algebra and l’Hˆ opital’s Rule we obtain

lim (M − K)

M→∞

Z



M

R∞

f (t) dt M→∞ (M − K)−1 −f (M ) = lim M→∞ −(M − K)−2

f (t) dt = lim

M

= lim (M − K)2 f (M ) M→∞

= 0. The last equality is true since f is a probability density function for a random variable whose variance is finite.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

368

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(19) The improper integral can be evaluated as follows. σ √ 2π

Z



2

te−t

/2

(K−µ)/σ

σ dt = lim √ M→∞ 2π

Z

M

2

te−t

/2

dt

(K−µ)/σ

! M σ −t2 /2 √ −e = lim M→∞ 2π (K−µ)/σ   2 2 σ −M 2 /2 = √ lim −e + e−(K−µ) /2σ 2π M→∞ σ −(K−µ)2 /2σ2 = √ e 2π

(20) Using the method of integration by substitution, where u = x2 /(4kt) we obtain r Z Z x2 kt x − 4kt √ e dx = e−u du π 2 kπt r kt −u =− e +C π r kt − x2 =− e 4kt + C π where C is a constant of integration. (21) Using the technique of integration by parts with x u= √ kπt 1 du = √ dx kπt we obtain Z

x2

v = −kte− 4kt dv =

x − x2 e 4kt dx 2

Z x2 x2 ktx − x2 kt − x2 √ √ e− 4kt dx = − √ e 4kt + e 4kt dx 2 kπt kπt kπt r   Z kt x2 x2 − 4kt − 4kt = −xe + e dx . π

(22) Assuming that k > 0 and t > 0 we have lim 2ktM e−M

M→∞

14:58:13.

2

/4kt

= lim

M→∞

2ktM . eM 2 /4kt

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

369

Solutions to Chapter Exercises

The limit on the right-hand side of the equation is indeterminate of the form ∞/∞ and thus we apply l’Hˆopital’s Rule. 2ktM = lim 2 M→∞ eM /4kt M→∞ lim

= lim

M→∞

2kt M M 2 /4kt 2kt e 2

(2kt) M eM 2 /4kt

=0 (23) The probability distribution function for the standard normal random variable is 2 1 f (x) = √ e−x /2 2π

and thus 1 (a) P (−1 < X < 1) = √ 2π 1 (b) P (−2 < X < 2) = √ 2π

Z

1

−1 Z 2

e−x

2

/2

dx ≈ 0.682689,

e−x

2

/2

dx ≈ 0.9545,

−2 Z 3

2 1 e−x /2 dx ≈ 0.9973, (c) P (−3 < X < 3) = √ 2π −3 Z 3 2 1 (d) P (1 < X < 3) = √ e−x /2 dx ≈ 0.157305. 2π 1

(24) Proof. Suppose X is a normally distributed random variable with mean µ and variance σ 2 . We will let z = g(x) = (x − µ)/σ. E [Z] = E [g(X)] Z ∞ 2 1 x − µ − (x−µ) = √ e 2σ2 dx σ 2π −∞ σ Making the substitution u = (x − µ)/σ yields 1 E [Z] = √ 2π = 0.

14:58:13.

Z



−∞

ue−

u2 2

du

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

370

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Therefore   2 V (Z) = E Z 2 − (E [Z]) h i 2 = E (g(X)) 2 Z ∞ (x−µ)2 1 x−µ = √ e− 2σ2 dx σ σ 2π −∞ Z ∞ u2 1 = √ u2 e− 2 du 2π −∞ = 1, where we made the substitution u = (x − µ)/σ.



(25) Think of the consecutive years as year 1 and 2. The rainfalls are normally distributed random variables, call them X1 and X2 in each year with means µ1 = µ2 = 14 and standard deviations σ1 = σ2 = 3.2. Let the random variable X be the sum of the rainfalls in two consecutive years, then µ = E [X] = E [X1 ] + E [X2 ] = µ1 + µ2 = 28. Likewise σ 2 = V (X) = V (X1 ) + V (X2 ) = σ12 + σ22 = (3.2)2 + (3.2)2 = 20.48, assuming that rainfalls in different years are independent. Now let Z = (X − µ)/σ and then   30 − 28 P (X > 30) = P Z > √ 20.48 = P (Z > 0.44192) = 1 − P (Z ≤ 0.44192)

= 1 − Φ (0.44192) ≈ 0.329266. (26) Let A and B be the arrival times of Alice and Bob respectively. A and B are independent and each uniformly distributed on the interval [0, 1]. Thus the ordered pair of arrival times (A, B) is uniformly distributed in the square S = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}. The pair will meet if |A − B| ≤ 1/4. The region in S corresponding to Alice and Bob meeting is the region where • 0 ≤ A ≤ 1, • 0 ≤ B ≤ 1, and • |A − B| ≤ 1/4 . 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

371

The area of this region is the probability they meet.       1 7 3 3 9 P (meeting) = 1 − 2 =1− = 2 4 4 16 16 (27) Let the random variable X represent the ratio of consecutive days selling prices of the security. P (X > 1) = P (ln X > 0)   0−µ =P Z> σ   0 − 0.01 =P Z> 0.05 = P (Z > −0.20)

= Φ (0.20) ≈ 0.57926

Thus the probability of a one-day increase in the price of the security is approximately 0.57926. Consequently the probability of a one-day decrease in the price of the security is approximately 1 − 0.57926 = 0.42074. The probability of a four-day decrease in the selling price of the security is  P X 4 < 1 = P (4 ln X < 0)   0 − 4µ =P Z< 2σ   0 − 0.04 =P Z< 0.10 = P (Z < −0.40) = Φ (−0.40) ≈ 0.344578.

(28) Since X is uniformly distributed on [a, b] its probability distribution function is  1 if a ≤ x ≤ b, f (x) = b−a 0 otherwise. We will make use of Theorem 3.9.  Z ∞  1 if x < a f (t) dt = b−x if a ≤ x ≤ b  b−a x 0 if x > b 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

372

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Thus   E (X − K)+ =

Z



K

Z



f (t) dt

x



dx

Ra R b b−x   RK 1 dx + a b−a dx if K < a b b−x = if a ≤ K ≤ b K b−a dx   0 if K > b  a+b   2 −2 K if K < a = (b−K) if a ≤ K ≤ b 2(b−a)   0 if K > b.

(29) The result follows graphically from the symmetry of the standard normal distribution about the y-axis. See Fig. 3.2. However, if we wish to establish this result using the definition of the standard normal distribution, we have 1 Φ (−x) = √ 2π

Z

−x

2

e−t

/2

dt.

−∞

Now we make the substitution u = −t. Z −x Z x 2 2 1 1 √ e−t /2 dt = − √ e−u /2 du 2π −∞ 2π ∞ Z ∞ 2 1 = √ e−u /2 du 2π x Z x 2 1 e−u /2 du = 1− √ 2π −∞ Φ (−x) = 1 − Φ (x) (30) Proof. Using the definition of variance and Eqs. (3.17) and (3.19) we have  V (X − K)+ h 2 i  2 = E (X − K)+ − E (X − K)+    µ − 2K (µ − 2K)σ −(µ−2K)2 /2σ2 2 2 √ = (µ − 2K) + σ Φ + e σ 2π   2 2 2 σ µ−K − √ e−(µ−K) /2σ + (µ − K)Φ . σ 2π

14:58:13.



May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

373

(31) Proof. Using the definition of variance and Eqs. (3.18) and (3.21) we have  V (X − K)+ h 2 i  2 = E (X − K)+ − E (X − K)+ 2

2

= e2(µ+σ ) Φ (w + 2σ) − 2Keµ+σ /2 Φ (w + σ) + K 2 Φ (w)  2 2 − eµ+σ /2 Φ (w + σ) − KΦ (w)

where w = (µ − ln K)/σ.



(32) Using the definition of expected value we find the following. (a)   E [X(X + 1)] = E X 2 + X   = E X 2 + E [X]   = E X2 + µ = σ 2 + µ2 + µ

(b)     E (X − C)2 = E X 2 − 2CX + C 2   = E X 2 − 2CE [X] + C 2

= σ 2 + µ2 − 2Cµ + C 2 = σ 2 + (µ − C)2

B.4

The Arbitrage Theorem

(1) Converting from the odds against to probabilities that the outcomes may occur produces a table of probabilities like that which follows. Outcome

Probability

A

1 3 1 4 1 2

B C

Since the probabilities sum to 13 12 > 1, the Arbitrage Theorem guarantees the existence of a betting strategy which yields a positive payoff

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

374

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

regardless of the outcome. Suppose we wager x, y, and z on outcomes A, B, and C respectively. The payoffs under different winning scenarios are listed in the next table. Winning Outcome

Payoff

A

2x − y − z

B C

−x + 3y − z −x − y + z

We need only find values for x, y, and z which make the three expressions in the column on the right positive. The graphic below shows the “cone” where all three payoffs are positive. Notice that each amount wagered must be negative (this just means we take the wagers from others). By inspection we can see that x = −3, y = −2.5, and z = −5 is one solution to this set of inequalities. 0 y -2

-4

0

-2 z

-4

-4

x

-2

0

(2) The region is convex and unbounded. There are three corners to the region located at (0, 6), (3/2, 3/2), and (6, 0). The region is sketched as follows.

14:58:13.

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

375

Solutions to Chapter Exercises 10

8

x2

6

4

2

0

0

2

4

6

8

10

x1

(3) The cost function will be minimized at the smallest value of k for which the graph of x1 + x2 = k intersects the feasible region sketched in exercise (2). This occurs when (x1 , x2 ) = (3/2, 3/2). Thus the minimum cost is k = 3. 10

8

6 x2

May 25, 2012

4

2

0

0

2

4

6 x1

14:58:13.

8

10

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

376

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(4) 0 ≤ x1 ≤ 2 and 0 ≤ x2 ≤ 3 4

3

x2

2

1

0

-1 -1

0

1 x1

2

3

(5) We can think of the solution as a system of inequalities: x1 ≥ 0 x2 ≥ 0 x3 ≥ 0 x4 ≥ 0

x1 + x3 = 2

x2 + x4 = 3 where x3 and x4 are the newly introduced slack variables. The last two equations can be re-written in matrix/vector form as   x1     1010  x2  2   Ax = = = b. 0 1 0 1  x3  3 x4

(6) Let vector b = h1, 1, 2i and y = hy1 , y2 , y3 i. The problem as stated can be thought of as a dual problem:   minimize bT y subject to AT y = 1 2 3 y ≥ [15] = c. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

377

This is equivalent to the primal problem: maximize cT x = 15x subject to Ax ≤ b, where x = hx1 i. From the inequality constraint in the primal problem we know that x1 ≤ 1/2. Thus the maximum value of 15x1 is 15/2. Thus the minimum value of bT y = 15/2. According to Theorem 4.3 this minimum occurs when y = h0, 15 2 , 0i. (7) Let vector b = h0, 7, 9, 0i and y = hy1 , y2 , y3 , y4 i. The problem as stated can be thought of as the dual problem:     1110 5 T T minimize b y subject to A y = y≥ = c. 0112 1 This is equivalent to the primal problem: maximize cT x = 5x1 + x2

subject to Ax ≤ b.

The constraint inequality for the primal problem is equivalent to the following system of linear inequalities. x1 ≤ 0

x1 + x2 ≤ 7 x1 + x2 ≤ 9 2x2 ≤ 0

The first and last of these inequalities imply strict inequality in the second and third. Thus by Theorem 4.3 y2 = y3 = 0. Therefore the dual problem can be restated as the following:      10 y1 5 minimize h0, 0i · hy1 , y4 i subject to ≥ . 02 y4 1 Thus the minimum value of the cost function is 0 and occurs where y = h5, 0, 0, 1/2i. (8) A feasible solution to a standard linear program is a non-negative vector, in this case x = hx1 , x2 i ≥ h0, 0i. However, plotting the constraint inequalities 3x1 + 5x2 ≤ 4

−2x1 + 4x2 ≤ −3 on the same set of axes reveals there is no vector with non-negative components which satisfies both inequalities simultaneously.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

378

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(9) The standard linear program is feasible since the region in the nonnegative quadrant of the x1 x2 -plane between the lines x2 = 2x1 − 3

and x2 =

1 3 x1 − 2 2

is non-empty and satisfies the inequality constraints. This feasible region is unbounded. For every k the line 2x1 − 3x2 = k passes through the feasible region, thus the cost function has no maximum and the linear program has no optimal solution. (10) The dual is the linear program: minimize −3y1 + 3y2 subject to the inequalities: −2y1 + y2 ≥ 2

y1 − 2y2 ≥ −3.

According to clause (iii) of Th. 4.4 the dual problem will have no feasible solutions. Plotting the system of inequalities for the dual reveals there is no ordered pair in the positive quadrant of the y1 y2 plane which satisfies both inequalities. Thus the dual problem has no feasible solution in accordance with the Duality Theorem. (11) In this case we can introduce a slack variable x4 to convert the inequality constraint in the problem to an equality constraint. Thus we want to think of the linear problem as: minimize x1 + x2 + x3 subject to 2x1 + x2 = 4 and x3 + x4 = 6 with xi ≥ 0 for i = 1, 2, 3, 4. If we define the following vectors and matrix   2100 c = h1, 1, 1, 0i, x = hx1 , x2 , x3 , x4 i, A = , 0011

and b = h4, 6i,

then the linear problem is the one of minimizing cT x subject to Ax = b and x ≥ 0. The dual to this problem is the one of maximizing bT y subject to AT y ≤ c where y = hy1 , y2 i. This is equivalent to maximizing 4y1 + 6y2 subject to the system of inequalities 2y1 ≤ 1 y1 ≤ 1 y2 ≤ 1

y2 ≤ 0. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

379

(12) Paying attention to the inequality constraints of the linear problem indicates that y1 ≤ 1/2 and y2 ≤ 0. Thus by Theorem 4.3, x2 = x3 = 0. We can also see that the maximum of bT y will be 2. This is also the minimum of the dual problem. However, we can also solve the dual problem directly after re-writing it as      20 x1 4 minimize h1, 0i · hx1 , x4 i subject to = . x4 01 6 From the constraint equation we see that now x = h2, 0, 0, 6i and the minimum value of the cost function is 2. (13) Written in matrix/vector form this linear problem becomes: maximize bT y = h2, 0, 4i · hy1 , y2 , y3 i subject to     y1   1 1 0 1 T   A y= = c. y2 ≤ 012 1 y3

Thus the dual problem is

minimize cT x = x1 + x2 subject to     2 10   x Ax =  1 1  1 =  0  = b. x2 4 02

(14) We can see from the constraint equation for the dual problem that x = hx1 , x2 i = h2, 2i. Thus the minimum value of the cost function is cT x = h1, 1i · h2, 2i = 4. Therefore the maximum of the dual problem is likewise 4. Since x1 + x2 = 4 > 0 then y2 = 0 by Theorem 4.3 and we may re-write the dual problem as “maximize 2y1 + 4y3 = 4 subject to y1 ≤ 1 and 2y3 ≤ 1.” This maximum occurs at y = h1, 0, 1/2i. (15) Assume that a < b and suppose a ≥ 0 and b ≤ 0. Then b ≤ 0 ≤ a which implies b ≤ a a contradiction. (16) By the Duality Theorem (Th. 4.4) and Eq. (4.6) we have ˆ = bT y ˆ=x ˆ AT y ˆ. cT x ˆ ) = cT x ˆ = bT y ˆ . Consequently Thus φ(ˆ x, y ˆ ) = cT x + bT y ˆ − xT AT y ˆ φ(x, y

ˆ ) + (c − AT y ˆ )T x = φ(ˆ x, y

ˆ) ≤ φ(ˆ x, y 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

380

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

ˆ ≥ c since y ˆ is feasible for the dual problem. since x ≥ 0 and AT y Likewise ˆ + bT y − x ˆ T AT y φ(ˆ x, y) = cT x

ˆ ) + (b − Aˆ = φ(ˆ x, y x)T y ˆ) ≥ φ(ˆ x, y

ˆ is feasible for the primal problem. since y ≥ 0 and Aˆ x ≤ b since x ˆ ) is a saddle point of φ(x, y) then (17) If (ˆ x, y ˆ ) ≤ φ(ˆ ˆ) φ(x, y x, y ˆ is optimal for the primal problem. for all x ≥ 0 which implies x Similarly ˆ) φ(ˆ x, y) ≥ φ(ˆ x, y ˆ is optimal for the dual problem. for all y ≥ 0 which implies y (18) Let x be the number of knives made and y be the number of scissors made. The problem is one of maximizing 2.25x + 3.50y subject to the constraints 0≥x 0≥y

2400 ≥ 15x + 25y 832 ≥ 6x + 8y.

The level sets of the function L(x, y) = 2.25x+3.50y intersect a vertex of the feasible region at (x, y) = (160/3, 64). Thus the maximum profit is $344. (19) Let x be the number of acres of corn planted and y be the number of acres of saw grass planted. The problem requires the maximization of 100x + 80y subject to the constraints 0≥x 0≥y

440 ≥ x + y

50000 ≥ 75x + 50y

30000 ≥ 110x + 30y. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

381

Solutions to Chapter Exercises

The level sets of the function L(x, y) = 100x + 80y intersect a vertex of the feasible region at (x, y) = (210, 230). Thus the maximum profit is $39,400. (20) Assign variables as follows. x1 : x2 : x3 : x4 :

number number number number

of of of of

7-inch tablet PCs made in China. 10-inch tablet PCs made in China. 7-inch tablet PCs made in Japan. 10-inch tablet PCs made in Japan.

The problem is that of finding the maximum of 65x1 + 85x2 + 65x3 + 85x4 subject to the constraints below. x1 ≥ 0 x2 ≥ 0 x3 ≥ 0 x4 ≥ 0

x1 + x2 ≤ 250

175x1 + 250x2 ≤ 50000 x3 + x4 ≤ 290

205x3 + 255x4 ≤ 60000 x1 + x3 ≤ 250

x2 + x4 ≤ 290 B.5

Random Walks and Brownian Motion

(1) Using the definition of the increments of the symmetric random walk " k # k2 2 X X E [Sk2 ,k1 ] = E Xi = E [Xi ] = 0 i=k1 +1

V (Sk2 ,k1 ) = V

k2 X

i=k1 +1

i=k1 +1

Xi

!

=

k2 X

i=k1 +1

V (Xi ) =

k2 X

k1 +1

1 = k2 − k1 .

The independence of the Bernoulli random variables Xi was used in the second equation. (2) Since there are initially a total of two marbles in the urn, at the nth iteration there will be n + 2 marbles in the urn. Let Bn denote the cumulative number of new black marbles added to the urn by the

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

382

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

completion of the nth iteration (so that B0 = 0). The fraction of black marbles at the nth iteration will be Mn = (Bn + 1)/(n + 2). The probability of choosing a black marble will be (Bn + 1)/(n + 2) and the probability of choosing a white marble will be 1 − (Bn + 1)/(n+ 2). Assuming B(n) = k then ( k+1 k + 1 with probability n+2 , and B(n + 1) = k+1 k with probability 1 − n+2 . Therefore E [Mn+1 | B1 B2 · · · Bn ]   Bn+1 + 1 =E | B1 B2 · · · Bn n+3 1 E [Bn+1 + 1 | B1 B2 · · · Bn ] = n+3 1 = (1 + E [Bn+1 | B1 B2 · · · Bn ]) n+3 1 = (1 + E [Bn+1 | Bn ]) n+3      k+1 k+1 1 = 1 + (k + 1) +k 1− n+3 n+2 n+2   1 (k + 1)(n + 3) = n+3 n+2 k+1 = n+2 = Mn . In the same manner we may show that E [Mn | B1 · · · Bn−1 ] = Mn−1 . Consequently E [Mn+1 | B1 B1 · · · Bk ] = Mk for all k ≤ n+1. Therefore the fraction of black marbles in the urn has the martingale property. (3) If f is a continuous, twice differentiable function whose second derivative is 0 on the interval [0, A], then f must be a linear function of the form f (x) = ax + b where a and b are constants. Since f (0) = 0 then b = 0. Since f (A) = 1 then a = 1/A. Thus we have f (x) = x/A. This agrees with our earlier derivation of exit probabilities for the discrete, symmetric random walk. (4) If g is a continuous, twice differentiable function whose second derivative is −2 on the interval [0, A], then g must be a quadratic function of the form g(x) = ax2 + bx + c where a, b, and c are constants.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

383

Since g(0) = 0 then c = 0. Since g 00 (x) = −2 then a = −1 which implies g(x) = x(b − x). Since g(A) = 0 then b = A. Thus we have g(x) = x(A − x). This agrees with our earlier derivation of the conditional stopping times for the discrete, symmetric random walk. (5) If f (x) is four times continuously differentiable at x = x0 then we can write f (x) as f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) + f 00 (x0 ) + f (4) (z)

(x − x0 )2 (x − x0 )3 + f 000 (x0 ) 2 6

(x − x0 )4 24

where z lies between x and x0 . Hence h2 h3 h4 + f 000 (x0 ) + f (4) (z1 ) 2 6 24 2 3 4 h h h f (x0 − h) = f (x0 ) − f 0 (x0 )h + f 00 (x0 ) − f 000 (x0 ) + f (4) (z2 ) . 2 6 24 f (x0 + h) = f (x0 ) + f 0 (x0 )h + f 00 (x0 )

Adding the two equations and solving for f 00 (x0 ) yields f 00 (x0 ) =

 h2 f (x0 + h) − 2f (x0 ) + f (x0 − h)  (4) (4) − f (z ) + f (z ) . 1 2 h2 24

Thus when h is small, f 00 (x0 ) ≈

f (x0 + h) − 2f (x0 ) + f (x0 − h) . h2

(6) This is an example of a symmetric random walk. Because of the spatial homogeneity property of the random walk, the desired probability is equivalent to determining the probability that a symmetric random walk initially in state S(0) = 50 will attain a value of 75 while avoiding the absorbing boundary at 0. Using Theorem 5.4, the probability is P75 (50) = 50/75 = 2/3. (7) As in exercise (6) the exit time is equivalent to the exit time of a symmetric random walk on the discrete interval [0, 75] which begins in the state S(0) = 50. Using Theorem 5.5, the probability is Ω0,75 (50) = 50(75 − 50) = 1250. (8) As in exercise (6) the conditional exit time is equivalent to the conditional exit time of a symmetric random walk on the discrete interval [0, 75] which begins in the state S(0) = 50. Using Eq. 5.11 with A = 75

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

384

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

and i = 50 we have Ω75 (50) =

1 (752 − 502 ) = 3125/3 ≈ 1041.67. 3

(9) Let P = P (0)eµt , then  d d µt  dP = P (0)eµt = P (0) e dt dt dt d = P (0)eµt (µt) = P (0)eµt µ = µP. dt (10) Various results are possible due to the fact that we are simulating a stochastic process. Using the data collected in Appendix A, the closing stock price on the last day for which data was collected was $42.86. From the data the estimated drift parameter and volatility are µ = −0.000555 day−1 and σ = 0.028139 day−1 respectively. If we substitute these values in Eq. (5.45) and set P (t0 ) = 42.86 with ∆t = 1 we can iteratively generate 252 (or any other amount for that matter) new days’ closing prices. The random walk generated is pictured below.

P 65 60 55 50 45 50

100

150

200

250

t

(11) From the website finance.yahoo.com the closing prices of stock for Continental Airlines, Inc. (symbol CAL) were captured for the time period of 06/15/2004 until 06/15/2005. A histogram of ∆X = ln P (ti+1 ) − ln P (ti ) is shown below. The mean and standard deviation of ∆X are µ∆X = 0.00121091 day−1 and σ∆X = 0.0332183 day−1 respectively.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

385

Solutions to Chapter Exercises

60 50 40 30 20 10 -0.1

-0.05

0

0.05

0.1

(12) Various results are possible due to the fact that we are simulating a stochastic process. The closing stock price on the last day for which data was collected was $10.21. From the data the estimated drift parameter and volatility are µ = 0.00121091 day−1 and σ = 0.0332183 day−1 respectively. If we substitute these values in Eq. (5.45) and set P (t0 ) = 10.21 with ∆t = 1 we can iteratively generate 252 (or any other amount for that matter) new days’ closing prices. The random walk generated is pictured below.

P

13 12 11 10

50

14:58:13.

100

150

200

250

t

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

386

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(13) If F ≡ F (y, z) and f (x) = F (y0 + xh, z0 + xk) then dy dz + Fz (y0 + xh, z0 + xk) dx dx d = Fy (y0 + xh, z0 + xk) [y0 + xh] dx d + Fz (y0 + xh, z0 + xk) [z0 + xk] dx = Fy (y0 + xh, z0 + xk)h + Fz (y0 + xh, z0 + xk)k

f 0 (x) = Fy (y0 + xh, z0 + xk)

f 0 (0) = Fy (y0 , z0 )h + Fz (y0 , z0 )k. Differentiating once again produces dz dy + Fyz (y0 + xh, z0 + xk)h dx dx dy dz + Fzy (y0 + xh, z0 + xk)k + Fzz (y0 + xh, z0 + xk)k dx dx d = hFyy (y0 + xh, z0 + xk) [y0 + xh] dx d + hFyz (y0 + xh, z0 + xk) [z0 + xk] dx d [y0 + xh] + kFzy (y0 + xh, z0 + xk) dx d + kFzz (y0 + xh, z0 + xk) [z0 + xk] dx = h2 Fyy (y0 + xh, z0 + xk) + 2hkFyz (y0 + xh, z0 + xk)

f 00 (x) = Fyy (y0 + xh, z0 + xk)h

+ k 2 Fzz (y0 + xh, z0 + xk) f 00 (0) = h2 Fyy (y0 , z0 ) + 2hkFyz (y0 , z0 ) + k 2 Fzz (y0 , z0 ). By definition the Taylor remainder of order 3 for f (x) is R3 (x) = f 000 (α) 3 x for some α between 0 and x. If we notice the binomial 3! coefficient pattern among the partial derivatives in the first and second derivatives of f (x) then we can readily see that f 000 (x) = h3 Fyyy (y0 + xh, z0 + xk) + 3h2 kFyyz (y0 + xh, z0 + xk) + 3hk 2 Fyzz (y0 + xh, z0 + xk) + k 3 Fzzz (y0 + xh, z0 + xk). From this expression we may directly determine the Taylor remainder of order 3.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

387

Solutions to Chapter Exercises

(14) If we apply Itˆ o’s Lemma with a(P, t) = µP and b(P, t) = σP and Y = P n , then   1 dY = µP nP n−1 + (σP )2 n(n − 1)P n−2 dt + σP nP n−1 dW (t) 2   n(n − 1)σ 2 n dt + nσP n dW (t). P = nµP n + 2 (15) If we apply Itˆ o’s Lemma with a(P, t) = µP and b(P, t) = σP and Y = ln P , then        1 1 1 2 −1 dY = µP + (σP ) dW (t) dt + σP P 2 P2 P   σ2 = µ− dt + σ dW (t). 2 (16) Z ∞ 2 2 1 √ x4 e−x /2σ dx σ 2π −∞ Z ∞   2 2 1 x3 xe−x /2σ dx = √ σ 2π −∞

  E X4 =

If we use integration by parts with u = x3 du = 3x2 dx

2

2

v = −σ 2 e−x /2σ 2 2 dv = xe−x /2σ dx

then the integral for the expected value above becomes Z ∞ 2 2 1  2 3 −x2 /2σ2  ∞ 3σ 2 √ x2 e−x /2σ dx −σ x e + √ −∞ σ 2π σ 2π −∞ Z ∞ 2 2 3σ 2 = √ x2 e−x /2σ dx σ 2π −∞ Z ∞   2 2 3σ 2 = √ x xe−x /2σ dx. σ 2π −∞

  E X4 =

Again using integration by parts with u=x du = dx

14:58:13.

2

2

v = −σ 2 e−x /2σ 2 2 dv = xe−x /2σ dx

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

388

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

then the integral for the expected value above becomes   Z ∞   2 2 3σ 2  2 −x2 /2σ2  ∞ + E X4 = √ −σ e σ 2 e−x /2σ dx −∞ σ 2π −∞   Z ∞ 2 2 1 √ e−x /2σ dx = 3σ 4 σ 2π −∞ = 3σ 4 . (17) Let 0 ≤ s < t then h i E [Z(t) | Z(τ ) 0 ≤ τ ≤ s] = E eW (t)−t/2 | W (τ ) 0 ≤ τ ≤ s h i = E eW (t)−W (s) eW (s)−t/2 | W (τ ) 0 ≤ τ ≤ s h i = eW (s)−t/2 E eW (t)−W (s) | W (τ ) 0 ≤ τ ≤ s . The last equality holds since t/2 is a deterministic (non-random) quantity and at time τ = s, the value of W (s) is known. Recall that the increment W (t) − W (s) is normally distributed with a mean of zero and a variance of t − s. Thus eW (t)−W (s) is lognormal with a mean of e(t−s)/2 by Lemma 3.1. Thus E [Z(t) | Z(τ ) 0 ≤ τ ≤ s] = eW (s)−t/2 e(t−s)/2 = eW (s)−s/2 = Z(s), and consequently the exponential random process Z(t) is a martingale. (18) Start by expanding 2

n X

k=1

=2

W (tk−1 ) [W (tk ) − W (tk−1 )]

n X

k=1

=2

n X

k=1

W (tk )W (tk−1 ) − 2 W (tk )W (tk−1 ) −

= W 2 (t) − = W 2 (t) −

n+1 X

k=1 n+1 X

n X

W 2 (tk−1 )

k=1

n X

k=1

W 2 (tk−1 ) + 2

W 2 (tk−1 ) − n X

k=1

W 2 (tk−1 ) + 2

k=2

14:58:13.

n X

k=1

n X

W 2 (tk−1 )

k=1

W (tk )W (tk−1 ) − W (tk )W (tk−1 ) −

n X

k=1 n X

k=1

W 2 (tk−1 ) W 2 (tk−1 )

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

389

Solutions to Chapter Exercises

= W 2 (t) − = W 2 (t) −

n X

k=1 n X

k=1

W 2 (tk ) + 2

n X

k=1

BC8495/Chp. B

W (tk )W (tk−1 ) −

n X

W 2 (tk−1 )

k=1

(W (tk ) − W (tk−1 ))2

which when divided on both sides by 2 yields the desired result. (19) If P is a partition of [0, T ] then 0 ≤ [f, f ] T n X = lim (f (ti ) − f (ti−1 ))2 kP k→0

= lim

kP k→0

= lim

kP k→0

i=1

n X

i=1 n X i=1

≤ M 2 lim

kP k→0

≤ M 2 lim

kP k→0

(f 0 (si )(ti − ti−1 ))

2

(by MVT)

2

(f 0 (si )) (ti − ti−1 )2 n X i=1

"

(ti − ti−1 )2

kP k

n X i=1

#

(ti − ti−1 )

= M 2 T lim kP k = 0. kP k→0

The quantity M is the maximum of the absolute value of f 0 (t) on the interval [0, T ]. By the Squeeze Theorem, [f, f ] T = 0. (20) According to Theorem 5.8  Z 2 3 t dW (t) = 0 E V

Z

0 2

0

 Z t3 dW (t) =

0

2

t6 dt =

128 . 7

(21) Let 0 ≤ s ≤ t then E [Z(t) | Z(τ ), 0 ≤ τ ≤ s] = E [Z(t) | W (τ ), 0 ≤ τ ≤ s]

= E [µt + σW (t) | W (τ ), 0 ≤ τ ≤ s] = E [µt | W (τ ), 0 ≤ τ ≤ s]

+ E [σW (t) | W (τ ), 0 ≤ τ ≤ s]

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

390

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

= µt + σW (s) 6= µs + σW (s) = Z(s) in general. (22) The limit     2 2 M e2µm/σ − 1 + m e−2µM/σ − 1  lim E [TB ] = lim µ→0 µ→0 µ e2µm/σ2 − e−2µM/σ2

is indeterminate of the form 0/0. Applying l’Hˆopital’s Rule lim E [TB ]

µ→0



 2 2 e2µm/σ − e−2µM/σ   = lim 2µm/σ2 2µm/σ2 + M e−2µM/σ2 µ→0 e − e−2µM/σ2 + 2µ σ2 me   4mM 2µm/σ2 −2µM/σ2 me + M e 4 σ   = lim 4 2 2µm/σ2 − M 2 e−2µM/σ2 µ→0 2 me2µm/σ2 + M e2µM/σ2 + 4µ σ σ4 m e 2mM σ2

=

4mM σ4 (m + M ) 4 σ2 (m + M )

=

mM . σ2

(23) The standard Brownian motion process obeys the stochastic differential equation dW (t) = (0) dt + (1) dW (t). Applying Itˆ o’s Lemma with Y (t) = 1 yields

W (t) , a(W, t) = 0, and b(W, t) = 1 + t2



 (0)(1) 2tW (t) 1 1 dY = − + (0) dt + dW (t) 2 2 2 1+t (1 + t ) 2 1 + t2 2tY (t) 1 =− + dW (t). 2 1+t 1 + t2 (24) Mean reverting Ornstein-Uhlenbeck equation:

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

391

Solutions to Chapter Exercises

(a) Let a(X, t) = k(µ − X(t)) and b(X, t) = σ and note that ∂Z = −ekt ∂X ∂2Z =0 ∂X 2 ∂Z = kekt (µ − X(t)). ∂t Applying Itˆ o’s Lemma yields   dZ(t) = k(µ − X(t))(−ekt ) + kekt (µ − X(t)) + 0 dt + σ(−ekt ) dW (t)

= −σekt dW (t). (b) Integrating both sides of the last equation yields Z

t

dZ(s) = −σ

0

Z(t) − Z(0) = −σ

Z

t

eks dW (s)

0

Z

t

eks dW (s)

0

Z

t

eks dW (s) Z t = µ − X(0) − σ eks dW (s).

Z(t) = Z(0) − σ

0

0

(c) Since X(t) = µ − e

−kt

Z(t) then   E [X(t)] = E µ − e−kt Z(t)

= E [µ] − e−kt E [Z(t)]

= µ − e−kt (µ − X(0))  V (X(t)) = V µ − e−kt Z(t)

n+1

(25) Let Y (t) = (W (t)) ential equation

= e−2kt V (Z(t))  σ2 = 1 − e−2kt . 2k

, then since W (t) satisfies the stochastic differ-

dW (t) = (0) dt + 1 dW (t),

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

392

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

if we let a(W, t) = 0 and b(W, t) = 1 and apply Itˆo’s Lemma, then Y (t) must satisfy the stochastic differential equation,   1 n−1 dY (t) = 0 + 0 + n(n + 1) (W (t)) dt 2 + (n + 1) (W (t))n dW (t)   1 n+1 n−1 n dt + (n + 1) (W (t)) dW (t). d (W (t)) = n(n + 1) (W (t)) 2 Integrating both sides of the equation and using the assumption that W (0) = 0 then we have Z t   Z t1 n−1 n+1 d (W (s)) = n(n + 1) (W (s)) ds 2 0 0 Z t n + (n + 1) (W (s)) dW (s) 0 Z t n n+1 n+1 n−1 (W (t)) − (W (0)) = (n + 1) (W (s)) ds 2 0 Z t n (W (s)) dW (s) + (n + 1) 0 Z t Z t n 1 n+1 n−1 n (W (t)) = (W (s)) ds + (W (s)) dW (s) n+1 2 0 0 Z t Z 1 n t n n+1 n−1 (W (s)) dW (s) = (W (t)) − (W (s)) ds. n+1 2 0 0

(26) Applying Itˆ o’s Lemma we get   1 1 dZ = − [f (t)]2 e−X(t) + 0 + [f (t)]2 e−X(t) dt 2 2   + f (t) −e−X(t) dW (t) = −f (t)Z(t) dW (t).

(27) Using the stochastic form of the product rule, dY = X2 dX1 + X1 dX2 + dX1 dX2 = X2 (a1 (t)X1 dt + b1 (t)X1 dW (t)) + X1 (a2 (t)X2 dt + b2 (t)X2 dW (t)) + (a1 (t)X1 dt + b1 (t)X1 dW (t))(a2 (t)X2 dt + b2 (t)X2 dW (t)) = (a1 (t) + a2 (t) + b1 (t)b2 (t))Y dt + (b1 (t) + b2 (t))Y dW (t),

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

393

Solutions to Chapter Exercises

since (dt)2 = dt dW (t) = 0 and (dW (t))2 = dt. (28) To solve the stochastic logistic equation: (a) Applying Itˆ o’s Lemma we derive  1 rP (K − P ) 2 + 0 + P   α2 = −r(1 − K/P ) − P

dY =

   1 2 2 −2 1 α P dt + αP dW (t) 2 P3 P2 1 dt + α dW (t) P

= [(α2 − rK)Y − r] dt − αY dW (t). (b) Once again applying Itˆo’s Lemma we derive     1 2 2 −1 1 1 2 +0+ α X dt − αX dW (t) dZ = (α − rK)X X 2 X2 X   1 2 = α − rK dt − α dW (t). 2 The initial condition for this differential equation is Z(0) = 0. Thus the solution can be written as Z(t) =



 1 2 α − rK t − αW (t). 2

Since X(t) = eZ(t) then 2

X(t) = e(α

/2−rK)t−αW (t)

.

(c) Yet another application of Itˆo’s Lemma yields   −1 1 2 2 2 2 dY0 = (α − rK)X 2 + 0 + α X dt X 2 X3   −1 + −αX 2 dW (t) X  = −(α2 − rK)Y0 + α2 Y0 dt + (αY0 ) dW (t) = rKY0 dt + αY0 dW (t).

(d) Since Z(t) = Y0 (t)Y (t) then using the result of stochastic form

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

394

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

the product rule dZ = (rKY0 dt + αY0 dW (t))Y    + (α2 − rK)Y − r dt − αY dW (t) Y0   + (rKY0 dt + αY0 dW (t)) (α2 − rK)Y − r dt − αY dW (t))

= (rKZ + (α2 − rK)Z − rY0 − α2 Z) dt + (αZ − αZ) dW (t) = −rY0 dt.

(e) Since Y (0) = X(0)Z(0) = Z(0) = −1/P0 then P (t) =

−1 h i Rt e(α2 /2−rK)t−αW (t) − P10 − r 0 e(rK−α2 /2)s+αW (s) ds 2

P0 e(rK−α /2)t+αW (t) = . Rt 1 + rP0 0 e(rK−α2 /2)s+αW (s) ds B.6

Forwards and Futures

(1) Let the bid price be B, then the ask price is A = B + 0.25. The net cost to the investor is 1000B − 1000A = −250. The round trip cost is $250. (2) Suppose the buyer pays the seller S < S(0), then the new owner may then instantly sell the stock for S(0) guaranteeing a profit of S(0) − S > 0. On the other hand the seller initially possesses a portfolio worth S(0). If the buyer is willing to pay the seller S > S(0), then the seller may borrow the stock from another investor and sell it to the buyer for S. Then the buyer immediately re-purchases the stock for S(0) and returns it to the other investor. The seller is left with a profit of S − S(0) > 0. (3) The future value of $17 one month hence at continuously compounded interest rate 5.05% is F = 17e0.0505/12 ≈ 17.0717. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

(4) The forward contract is worth F = 23e0.0475(3/12) ≈ 23.2748. (5) The value of the prepaid forward is F = 97 − 2.50e−0.0365(6/12) − 2.75e−0.0365(12/12) ≈ 91.8938. The value of a forward contract on the dividend paying stock is F = 91.8938e0.0365(12/12) ≈ 95.3099. (6) The value of three-month prepaid forward on the investment is F = 195e−0.0195(3/12) ≈ 194.052. The value of a three-month forward contract on the investment is F = 195e(0.0455−0.0195)(3/12) ≈ 196.272. (7) The equation 990 = 1000e(0.0505−r)(1/2) implies r ≈ 0.0706. Thus the continuous dividend rate is 7.06%. (8) Using inequality (6.1) (74 − 2(2))e0.03(6/12) ≤ F ≤ (75 + 2(2))e0.04(6/12) 71.0579 ≤ F ≤ 80.5959.

(9) Let S = 80, Di = 2 for i = 1, 2, 3, 4 and r = 0.045. (a) Prepaid forward contract: F = 80 −

4 X i=1

2e−(0.045)(i/4) ≈ 72.2213.

(b) Forward contract: F = 80e0.045 −

4 X i=1

2e−(0.045)((i/4)−1) ≈ 75.5454.

(10) Let S = 75, r = 0.0475, and d = 0.08. (a) Prepaid forward contract: F = 75e−0.08 ≈ 69.2337. 14:58:13.

BC8495/Chp. B

395

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

396

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(b) Forward contract: F = 75e0.0475−0.08 ≈ 72.6017. (11) Let S = 85, r = 0.0625, and F = 89. F = Se(r−d)T 89 = 85e0.0625−d d = 0.0625 − ln

89 ≈ 0.0165149 85

(12) The initial margin deposit is (875)(1000)(0.25) = $218, 750. After one week, suppose the price of the contracts is now S, the margin balance will be 218750e0.055(7/365) + 1000(S − 875). A margin call is triggered if this balance is smaller than 80% of $218,750 which is $175,000. 218750e0.055(7/365) + 1000(S − 875) < 175, 000

1000(S − 875) < −43974.68866

S − 875 < −43.97469866 S < 831.0253013

Thus the greatest price which will trigger a margin call is $831.02. (13) The initial margin is in the amount of (100)(950)(0.125) = 11875, which after one month of interest compounded continuously at 6% per annum will be 11934.52. A margin call will be issued only if the price of the security at the one-month mark satisfies 11934.52 + (S(1) − 950)(100) ≤ S(1)(100)(0.125) S(1) ≤ 949.32.

(14) If the current price of the stock is S(0) then the value of the forward contract is S(0)erT . However, the price of a forward contract is not paid until it matures. Hence the present value of the forward contract

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

397

Solutions to Chapter Exercises

is S(0)erT e−rT = S(0). If this amount is lent at interest rate r continuously compounded for time T , then the payoff of entering into the long forward contract and lending the present value of the contract is S(T ) − S(0)erT + S(0)ert − S(0) = S(T ) − S(0). | {z } | {z } forward loan

If the stock is purchased at t = 0 and sold at time t = T the payoff is S(T ) − S(0). Hence the two investments have the same payoff. (15)

Day 0 1 2 3 4 5 6 7 8 9 10

No. of Contracts 1500 1500 1500 1500 1500 1500 1500 1500 1500 1500 1500

Futures Price 850.00 774.67. 779.39 778.42 749.56 742.87 735.64 741.59 759.88 766.25 805.36

Price Change — −75.33 4.73 −0.97 −28.86 −6.69 −7.23 5.95 18.29 6.38 39.11

Margin Balance 191, 250.00 78, 302.40 181, 438.67 180, 027.86 136, 791.47 158, 659.45 156, 345.58 174, 489.36 201, 965.75 211, 583.59 270, 308.23

Margin Call — 95, 997.60 0.00 0.00 31, 860.23 8, 486.48 9, 173.43 0.00 0.00 0.00 0.00

The profit to the holder of the long position in the futures contract is the difference between the final margin balance and the future value of the initial margin. 270, 308.23 − 191, 250e10(0.10)/365 = 78, 533.54 (16) The price of the forward contract should be the future value of the copper and its storage costs. Thus F =e

0.035(0.5)

"

4.10 +

5 X i=0

0.10e

−0.035(i/12)

#

≈ 4.77855.

(17) Consider the vector of differences between the forward prices given in

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

398

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

the table and the formula for the forward price. 

  4.0945 − c −  x=  4.1215 − c −  4.1235 − c −



P1

−ri/12 i=0 e  P2 0.0053 i=0 e−ri/12    P3 −ri/12  0.0053 i=0 e  P6 0.0053 i=0 e−ri/12

4.0995 − c − 0.0053

The symbol c represents the price of copper at the beginning of June and r is the annual interest rate. The best fitting values for c and r will be the values which minimize the magnitude of x or equivalently minimize f (c, r) = xT x. These values can be found numerically to be c ≈ 4.0887 and r ≈ 0.0366. (18) The present value is P = 1500(4.0995 − 1.25)e−0.03(2) + 1575(4.0945 − 1.25)e−0.03(3)

+ 1625(4.1215 − 1.25)e−0.03(4) + 1500(4.1235 − 1.25)e−0.03(7)

≈ 15752.20.

(19) The delivery price of the bond is the future value of the selling price minus the present value of its dividends. "

F = 1050 −

50 1+

 − 0.0325 2 2

50 1+

 0.0275 2

#

1+

0.0325 2

2

≈ 983.464

The current value of the investor’s forward contract is the present value of the difference between the delivery price of the forward and the delivery price of $1000. 983.464 − 1000 2 ≈ −16.011 1 + 0.0325 2

(20) In the absence of arbitrage, the future value of a euro earning continuously compounded interest at rate re should be the same as the value of an equivalent exchange of euros for dollars earning continuously compounded interest at rate r. At time T the future value of 1 euro will be ere T . If the euro is exchanged for dollars and then begins to earn interest, its future value will be SerT . Applying the forward exchange rate to convert the dollars back to euros yields, (S/F )erT .

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

399

Solutions to Chapter Exercises

Hence, S rT e F F = Se(r−re )T .

e re T =

B.7

Options

(1) Suppose the value of an American put option is P a and that P a < P e , the value of a European put option on the same underlying security with identical strike price and exercise time. An investor could sell the European put option and buy the American put option and invest the net positive cash flow of P e − P a in a risk-free bond at interest rate r. At expiry, in either case whether the owner of the European put exercises the option or not, and the investor does likewise, the investor has a portfolio with value (P e − P a )erT > 0. (2) The net payoff will be the quantity (S(T ) − K)+ − C where K = 60 and C = 10. The payoff is plotted below. payoff 30

20

10

20

40

60

80

100

SHTL

-10

(3) We have already established that C e ≤ C a for European and American style options. Thus we need only show that C a (t) ≤ S(t) for 0 ≤ t ≤ T , where the expiry time is T . Let the strike price be K. If we suppose there is a time t∗ at which C a (t∗ ) > S(t∗ ) then an investor could create a portfolio consisting of a short call and a long position in one share of the security. This portfolio generates a positive cash flow to the investor of C a (t∗ )−S(t∗ ) which will be placed in a risk-free

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

400

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

savings account earning interest at rate r compounded continuously. For any time t∗ ≤ t ≤ T the investor can close out their long position in the security and make a profit of ∗

(C a (t∗ ) − S(t∗ )) er(t−t ) + min{K, S(t)} > 0. (4) Since C e ≥ S − Ke−rT then when S = 29, K = 26, r = 0.06, and T = 1/4, C e ≥ 29 − 26e−(0.06)(0.25) ≥ 3.39.

(5) Here S = 31, T = 1/4, C e = 3, K = 31, and r = 0.10 then according to the Put-Call Parity formula in Eq. (7.1): P e + S = C e + Ke−rT P e + 31 = 3 + 31e−(0.10)(0.25) P e = 2.23. (6) Here S = 31, T = 1/4, C e = 3, P e = 2.25, and r = 0.10 then according to the Put-Call Parity formula in Eq. (7.1): P e + S = C e + Ke−rT 2.25 + 31 = 3 + Ke−(0.10)(0.25) K = 31.02. (7) Here T = 1/6, K = 14, S = 11, and r = 0.07 then according to the Put-Call Parity formula in Eq. (7.1): P e = C e − S + Ke−rT ≥ −S + Ke−rT

= −11 + 14e−(0.07)/6 = 2.84.

(8) An investor could take a long position in the put option and the security and short position in the call option while borrowing, at the risk-free rate, the strike price of the options. This generates an initial cash flow of K + C e − P e − S = 30 + 3 − 1 − 31 = $1. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

401

This amount could be invested at the risk-free rate until the strike time arrives. At t = 3 months if S(3) > 30 = K, then the call option will be exercised and the investor cancels all positions with a portfolio worth 1e(0.10)(0.25) + 30 − 30e(0.10)(0.25) = $0.27 > 0. If the call option finishes out of the money, the investor exercises the put option and finishes with a portfolio worth again 1e(0.10)(0.25) + 30 − 30e(0.10)(0.25) = $0.27 > 0. (9) Using the Put-Call Parity formula P + 36 = 2.25 + 38e−0.0475(4/12) P ≈ 3.65307. (10) If f1 (S, t) and f2 (S, t) both satisfy Eq. (8.5) and if c1 and c2 are real numbers then (c1 f1 + c2 f2 )t + rS(c1 f1 + c2 f2 )S 1 + σ 2 S 2 (c1 f1 + c2 f2 )SS − r(c1 f1 + c2 f2 ) 2   1 = c1 f1,t + rSf1,S + σ 2 S 2 f1,SS − rf1 2   1 + c2 f2,t + rSf2,S + σ 2 S 2 f2,SS − rf2 2 = c1 (0) + c2 (0) = 0. Thus f = c1 f1 + c2 f2 is a solution to Eq. (8.5) as well. (11) At time t = T the payoff of the European put option will be max{0, K − S(T )} = (K − S(T ))+ , thus F (S, T ) = (K − S(T ))+ . If the stock is worthless (i.e., along the boundary where S = 0) then the put option will be exercised and F (0, t) = Ke−r(T −t) . As the price of the stock increases toward infinity the put option will not be exercised and thus limS→∞ F (S, t) = 0. (12) Suppose K1 < K2 . (a) Suppose C(K1 ) < C(K2 ). Purchase C(K1 ) and sell C(K2 ). This generates a cash flow C(K2 ) − C(K1 ) > 0. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

402

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

case S(t) < K1 < K2 for 0 ≤ t ≤ T : neither option will be exercised and your total profit is C(K2 ) − C(K1 ) > 0. case K1 < S(t) < K2 for 0 ≤ t ≤ T : the owner of C(K2 ) will not exercise their option, but you will exercise C(K1 ) purchasing the stock for K1 and selling it for S(t). Your total profit is C(K2 ) − C(K1 ) + S(t) − K1 > 0. case K1 < K2 < S(t) for 0 ≤ t ≤ T : both options will be exercised. You will purchase the option for K1 and sell it for K2 . Your total profit is C(K2 ) − C(K1 ) + K2 − K1 > 0. This implies arbitrage is present. (b) Suppose P (K2 ) < P (K1 ). Purchase P (K2 ) and sell P (K1 ). This generates a cash flow P (K1 ) − P (K2 ) > 0.

case S(t) < K1 < K2 for 0 ≤ t ≤ T : both options will be exercised. You will purchase the stock for K1 and sell it for K2 . Your total profit is P (K1 ) − P (K2 ) + K2 − K1 > 0. case K1 < S(t) < K2 for 0 ≤ t ≤ T : the owner of P (K1 ) will not exercise their option, but you will exercise P (K2 ) purchasing the stock for S(t) and selling it for K2 . Your total profit is P (K1 ) − P (K2 ) + K2 − S(t) > 0. case K1 < K2 < S(t) for 0 ≤ t ≤ T : neither option will be exercised. Your total profit is P (K1 ) − P (K2 ) > 0.

This implies arbitrage is present. (c) Suppose C(K1 )−C(K2 ) > K2 −K1 > 0. Sell C(K1 ) and purchase C(K2 ). The initial cash flow is C(K1 ) − C(K2 ) > 0. case S(t) < K1 < K2 for 0 ≤ t ≤ T : neither option will be exercised and your total profit is C(K2 ) − C(K1 ) > 0. case K1 < S(t) < K2 for 0 ≤ t ≤ T : the owner of C(K1 ) will exercise their option. You will purchase the stock for S(t) and sell it for K1 . Your total profit is

C(K1 )−C(K2 )+K1 −S(t) > K2 −K1 +K1 −S(t) = K2 −S(t) > 0. case K1 < K2 < S(t) for 0 ≤ t ≤ T : both options will be exercised. You will purchase the stock for K2 and sell it for K1 . Your total profit is C(K1 ) − C(K2 ) + K1 − K2 > K2 − K1 + K1 − K2 = 0. This implies arbitrage is present.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

403

Solutions to Chapter Exercises

(d) Suppose P (K2 )−P (K1 ) > K2 −K1 > 0. Sell P (K2 ) and purchase P (K1 ). The initial cash flow is P (K2 ) − P (K1 ) > 0. case S(t) < K1 < K2 for 0 ≤ t ≤ T : both options will be exercised. You will buy the stock for K2 and sell it for K1 . Your total profit is P (K2 ) − P (K1 ) + K1 − K2 > K2 − K1 + K1 − K2 > 0. case K1 < S(t) < K2 for 0 ≤ t ≤ T : the owner of P (K2 ) will exercise their option. You will purchase the stock for K2 and sell it for S(t). Your total profit is P (K2 ) − P (K1 ) + S(t) − K2 > P (K2 ) − P (K1 ) + K1 − K2 > 0. case K1 < K2 < S(t) for 0 ≤ t ≤ T : neither option will be exercised. Your total profit is P (K2 ) − P (K1 ) > 0.

This implies arbitrage is present.

(13) Given the descriptions of the options and strike prices the following “no arbitrage” arguments can be made. (a) The given inequality is equivalent to C(K2 ) ≤

K3 − K2 K2 − K1 C(K1 ) + C(K3 ). K3 − K1 K3 − K1

For the sake of compactness we will let λ=

K3 − K2 K3 − K1

which implies

Note that 0 < λ < 1. Now suppose

1−λ=

K2 − K1 . K3 − K1

C(K2 ) > λC(K1 ) + (1 − λ)C(K3 ). We can sell C(K2 ) and buy λC(K1 ) calls and (1 − λ)C(K3 ) calls. At time t = 0 we have a cash flow of C(K2 ) − λC(K1 ) − (1 − λ)C(K3 ) > 0. • If S(t) < K1 for the life of the options, then no options will be exercised and we keep our positive profit. • If K1 ≤ S(t) < K2 , then we may exercise the C(K1 ) option profiting λ(S(t)−K1 ). The other options will not be exercised and our total profit is C(K2 ) − λC(K1 ) − (1 − λ)C(K3 ) + λ(S(t) − K1 ) > 0. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

404

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

• If K2 ≤ S(t) < K3 , then we may exercise the C(K1 ) option profiting λ(S(t) − K1 ) and the owner of C(K2 ) will exercise that option costing us K2 − S(t). The remaining option will not be exercised. The balance of the transactions at the time of exercise are λ(S(t) − K1 ) + K2 − S(t)

= λ(S(t) − K1 ) + λK1 + (1 − λ)K3 − S(t) = λS(t) + (1 − λ)K3 − S(t) = (1 − λ)K3 − (1 − λ)S(t) = (1 − λ)(K3 − S(t)) > 0.

Added to our initial positive cash flow, our profit remains positive. • If K3 ≤ S(t), then we may exercise the C(K1 ) option profiting λ(S(t) − K1 ), the owner of C(K2 ) will exercise that option costing us K2 − S(t), and we will exercise the C(K3 ) option earning (1 − λ)(S(t) − K3 ). At the time of exercise our net transaction is λ(S(t) − K1 ) + K2 − S(t) + (1 − λ)(S(t) − K3 ) = K2 − λK1 − (1 − λ)K3 = 0.

We keep our initial positive cash flow. Thus arbitrage is present. (b) Define λ as above and suppose P (K2 ) > λP (K1 ) + (1 − λ)P (K3 ). We can sell P (K2 ) and buy λP (K1 ) puts and (1 − λ)P (K3 ) puts. At time t = 0 we have a cash flow of P (K2 ) − λP (K1 ) − (1 − λ)P (K3 ) > 0. • If S(t) < K1 , then we may exercise the P (K1 ) option profiting λ(K1 − S(t)), the owner of P (K2 ) will exercise that option costing us S(t) − K2 , and we will exercise the P (K3 ) option earning (1 − λ)(K3 − S(t)). At the time of exercise our net 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

405

transaction is λ(K1 − S(t)) + S(t) − K2 + (1 − λ)(K3 − S(t)) = −K2 + λK1 + (1 − λ)K3 = 0.

We keep our initial positive cash flow. • If K1 ≤ S(t) < K2 , then we will not exercise the P (K1 ) option, the owner of P (K2 ) will exercise that option costing us S(t) − K2 , and we will exercise the P (K3 ) option earning us (1 − λ)(K3 − S(t)), The balance of the transactions at the time of exercise are (1 − λ)(K3 − S(t)) + S(t) − K2

= (1 − λ)(K3 − S(t)) + S(t) − λK1 − (1 − λ)K3 = λ(S(t) − K1 ) > 0.

Added to our initial positive cash flow, our profit remains positive. • If K2 ≤ S(t) < K3 , then we will exercise the P (K3 ) option profiting (1 − λ)(K3 − S(t)). The other options will not be exercised and our total profit is P (K2 ) − λP (K1 ) − (1 − λ)P (K3 ) + (1 − λ)(K3 − S(t)) > 0. • If K3 ≤ S(t) for the life of the options, then no options will be exercised and we keep our positive profit. Thus arbitrage is present. (14) Let r = 0.0325, S(0) = 500, P = 40, K = 495, and t = 2/12. (a) max{450, 495} − (500 + 40)e0.0325(2/12) ≈ −47.93 (b) max{550, 495} − (500 + 40)e0.0325(2/12) ≈ 7.07

(15) Let r = 0.0325, S(0) = 498, C = 35, K = 500, and t = 6/12. (a) min{490, 500} + (35 − 498)e0.0325(6/12) ≈ 19.41 (b) min{510, 500} + (35 − 498)e0.0325(6/12) ≈ 29.41

(16) Let r = 0.0295, S(0) = 525, C = 50, K = 530, and t = 2/12. (a) (525 − 50)e0.0295(2/12) − min{530, 500} ≈ −22.66 (b) (525 − 50)e0.0295(2/12) − min{530, 555} ≈ −52.66

(17) Let r = 0.0295, S(0) = 475, P = 45, K = 500, and t = 3/12. (a) (475 + 45)e0.0295(3/12) − max{500, 485} ≈ 23.85 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

406

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(b) (475 + 45)e0.0295(3/12) − max{500, 515} ≈ 8.85 (18) Let r = 0.0375, C(K1 ) = 7.57, K1 = 100, C(K2 ) = 4.75, K2 = 110, and T = 2/12. (a) When S(2) = 98, the profit is (4.75 − 7.57)e0.0375(2/12) ≈ −2.84. (b) When S(2) = 107, the profit is (4.75 − 7.57)e0.0375(2/12) + 107 − 100 ≈ 4.16. (c) When S(2) = 115, the profit is (4.75 − 7.57)e0.0375(2/12) + 110 − 100 ≈ 7.16. (19) Let r = 0.0425, P (K1 ) = 10.25, K1 = 400, P (K2 ) = 15.75, K2 = 425, and T = 3/12. (a) When S(3) = 375, the profit is (10.25 − 15.75)e0.0425(3/12) + 425 − 400 ≈ 19.44. (b) When S(3) = 410, the profit is (10.25 − 15.75)e0.0425(3/12) + 425 − 410 ≈ 9.44. (c) When S(3) = 450, the profit is (10.25−15.75)e0.0425(3/12) ≈ −5.56. (20) Given the premiums charged for the options: (a) The premium to set up the straddle is the sum of the premiums of the put and call with the common strike price of $50. $3.64 + $2.72 = $6.36 (b) The premium to set up the strangle is the sum of the premiums of the put with a strike price of $45 and the call with the strike price of $55. $0.94 + $1.70 = $2.64 (c) The straddle generates more profit when |S(t) − 50| ≥ 2.64

⇐⇒

S(t) ≤ 47.36 or S(t) ≥ 52.64.

(21) The net premium for creating the butterfly spread is 2(2.72) − 0.94 − 5.68 = −1.18. (a) If S(t) = 40, all options are exercised and the profit is (45 − 40) + (55 − 40) − 2(50 − 40) − 1.18 = −1.18. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

407

Solutions to Chapter Exercises

(b) If S(t) = 47, all options except the one with a strike price of $45 are used. The profit is (55 − 47) − 2(50 − 47) − 1.18 = 0.82. (c) If S(t) = 52 only the option with a strike price of $55 is exercised. In this case the profit is (55 − 52) − 1.18 = 1.82. (d) If S(t) = 57 none of the put options are exercised and the profit is −1.18.

(22) The table of payoffs is shown below. Asset Price S(t) < K1 K1 < S(t) < K2 K2 < S(t) < K3 K3 < S(t)

Payoff from Purchased Call 0 0 0 S(t) − K3

Payoff from Purchased Put K1 − S(t) 0 0 0

Payoff from Sold Call 0 0 −(S(t) − K2 ) −(S(t) − K2 )

Payoff from Sold Put −(K2 − S(t)) −(K2 − S(t)) 0 0

Total Payoff K1 − K2 S(t) − K2 K2 − S(t) K2 − K3

Note that K1 − K2 = K2 − K3 under the assumption that K2 = (K1 + K3 )/2. B.8

Solution of the Black-Scholes Equation

(1) fˆ(ω) = =

Z



−∞ ∞

Z

f (x)e−iωx dx e−ax e−iωx dx

0

= lim

M→∞

Z

M

e−(a+iω)x dx

0

M −1 lim e−(a+iω)x a + iω M→∞ 0   −1 = lim e−(a+iω)M − 1 a + iω M→∞

=

Since 0 ≤ |e−iωM |e−aM ≤ e−aM then by the Squeeze Theorem (Theorem 2.7 of [Smith and Minton (2002)]),   1 . lim e−(a+iω)M − 1 = −1 =⇒ fˆ(ω) = M→∞ a + iω 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

408

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(2) fˆ(ω) = =

Z



f (x)e−iωx dx

−∞ Z π

sin xe−iωx dx

−π −iωx

π e = 2 (cos x + iω sin x) ω −1 −π

eiωπ e−iωπ + ω2 − 1 ω2 − 1 −(cos(−ωπ) + i sin(−ωπ)) + (cos(ωπ) + i sin(ωπ)) = ω2 − 1 2i sin(ωπ) = ω2 − 1 =−

(3) fˆ(ω) = =

Z



2

xe−x e−iωx dx

−∞ ∞

Z

2

xe−(x −∞ Z ∞

= e−ω

2

+iωx)

dx 2

/4

xe−(x+iω/2) dx −∞  Z ∞ Z ∞ 2 2 iω −(x+iω/2)2 iω −ω 2 /4 =e x+ e−(x+iω/2) dx − e−ω /4 e dx 2 −∞ −∞ 2 Z ∞ Z ∞ 2 2 2 2 iω e−z dz = e−ω /4 ze−z dz − e−ω /4 2 −∞ −∞ Z iω −ω2 /4 ∞ −z2 =− e e dz 2 −∞ √ iω π −w2 /4 =− e 2 (4) Z ∞ 1 f (x) = fˆ(ω)eiωx dω 2π −∞ Z ∞ 1 = e−aω eiωx dω 2π 0 Z M 1 = lim e−(a−ix)ω dω 2π M→∞ 0 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

409

Solutions to Chapter Exercises

M 1 −1 lim e−(a−ix)ω 2π M→∞ a − ix 0   −1 = lim e−(a−ix)M − 1 2π(a − ix) M→∞

=

(5)

Since 0 ≤ |e−ixM |e−aM ≤ e−aM then by the Squeeze Theorem (Theorem 2.7 of [Smith and Minton (2002)]),   1 . lim e−(a−ix)M − 1 = −1 =⇒ f (x) = M→∞ 2π(a − ix)

f (x) = = = = =

Z ∞ 1 fˆ(ω)eiωx dω 2π −∞ Z ∞ 2 1 e−aω eiωx dω 2π −∞ Z ix 2 1 −x2 /4a ∞ −a(ω− 2a ) dω e e 2π −∞ Z 1 −x2 /4a ∞ −az2 e e dz 2π −∞ 2 1 √ e−x /4a 2 πa

The last step follows since Z ∞

2

e−az dz =

−∞

r

π . a

(6) f (x) =

1 2π

1 = 2π =

i 2π

Z



−∞ Z 0 −∞ Z 0 −∞

fˆ(ω)eiωx dω ie

aω iωx

e

1 dω − 2π

e(a+ix)ω dω −

i 2π

Z

0 Z ∞

i i + 2π(a + ix) 2π(−a + ix) x = 2 π(x + a2 )

=

14:58:13.



0

ie−aω eiωx dω e(−a+ix)ω dω

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

410

Juliet

Undergrad Introd to... 3rd edn

An Undergraduate Introduction to Financial Mathematics

(7) ∂ ∂ (F (S, t)) = K (v(x, τ )) ∂t ∂t   ∂v dx ∂v dτ + =K ∂x dt ∂τ dt    ∂v ∂v −σ 2 =K ·0+ · ∂x ∂τ 2 2 Kσ ∂v =− 2 ∂τ (8)   ∂2F ∂ ∂F = ∂S 2 ∂S ∂S  ∂ = e−x vx (by Eq. (8.12)) ∂S   dx dτ dx −x −x = −e vx +e + vxτ vxx dS dS dS   1 1 −x −x = −e vx · + e vxx · + vxτ · 0 S S −x e =− (vx − vxx ) S −2x e = (vxx − vx ) K (9) 1 rF = Ft + σ 2 S 2 FSS + rSFS 2 1 e−2x Kσ 2 rKv = − vτ + σ 2 S 2 (vxx − vx ) + rSe−x vx 2 2 K σ2 1 rv = − vτ + σ 2 (vxx − vx ) + rvx 2 2 σ2 1 2 vτ = −rv + σ (vxx − vx ) + rvx 2 2 vτ = vxx + (k − 1)vx − kv

14:58:13.

BC8495/Chp. B

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

411

Solutions to Chapter Exercises

(10) lim u(x, τ ) =

x→∞

h

lim e(k−1)x/2+(k+1)

2

τ /4

x→∞

= lim e(k−1)x/2+(k+1)

2

τ /4

x→∞

ih

i lim v(x, τ ) x→∞  ex − e−kτ

Dropping the limit we see that e(k−1)x/2+(k+1)

2

= e(k+1)x/2+(k+1) =e =e =e

ex − e−kτ

τ /4 2

τ /4

(k+1) [x+(k+1)τ /2] 2 (k+1) [x+(k+1)τ /2] 2 (k+1) [x+(k+1)τ /2] 2



− e(k−1)x/2+(k+1)

− e(k−1)x/2+[(k+1) − e(k−1)x/2+(k−1) −e

2

2

2

τ /4−kτ

−4k]τ /4

τ /4

(k−1) [x+(k−1)τ /2 2

.

(11) 

e(k+1)(x+

√ 2τ y)/2

− e(k−1)(x+

+ √ 2τ y)/2

>0

if and only if e(k+1)(x+



e

2τ y)/2 √ x+ 2τ y

> e(k−1)(x+



2τ y)/2

>1 √ x + 2τ y > 0 x y > −√ . 2τ (12) By the result of exercise (11) we know that 1 √ 2π

Z



−∞

 + √ √ 2 e(k+1)(x+ 2τ y)/2 − e(k−1)(x+ 2τ y)/2 e−y /2 dy

Z ∞   √ √ 2 1 (k+1)(x+ 2τ y)/2 (k−1)(x+ 2τ y)/2 = √ e − e e−y /2 dy √ 2π −x/ 2τ Z ∞ √ √ 2 2 1 = √ e[(k+1)(x+ 2τ y)−y ]/2 − e[(k−1)(x+ 2τ y)−y ]/2 dy. √ 2π −x/ 2τ 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

412

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Completing the square in the exponents present in the integrand yields √ y 2 − 2τ (k ± 1)y − (k ± 1)x   √ (k ± 1)2 τ (k ± 1)2 τ = y 2 − 2τ (k ± 1)y + − − (k ± 1)x 2 2 √ !2 (k ± 1) 2τ (k ± 1)2 τ = y− − − (k ± 1)x. 2 2 Thus the integral in Eq. (8.33) becomes "Z √ ∞ (k+1) 2τ 2 (k+1)2 τ 1 √ e−(y− 2 ) /2 e 4 +(k+1)x/2 dy √ 2π −x/ 2τ # Z ∞ √ 2τ 2 (k−1)2 τ −(y− (k−1) ) /2 +(k−1)x/2 2 − e e 4 dy √ −x/ 2τ

=

e(k+1)



2

τ /4+(k+1)x/2

√ 2π

e(k−1)

2

Z

√ −x/ 2τ

τ /4+(k−1)x/2







Z

e−(y−



√ −x/ 2τ

√ (k+1) 2τ 2 ) /2 2

e−(y−

dy

√ (k−1) 2τ 2 ) /2 2

dy.

√ √ (13) If we let w =√−y + 21 (k + 1) √2τ , then dw = −dy. When y = −x/ 2τ then w = x/ 2τ + 12 (k + 1) 2τ and Z ∞ √ 1 −(y− 21 (k+1) 2τ )2 /2 √ e dy 2π −x/√2τ Z x/√2τ + 12 (k+1)√2τ 2 1 = √ e−w /2 dw 2π −∞   √ 1 x = Φ √ + (k + 1) 2τ . 2 2τ √ √ (14) If we let w0 =√−y + 12 (k −1) √ 2τ , then dw0 = −dy. When y = −x/ 2τ then w0 = x/ 2τ + 12 (k − 1) 2τ and Z ∞ √ 2 1 1 √ e−(y− 2 (k−1) 2τ ) /2 dy √ 2π −x/ 2τ Z x/√2τ + 12 (k−1)√2τ 02 1 = √ e−w /2 dw0 2π −∞   √ x 1 = Φ √ + (k − 1) 2τ . 2 2τ 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

413

Solutions to Chapter Exercises

(15) We verify the initial condition by evaluating a limit.    √ (k+1) x 1 lim+ u(x, τ ) = lim+ e 2 [x+(k+1)τ /2] Φ √ + (k + 1) 2τ 2 τ →0 τ →0 2τ   √ (k−1) 1 x − e 2 [x+(k−1)τ /2] Φ √ + (k − 1) 2τ 2 2τ     x = e(k+1)x/2 − e(k−1)x/2 lim Φ √ τ →0+ 2τ √  If x < 0 then limτ →0+ Φ x/ 2τ = 0. If√x = 0 then e(k+1)x/2 − e(k−1)x/2 = 0. If x > 0 then limτ →0+ Φ x/ 2τ = 1. Hence we see that for any x,  + lim u(x, τ ) = e(k+1)x/2 − e(k−1)x/2 .

τ →0+

(16) √ 1 ln(S/K) x 1 √ + (k + 1) 2τ = p + 2 2 2τ σ (T − t) 2 =



 p 2r + 1 σ 2 (T − t) 2 σ

ln(S/K) + (r + σ 2 /2)(T − t) √ σ T −t

√ √ √ 1 x 1 x √ + (k − 1) 2τ = √ + (k + 1) 2τ − 2τ 2 2 2τ 2τ √ ln(S/K) + (r + σ 2 /2)(T − t) √ = −σ T −t σ T −t (17) Using the parameter values specified w ≈ 0.536471 and thus C e ≈ $5.05739. (18) Using the parameter values specified w ≈ 0.16662 and thus P e ≈ $6.40141. (19) One way to find the volatility is to approximate it using Newton’s Method (Sec. 3.2 of [Smith and Minton (2002)]). We wish to find a root of the equation C e (σ) = 2.50. The graph below shows that the solution is near 0.40 which we will use as the initial approximation to the solution for Newton’s Method.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

414

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

C 4.0

3.5

3.0

2.5

Σ 0.2

0.4

0.6

0.8

1.0

Newton’s Method approximates the solution as σ ≈ 0.396436. (20) If f (S, t) = S, then fS = 1, fSS = 0, and ft = 0. Thus 1 rf = ft + σ 2 S 2 fSS + rSfS 2 1 2 2 rS = 0 + σ S · 0 + rS 2 rS = rS. Therefore the price of the security itself solves the Black-Scholes partial differential equation Eq. (8.5). (21) If f (S, t) = ert , then fS = 0, fSS = 0, and ft = rert . Thus 1 rf = ft + σ 2 S 2 fSS + rSfS 2 1 rt rt re = re + σ 2 S 2 · 0 + rS · 0 2 rert = rert . Therefore a unit of currency earning the risk-free interest rate r compounded continuously solves the Black-Scholes partial differential Eq. (8.5). (22) If Y = ln S then by Itˆ o’s Lemma 5.4, Y obeys the stochastic differential equation,   σ2 dt + σ dW (t). dY = µ − 2 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

415

Solutions to Chapter Exercises

Integrating both sides with respect to t produces   σ2 Y (t) = Y (0) + µ − t + σW (t) 2 where Y (0) = ln S(0). Then exponentiating both sides yields S(t) = S(0)e(µ−σ

2

/2)t+σW (t)

.

(23) Constructing a binomial lattice of stock values using n = 4 and u = 1.07484,

d = 0.930374,

and p = 0.545466,

we can summarize the values of the stock, the payoffs of the option, and binomial probabilities of achieving those payoffs. S 32.218 37.2206 43 49.6768 57.3903

(S − K)+ 0 0 1 7.6768 15.3903

P 0.0426839 0.204893 0.368824 0.295073 0.0885262

Thus the value of the call option is approximated as (1)(0.368824) + (7.6768)(0.295073) + (15.3903)(0.088562) e(0.11)(4/12) ≈ 3.8526.

C≈

(24) Constructing a binomial lattice of stock values using n = 3 and u = 1.09995,

d = 0.909134,

and p = 0.502427,

we can summarize the values of the stock, the payoffs of the option, and binomial probabilities of achieving those payoffs. S 72.1365 87.2769 105.595 127.758

14:58:13.

(K − S)+ 27.8635 12.7231 0 0

P 0.123188 0.373171 0.376812 0.126829

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

416

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Notice that the middle column is the payoff for a European put option. Thus the value of the put option is approximated as P ≈ B.9

(27.8635)(0.123188) + (12.7231)(0.373171) = 8.05857. e(0.06)(3/12)

Derivatives of Black-Scholes Option Prices

(1) Using Eq. (9.3) with S = 300, K = 310, T = 1/4, t = 0, r = 0.03, and σ = 0.25 we have w = −0.139819

Θ = −33.281.

(2) The graph below contains the three curves. Q 80

100

120

140

S

-5

-10

-15

-20

(3) To find Θ for a European put option we will make use of the Put-Call Parity formula (Eq. (7.1)) and Eq. (9.3). According to the Put-Call Parity formula, ∂P ∂C = + Kre−r(T −t) . ∂t ∂t Substituting the expression already found for side of this equation yields

14:58:13.

∂C ∂t

in the right-hand

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

417

Solutions to Chapter Exercises

∂P Se−w = ∂t

2

/2

− Ke−r(T −t)−(w−σ p 2σ 2π(T − t)

− Ke−r(T −t)

+ Kre−r(T −t)

√ T −t)2 /2

BC8495/Chp. B



ln(S/K) − r − σ 2 /2 T −t !   σe−(w−σ√T −t)2 /2 √ p rΦ w − σ T − t + 2 2π(T − t)



√   2 − Ke−r(T −t)−(w−σ T −t) /2 ln(S/K) 2 p − r − σ /2 = T −t 2σ 2π(T − t)    √ + Kre−r(T −t) 1 − Φ w − σ T − t

Se−w

2

/2



2

σKe−r(T −t)−(w−σ T −t) /2 p − 2 2π(T − t)   2 Se−w /2 ln(S/K) 2 p = − r − σ /2 T −t 2σ 2π(T − t)  √  + Kre−r(T −t) Φ σ T − t − w

√   2 Ke−r(T −t)−(w−σ T −t) /2 ln(S/K) p − − r/σ − σ/2 + σ σ(T − t) 2 2π(T − t)   2 Se−w /2 ln(S/K) p = − r − σ 2 /2 T −t 2σ 2π(T − t)   √ + Kre−r(T −t) Φ σ T − t − w √   2 Ke−r(T −t)−(w−σ T −t) /2 ln(S/K) 2 p − − r + σ /2 . T −t 2σ 2π(T − t)

(4) Using Eq. (9.5) with S = 275, K = 265, T = 1/3, t = 0, r = 0.02, and σ = 0.20 we have w = 0.436257 Θ = −15.3073. (5) According to Eq. (8.36) w=

14:58:13.

ln(S/K) + (r + σ 2 /2)(T − t) √ , σ T −t

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

418

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

which implies that   ∂w ∂ ln(S/K) √ = ∂S ∂S σ T − t 1 1 = √ · σ T −t S 1 √ = . σS T − t (6) Using Eq. (9.8) with S = 150, K = 165, T = 5/12, t = 0, r = 0.025, and σ = 0.22 we have w = −0.526797

∆ = 0.299167.

(7) Using Eq. (9.8) with S = 50, K = 55, T = 4/12, t = 0, r = 0.0325, and σ = 0.27 we have w = −0.463977

∆ = 0.321332.

(8) Using Eq. (9.9) with S = 125, K = 140, T = 2/3, t = 0, r = 0.055, and σ = 0.15 we have w = −0.564706

∆ = −0.713863. (9) Using Eq. (9.9) with S = 75, K = 81, T = 7/12, t = 0, r = 0.065, and σ = 0.33 we have w = −0.0288916

∆ = −0.511524.

(10) Using Eq. (9.11) with S = 180, K = 175, T = 1/3, t = 0, r = 0.0375, and σ = 0.30 we have w = 0.321416 Γ = 0.0121519. (11) Using Eq. (9.10) with S = 205, K = 195, T = 2/12, t = 0, r = 0.0465,

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

419

and σ = 0.45 we have w = 0.406264 Γ = 0.00975392. (12) According to Eq. (8.36) w=

ln(S/K) + (r + σ 2 /2)(T − t) √ , σ T −t

which implies that   ∂w ∂ ln(S/K) + (r + σ 2 /2)(T − t) √ = ∂σ ∂σ σ T −t √ √ σ(T − t)σ T − t − (ln(S/K) + (r + σ 2 /2)(T − t)) T − t = σ 2 (T − t) 2 σ (T − t) − (ln(S/K) + (r + σ 2 /2)(T − t)) √ = σ2 T − t   √ 1 ln(S/K) + (r + σ 2 /2)(T − t) √ = T −t− σ σ T −t √ w = T −t− . σ (13) Using Eq. (9.14) with S = 300, K = 305, T = 1/2, t = 0, r = 0.0475, and σ = 0.25 we have w = 0.129235 V = 83.9247. (14) Using Eq. (9.14) with S = 123, K = 125, T = 3/12, t = 0, r = 0.0515, and σ = 0.35 we have w = 0.0689035 V = 24.4768. (15) Without loss of generality we may assume the option is a European call option. Using the relationship between the differentials, then dC e = V dσ

= (24.4768)(0.05) = 1.22384.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

420

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(16) Using Eq. (9.19) with S = 270, K = 272, T = 1/6, t = 0, r = 0.0375, and σ = 0.15 we have w = 0.012164 ρ = −23.4071. (17) Using Eq. (9.19) with S = 305, K = 325, T = 1/3, t = 0, r = 0.0255, and σ = 0.35 we have w = −0.171209 ρ = 38.0758.

(18) Using S = 425, K = 435, T = 6/12, t = 0, r = 0.0525, and σ = 0.17 we have w = 0.0850037 C = 20.98 ∆ = 0.533871 (425)(0.533871) Ω= = 10.8148. 20.98 (19) Using S = 315, K = 310, T = 4/12, t = 0, r = 0.0475, and σ = 0.24 we have w = 0.299022 P = 12.6408 ∆ = −0.382462 (315)(−0.382462) Ω= = −9.53069. 12.6408 (20) Using the formula σoption = |Ω|σstock we have σoption = (10.8148)(0.17) = 1.8385 or 183.85%.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

421

Solutions to Chapter Exercises

B.10

BC8495/Chp. B

Hedging

(1) If the call option is in the money then S > K. Start by considering the following limit.

t→T −



 ln(S/K) + (r + σ 2 /2)(T − t) √ t→T − σ T −t   (r + σ 2 /2) √ ln(S/K) √ = lim + T −t σ t→T − σ T −t = +∞

lim w = lim

We know the limit is +∞ since ln(S/K) > 0 for an in the money call option. Now since ∆C = Φ (w) and Φ (w) is a continuous function then lim ∆C = lim Φ (w) = 1. w→∞

t→T −

(2) If the call option is out of the money then S < K. Start by considering the following limit.  ln(S/K) + (r + σ 2 /2)(T − t) √ t→T σ T −t   ln(S/K) (r + σ 2 /2) √ √ T −t = lim− + σ t→T σ T −t = −∞

lim− w = lim−

t→T



We know the limit is −∞ since ln(S/K) < 0 for an out of the money call option. Now since ∆C = Φ (w) and Φ (w) is a continuous function then lim ∆ = lim Φ (w) = 0. w→−∞

t→T −

(3) If the call option is at the money then S = K. Start by considering the following limit.

t→T −



ln(S/S) + (r + σ 2 /2)(T − t) √ t→T − σ T −t r σ√ = lim− + T −t σ 2 t→T =0

lim w = lim

14:58:13.



May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

422

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Now since ∆C = Φ (w) and Φ (w) is a continuous function then lim− ∆ = lim Φ (w) =

t→T

w→0

1 . 2

(4) Using the values S = 54, K = 57, r = 0.045, T = 5/12, σ = 0.23, and t = 0 we have w = −0.163651

∆ = 0.435003 C1 = 2.37367.

Thus the linear approximation to the value of the second option is C2 = C1 + (∆)(54.75 − 54.00) = 2.69992. Calculating the value of the second option directly using the values S = 54.75, K = 57, r = 0.045, T = 5/12, σ = 0.23, and t = 1/12 we have w = −0.123934

C2 = 2.28762.

(5) Using Eq. (9.3) we have Θ = −4.73721. Now the linear approximation can be written as C2 = C1 + (∆)(54.75 − 54.00) + (Θ)(1/12 − 0) = 2.30516.

(6) Using Eq. (9.10) we have Γ = 0.0490997. Now the approximation can be written as 1 C2 = C1 + (∆)(54.75 − 54.00) + (54.75 − 54.00)2 (Γ) + (Θ)(1/12 − 0) 2 = 2.31897. (7) These calculations are similar to those used to create Table 10.1.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

423

Solutions to Chapter Exercises

Week 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

S 45.00 44.58 46.55 47.23 47.62 47.28 49.60 50.07 47.79 48.33 48.81 51.36 52.06 51.98 54.22 54.31

∆ 0.408940 0.366642 0.526428 0.581970 0.614837 0.583262 0.782988 0.824866 0.635479 0.698867 0.761174 0.954430 0.986327 0.995748 1.000000 1.000000

Shares Held 2044.70 1833.21 2632.14 2909.85 3074.18 2916.31 3914.94 4124.33 3177.39 3494.34 3805.87 4772.15 4931.64 4978.74 5000.00 5000.00

Share Cost 92011.6 81724.6 122526 137432 146393 137883 194181 206505 151848 168881 185765 245098 256741 258795 271100 271550

BC8495/Chp. B

Interest Cost 0.000000 79.6599 71.5663 103.826 115.271 122.146 115.790 158.773 167.987 128.953 142.326 155.614 198.715 206.075 208.374 209.552

Cumulative Cost 92011.6 82663.0 119924 133144 141085 133743 183391 194034 148948 164395 179743 229527 238029 240683 242044 242254

(8) These calculations are similar to those used to create Table 10.2. Week 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

S 45.00 44.58 45.64 44.90 43.42 42.23 41.18 41.52 41.94 42.72 44.83 44.93 44.19 41.77 39.56 40.62

∆ 0.408940 0.366642 0.447836 0.374621 0.238290 0.140594 0.073032 0.073130 0.075921 0.097654 0.254169 0.235189 0.114050 0.001621 0.000000 0.000000

Shares Held 2044.70 1833.21 2239.18 1873.11 1191.45 702.968 365.159 365.651 379.604 488.269 1270.84 1175.94 570.252 8.10264 0.00000 0.00000

Share Cost 92011.6 81724.6 102196 84102.5 51732.7 29686.3 15037.3 15181.8 15920.6 20858.9 56971.9 52835.1 25199.4 338.447 0.00007 0.00000

Interest Cost 0.00000 79.6599 71.5663 87.6693 73.5151 47.9543 30.1364 18.1190 18.1524 18.6747 22.7099 53.1028 49.4573 26.3277 6.02157 5.74928

Cumulative Cost 92011.6 82663.0 101263 84914.0 55389.9 34809.3 20928.5 20967.0 21570.4 26231.2 61336.7 57125.9 30409.9 6955.25 6640.73 6645.48

(9) In this case the put option will not be exercised since the final price of the stock exceeds the strike price. In other words, the put option finishes out of the money. At the time the Put option is issued its price is $2.73256. Delta for a European Put option is always negative, thus the financial institution will take a short position in the stock in creating the hedge. The revenue generated by the sale of the Put

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

424

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

option and the short position in the stock will be invested in a bond at the risk-free interest rate r = 0.045 yr−1 . Week 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

S 45.00 44.58 46.55 47.23 47.62 47.28 49.60 50.07 47.79 48.33 48.81 51.36 52.06 51.98 54.22 54.31

∆ −0.591060 −0.633358 −0.473572 −0.418030 −0.385163 −0.416738 −0.217012 −0.175134 −0.364521 −0.301133 −0.238826 −0.045570 −0.013673 −0.004252 0.000000 0.000000

Shares Shorted 2955.30 3166.79 2367.86 2090.15 1925.82 2083.69 1085.06 875.670 1822.61 1505.66 1194.13 227.848 68.3635 21.2594 0.00051 0.00000

Share Value 132988 141175 110224 98717.8 91707.4 98516.8 53819.1 43844.8 87102.3 72768.7 58285.4 11702.3 3559.01 1105.06 0.02738 0.00000

Interest Earned 0.00000 115.136 123.398 3073 80.0309 73.3252 79.8508 37.0372 27.9925 67.1958 53.9924 40.8744 −2.05618 −9.24617 −11.3740 −12.3817

Cumulative Earnings 132988 142532 105465 92440.2 84694.6 92232.2 42780.1 32332.9 77614.9 62364.3 47212.3 −2375.01 −10679.8 −13137.6 −14301.6 −14314.0

Since the buyer of the option pays for the option at the beginning of week 0, the net cost to the financial institution is (5000)(2.73256)e0.045(15/52) − 14314.0 = −472.69. (10) In this exercise the Put option will be exercised since the final stock price is smaller than the strike price. Week 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

S 45.00 44.58 45.64 44.90 43.42 42.23 41.18 41.52 41.94 42.72 44.83 44.93 44.19 41.77 39.56 40.62

∆ −0.591060 −0.633358 −0.552164 −0.625379 −0.761710 −0.859406 −0.926968 −0.926870 −0.924079 −0.902346 −0.745831 −0.764811 −0.885950 −0.998379 −1.000000 −1.000000

14:58:13.

Shares Shorted 2955.30 3166.79 2760.82 3126.89 3808.55 4297.03 4634.84 4634.35 4620.40 4511.73 3729.16 3824.06 4429.75 4991.90 5000.00 5000.00

Share Value 132988 141175 126004 140397 165367 181464 190863 192418 193779 192741 167178 171815 195751 208512 197800 203100

Interest Earned 0.00000 115.136 123.398 107.464 121.787 147.517 165.504 177.691 177.827 177.473 173.609 143.386 147.202 170.502 190.987 191.421

Cumulative Earnings 132988 142532 124127 140671 170390 191166 205253 205400 204993 200528 165619 170026 196939 220590 221102 221293

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

425

At the strike time the financial institution must honor the put option and cancel its short position in the stock. Thus the financial institution purchases 5000 shares of the stock at the strike price of $47 per share. This final transaction changes the holdings of the financial institution by 221293 − (5000)(47) + (5000)(2.73256)e0.045(15/52) = $134.31. (11) Let S = 85, K = 88, r = 0.055, and σ = 0.17. We will let the number of four-month options be w4 and the number of six-month options be w6 . If P represents the value of the portfolio consisting of a short position in the four-month options and a long position in the six-month options, then the value of the portfolio is P = w4 C(S, 13 ) − w6 C(S, 12 ). Note that C(85, 31 ) = 1.2972 and C(85, 12 ) = 2.59313. Gamma for the portfolio is Γ = Γ4 w4 − Γ6 w6

= 0.0474901w4 − 0.0390443w6.

Thus the portfolio is Gamma neutral whenever w4 = 0.822155w6. (12) Using the results of exercise (11) and including a position of x shares of the security, the value of the portfolio is now P = Sx + w4 C(S, 31 ) − w6 C(S, 21 ). Delta for the portfolio is ∆ = x + ∆4 w4 − ∆6 w6

= x + 0.45322w4 − 0.500131w6

= x + (0.45322)(0.822155)w6 − 0.500131w6

= x − 0.127514w6.

Thus the portfolio is both Gamma and Delta neutral whenever x = 0.127514w6. (13) Let S = 95, r = 0.045, and σ = 0.23. We will let the number of threemonth options be w3 and the number of five-month options be w5 . If P represents the value of the portfolio consisting of a short position in the three-month options and a long position in the five-month options, 5 ). Note then the value of the portfolio is P = w3 C(S, 14 ) − w5 C(S, 12 1 5 that C(95, 4 ) = 1.74863 and C(95, 12 ) = 3.08418. Gamma for the portfolio is Γ = Γ3 w3 − Γ5 w5

= 0.0365043w3 − 0.0282844w5.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

426

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Thus the portfolio is Gamma neutral whenever w3 = 0.774825w5. (14) Using the results of exercise (13) and including a position of x shares of the security, the value of the portfolio is now P = Sx + w3 C(S, 41 ) − 5 w5 C(S, 12 ). Delta for the portfolio is ∆ = x + ∆3 w3 − ∆5 w5

= x + 0.489693w3 − 0.496454w5

= x + (0.489693)(0.774825)w5 − 0.496454w5

= x − 0.117028w5.

Thus the portfolio is both Gamma and Delta neutral whenever x = 0.117028w5. (15) Using the values S = 60, r = 0.0565, T = 3/12, t = 0, and σ = 0.45 we have w62 = 0.0295452 w65 = −0.180468 Γ62 = 0.0295384 Γ65 = 0.029074. Thus the portfolio is Gamma neutral when 0 = Γ62 − xΓ65

=⇒

x = 1.01597.

(16) We can use the values above to determine ∆62 = 0.511785 ∆65 = 0.428393. Let y be the position in the underlying then the portfolio becomes Delta neutral when 0 = ∆62 − 1.01597∆65 + y

=⇒

y = −0.0765509.

Hence the investor should purchase −0.0765509 shares of the underlying security. (17) The one-day profit/loss will be the (a) interest cost associated with the Delta neutral portfolio, (b) the gain/loss on the shares of the security in the portfolio, (c) the gain/loss on the sold call option.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

427

The day 0 value of the call option is C0 = 5.72598 and its Delta is ∆0 = 0.569096. Thus to set up the Delta neutral portfolio an investor must borrow ∆0 S0 − C0 = (0.569096)(60) − 5.725598 = 28.4198. One day’s worth of interest on this amount is (∆0 S0 − C0 )(er/365 − 1) = 28.4198(e0.0565/365 − 1) = 0.00439957. On day 1 the value of the security has increased to $60.50, thus the gain in value of the shares owned is ∆0 (S1 − S0 ) = 0.569096(60.50 − 60) = 0.284548. Using the day 1 price of the security and the fact that the call option expires in 89 days the call is worth C1 = 5.98025. The loss on the value of the option is C1 − C0 = 5.98025 − 5.72598 = 0.254272. Thus the one-day profit/loss becomes ∆0 (S1 − S0 ) − (C1 − C0 ) − (∆0 S0 − C0 )(er/365 − 1) = 0.284548 − 0.254272 − 0.00439957 = 0.0258769.

(18) As in the previous exercise the one-day profit/loss can be expressed using the formula: Profit/Loss = ∆0 (S − S0 ) − (C(S) − C0 ) − (∆0 S0 − C0 )(er/365 − 1) where S is the day 1 price of the security and this price is used to calculate the value of the call option on day 1. The plot of profit/loss resembles the following.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

428

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

ProfitLoss 55

60

65

70

S

-0.5

-1.0

-1.5

(19) Approximate a solution to the equation 0 = ∆0 (S − S0 ) − (C(S) − C0 ) − (∆0 S0 − C0 )(er/365 − 1) using Newton’s Method. The two solutions are approximately S = 58.6106 and S = 61.435. (20) The one-day profit/loss will be the (a) (b) (c) (d)

interest cost associated with the Delta/Gamma neutral portfolio, the gain/loss on the shares of the security in the portfolio, the gain/loss on the sold call option, the gain/loss on the purchased call option.

The day 0 value of the sold call option is C0s = 4.87838 and its Delta is ∆s0 = 0.511785. The day 0 value of the purchased call option is C0p = 3.74902 and its Delta is ∆p0 = 0.511785. To create the Delta/Gamma neutral portfolio an investor must purchase 1.01597 of the 65-strike call options and must purchase 0.0765509 shares of the security. Thus to set up the Delta neutral portfolio an investor must borrow 4.87838 − (1.01597)(3.74902) − (0.0765509)(60) = −3.52357, the negative sign indicates the amount is borrowed. One day’s worth of interest on this amount is (3.52357)(e0.0565/365 − 1) = 0.000545471. On day 1 the value of the security has increased to $60.50, thus the

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

429

Solutions to Chapter Exercises

gain in value of the shares owned is 0.0765509(S1 − S0 ) = 0.0765509(60.50 − 60) = 0.0382755. Using the day 1 price of the security and the fact that the call options expire in 89 days, the sold call is worth C1s = 5.10405 while the purchased call is worth C1p = 3.93379. The loss on the value of the sold option is C1s − C0s = 5.10405 − 4.87838 = 0.22567. The loss of value of the purchased option is C1p − C0p = 3.93379 − 3.74902 = 0.18477. Thus the one-day profit/loss becomes 0.0382755−0.22567+(1.01597)(0.18477)−0.000545471 = −0.000219193. (21) As in the previous exercise the one-day profit/loss can be expressed using the formula: Profit/Loss = 0.0765509(S − S0 ) − (C1s (S) − C0s )

+ 1.10597(C1p(S) − C0p ) − 0.00054547

where S is the day 1 price of the security and this price is used to calculate the value of the call options on day 1. The plot of profit/loss resembles the following. Profit  Loss 0.06 0.04 0.02

55 - 0.02 - 0.04 - 0.06

14:58:13.

60

65

70

S

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

430

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

B.11

Extensions of the Black-Scholes Model

(1) The sum of the present values of the dividends is D D(1 + g) D(1 + g)k−1 + + · · · + 1+r (1 + r)2 (1 + r)k   (1 + g) (1 + g)k−1 + ··· + (1 + r)S = D 1 + (1 + r) (1 + r)k−1   k  D 1 − 1+g 1+r = (using Eq. (1.8)) 1+g 1 − 1+r   k  1+g D 1 − 1+r S= r−g S=

assuming 1 6= (1 + g)/(1 + r). (2) If 0 < g < r then 0 < (1 + g)/(1 + r) < 1 and

lim S = lim

k→∞

k→∞

 k   D 1 − 1+g 1+r r−g

=

D . r−g

(3) According to the Put-Call Parity formula for stocks paying dividends continuously, P e,δ + Se−δ(T −t) = C e,δ + Ke−r(T −t) . Substituting Eq. (11.13) in this equation and re-arranging terms yields the following. P e,δ = −Se−δ(T −t) + C e,δ + Ke−r(T −t)

= −Se−δ(T −t) + e−δ(T −t) SΦ (w)   √ − Ke−r(T −t)Φ w − σ T − t + Ke−r(T −t)  √  = Ke−r(T −t) Φ σ T − t − w − e−δ(T −t) SΦ (−w)

(4) Using Eqs. (11.11) and (11.13) w ≈ −0.0516528 and C e ≈ $6.74248. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

431

Solutions to Chapter Exercises

(5) From Eq. (7.8) P e + Se−δ(T −t) = i h ∂ P e + Se−δ(T −t) = ∂δ ρδP − S(T − t)e−δ(T −t) =

C e + Ke−r(T −t) i ∂ h e C + Ke−r(T −t) ∂δ ρδC

ρδP = ρδC + S(T − t)e−δ(T −t)

= S(T − t)e−δ(T −t) (1 − Φ (w))

using Eq. (11.15). (6) Letting S = 110, δ = 0.01, T = 3/12, t = 0, r = 0.055, σ = 0.25, and K = 120 we have w = −0.543591

C = 2.37656.

(7) Using the results of exercises (3) and (6) we have P = 11.0125. (8) Using Eqs. (11.11) and (11.14) w ≈ −0.185844 and P e ≈ $4.91924. (9) Using the definition of d (t), Z



d (t) dt =

−∞

Z



−

 t 1 = 1. dt = 2 2 −

(10) Consider the Dirac delta function D (t). (a) Let t 6= 0 be fixed, then |t| > 0 and

D (t) = lim d (t) = 0 →0+

for all 0 <  < |t|/2. (b) Evaluating the limit of the integral, Z



D (t) dt =

−∞

Z



−∞

14:58:13.



 Z lim d (t) dt = lim

→0+

→0+



−∞

d (t) dt = 1.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

432

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(11) Let f (t) be continuous, then Z ∞ δ(t − t0 ) f (t) dt −∞

=

Z



−∞

= lim+ →0



 lim+ d (t − t0 ) f (t) dt

→0

Z



−∞

1 = lim + →0 2

Z

d (t − t0 ) f (t) dt t0 +

f (t) dt

t0 −

1 f (c)(2) (for some t0 −  ≤ c ≤ t0 +  by IMVT) →0 2 = lim f (c) = lim+

→0+

= f (t0 ). (12) Letting Y = ln S then ∂Y 1 = , ∂S S

∂2Y 1 = − 2, 2 ∂S S

and

∂Y = 0. ∂t

Applying Lemma 5.4 with a(S, t) = (µ−δD (t − td ))S and b(S, t) = σS we obtain      1 1 1 dY = (µ − δD (t − td ))S + 0 + σ2 S 2 − 2 dt S 2 S   1 + σS dW (t) S   1 = µ − δD (t − td ) − σ 2 dt + σ dW (t). 2 (13) From Eq. (11.16) we know that S(t+ d) − = 1 − δ. S(td ) Likewise from Eq. (11.20) we see that S(t+ d) = e−dy . S(t− ) d

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

433

Equating these two expressions reveals that dy = − ln(1 − δ). (14) Suppose that − − + e + C − ≡ C e (S(t− d ), td ) 6= C (S(td )(1 − δ), td ) ≡ C

then there are two cases to consider. If C − < C + then immediately prior to the dividend date, an investor could purchase the call option for C − and sell it for C + immediately after the dividend date. This generates a risk-free profit of C + − C − > 0 and hence arbitrage exists. If C − > C + then an investor could write and sell a call option for C − just before the dividend date and purchase a call option on the same underlying stock and with the same strike price and expiry immediately after the dividend date for C + . The purchased option can be exercised if the sold option is exercised. This situation results in a risk-free profit of C − − C + > 0 and once again arbitrage exists. Therefore in the absence of arbitrage C − = C + . (15) First calculate the value of the option immediately after the dividend is paid. Letting td = 2/12, T = 5/12, r = 0.025, K = 50, S(t+ d) = 50(1 − 0.03) = 48.50, and σ = 0.35 then w = −0.0508383

e

C (2/12) = 2.86152. At time t = 0

w = 0.0242484 e

C (0) = 3.93253. (16) Using the results of exercise (15) and the Put-Call Parity formula given in Eq. (11.27) we obtain P e (0) = 4.90816 P e (2/12) = 4.05. (17) Letting T = 5/12, r = 0.025, K = 50, S = 50 − 1.50e−0.025(2/12) = 48.5062, and σ = 0.35 then w = 0.0248176 C e = 3.93571.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

434

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

The values are very close. (18) Letting T = 5/12, r = 0.025, K = 50, σ = 0.35, and S = 50 − 0.75e−0.025(2/12) − 0.75e−0.025(4/12) = 48.5093 then w = 0.0251009 C e = 3.93729. (19) Letting T = 5/12, r = 0.025, K = 50, S = 50 − 1.50e−0.025(2/12) = 48.5062, and σ = 0.35 then w = 0.0248176 P e = 4.91134. The values are very close. (20) Letting T = 5/12, r = 0.025, K = 50, σ = 0.35, and S = 50 − 0.75e−0.025(2/12) − 0.75e−0.025(4/12) = 48.5093 then w = 0.0251009 P e = 4.90982. B.12

Optimizing Portfolios

(1) Proof of statement 1:   2 Cov (X, X) = E X 2 − (E [X]) = V (X)

(by Theorem (2.6))

Proof of statement 2: Cov (X, Y ) = E [XY ] − E [X] E [Y ]

(by Eq. (12.2))

= E [Y X] − E [Y ] E [X] = Cov (Y, X)

(by Eq. (12.2))

(2) Let X and Y be random variables, then by Theorem 12.1 V (X + Y ) = Cov (X + Y, X + Y )   = E X 2 + 2XY + Y 2 − (E [X] + E [Y ]) (E [X] + E [Y ]) 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

435

Solutions to Chapter Exercises

    = E X 2 + 2E [XY ] + E Y 2 2

BC8495/Chp. B

2

− (E [X]) − 2E [X] E [Y ] − (E [Y ])     2 2 = E X 2 − (E [X]) + E Y 2 − (E [Y ]) + 2 (E [XY ] − E [X] E [Y ])

= V (X) + V (Y ) + 2Cov (X, Y ) . Recall that if X and Y are independent random variables E [XY ] = E [X] E [Y ] and thus E [XY ] − E [X] E [Y ] = 0. Therefore, if X and Y are independent random variables V (X + Y ) = V (X) + V (Y ) .   (3) Suppose that E X 2 = 0, then by definition

n   X E X2 = x2i · P (X = xi ) . i=1

If xi = 6 0 for some i, then P (X = xi ) = 0. Therefore we must have P (X = 0) = 1. In other words X = 0 which implies XY = 0 and thus     2 0 = (E [XY ]) ≤ E X 2 E Y 2 = 0.

 2 A similar statement  2 can be made when E Y = 0. Suppose that E X = ∞, then     (E [XY ])2 < ∞ = E X 2 E Y 2 .

  A similar statement can be made when E Y 2 = ∞. (4) Let X be the golf handicap of the CEO and let Y be the stock rating of the CEO’s corporation. Then we may calculate the following quantities: E [X] = 11.9923 E [Y ] = 56.6923 E [XY ] = 671.069 V (X) = 6.8741 V (Y ) = 45.5641

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

436

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Thus we have Cov (X, Y ) = 671.069 − (11.9923)(56.6923) = −8.80207 −8.80207 ρ (X, Y ) = p = −0.497354. (6.8741)(45.5641)

(5) Let X be the height of the model and let Y be the weight of the model. Then we may calculate the following quantities: E [X] = 67.05 E [Y ] = 117.6 E [XY ] = 7887.85 V (X) = 1.46944 V (Y ) = 41.8222 Thus we have Cov (X, Y ) = 7887.75 − (67.05)(117.6) = 2.77 2.77 = 0.353346. ρ (X, Y ) = p (1.46944)(41.8222)

(6) From the data we calculate the following quantities: E [X] = E [Y ] = E [XY ] = V (X) = V (Y ) =

9 10 y + 45 10 9 2 1 10 y2 55 −y+ 10 6

Thus we have 9 Cov (X, Y ) = − 2 ρ (X, Y ) = r 14:58:13.



1 10

9 10



9 100 (5



y2 10

y + 45 10

− y) −y+



=

9 (5 − y) 100

9 5−y  = 10 q 2 55 y − 10y + 6

275 3

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

437

As y → −∞, ρ (X, Y ) → 9/10. Not quite 1, but by increasing the sample size, the limit can be increased as well. (7) If n is the ratio of the number of shares of security D purchased compared to the number of call options C sold, then the value of the portfolio is P = C − nD which implies the variance in the value of the portfolio is V (P) = n2 V (D) − 2nCov (C, D) + V (C)

= (0.037)2 n2 − 2n(0.86)(0.045)(0.037) + (0.045)2 = 0.001369n2 − 0.0028638n + 0.002025.

This quadratic function is minimized when n = 1.04595. (8) From Eq. (12.8) we know that n=

Cov (∆C, ∆S) . V (∆S)

Assuming that S is governed by an Itˆo process of the form dS = µS dt + σS dW then over a short time interval of length ∆T   2 V (∆S) = E (∆S)2 − E [∆S]   2 = E (µS ∆T + σS ∆W )2 − E [(µS ∆T + σS ∆W )] ≈ σ 2 S 2 ∆T.

Note that we have retained expressions containing powers of ∆T that are 1 or less. If C is described by an Itˆo process of the form   1 2 2 dC = µS∆ + σ S Γ + Θ dt + (σS∆) dW 2 then over a short time interval of length ∆T Cov (∆C, ∆S) = E [(∆C)(∆S)] − E [∆C] E [∆S]   1 2 2 2 2 = (σ S ∆)∆T − µS µS∆ + σ S Γ + Θ (∆T )2 2 ≈ (σ 2 S 2 ∆)∆T.

Again, we have kept only the expressions containing the powers of ∆T which are 1 or less. Now substituting the expressions just found for

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

438

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

the variance and covariance into the formula for n we obtain n=

(σ 2 S 2 ∆)∆T = ∆. σ 2 S 2 ∆T

(9) According to Theorem 12.4, a twice differentiable function u(x), is concave when u00 (x) ≤ 0. (a) u(x) = ln x u0 (x) =

1 x

u00 (x) = −

1 0 then T =

XP(x))

Q

≥ eln(

X

X P(X) )

= G.

> 0. By the previous result

Y X

X

T P(X) ≤ E [T ] =

X

T P (X)

X

X P (X) 1 ≤ P(X) X XX X Y 1 X P(X) ≥ P P(X)

Q

X

X

X

G ≥ H.

Thus we have shown that H ≤ G ≤ E [X]. Now suppose that all the random variable X takes on only one value (with probability 1). Then we have X = H = G = E [X] . Before finishing the exercise we must establish another inequality, an obvious one. X E [X] = XP (X) ≤ max{X} X

X

Equality holds if and only if X takes on only one value. From this inequality follows X P (X) X

X

1

P

X

P(X) X

≤ max{ X

1 } X

≥ min{X}. X

Again equality holds if and only if X takes on only one value. Thus the following string of inequalities is true. min{X} ≤ H ≤ G ≤ E [X] ≤ max{X} X

14:58:13.

X

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

440

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Now if minX {X} = H then X takes on only one value and we have minX {X} = maxX {X} and thus min{X} = H = G = E [X] = max{X}. X

X

(11) If g(t) = t, Z

1

0

f (g(t)) dt =

Z

1

tanh t dt 0

= ln(cosh 1) ≈ 0.433781

< 0.462117   1 ≈ tanh 2 Z 1  = tanh t dt 0

=f

Z

0

1



g(t) dt .

(12) The certainty equivalent is the solution C of the equation 1 (f (10) + f (−2)) = f (C) 2 74 C2 =C− 25 50 √ C = 25 ± 3 53. The certainty equivalent will be the√smaller of the two roots of the quadratic equation, i.e., C = 25 − 3 53 ≈ 3.15967. (13) The certainty equivalent is the solution C of the equation 1 1 f (15) + f (−15) = f (C) 2 2 225 C2 − =C− 2 √2 C = 1 ± 226. The certainty equivalent will be √ the smaller of the two roots of the quadratic equation, i.e., C = 1 − 226 ≈ −14.0333. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

441

(14) If we let X represent the random variable of the investor’s return on the investment, then X=



2αx + (1 + r)(1 − α)x with probability p, (1 + r)(1 − α)x with probability 1 − p.

If the investor’s utility function is u(x) = ln x, then the expected value of the utility function is E [u(X)] = pu(2αx + (1 + r)(1 − α)x) + (1 − p)u((1 + r)(1 − α)x). This function is maximized when d E [u(X)] dα p−1 p(1 − r) = + 1 − α α(1 − r) + 1 + r 2p − r − 1 α= . 1−r 0=

(15) Let u(x) = 1 − e−bx where b > 0, then u0 (x) = be−bx and u00 (x) = −b2 e−bx . Thus u00 (x) ≤ 0 for all x. Hence u(x) is concave. The function u(x) is monotone increasing since u0 (x) > 0 for all x. Therefore if 0 ≤ x1 < x2 then u(x1 ) < u(x2 ). (16) Let W represent the wealth returned to the investor and let y represent the amount of money invested in the first investment (0 ≤ y ≤ 1000). E [W ] = 1000 + 0.08y + 0.13(1000 − y) = 1130 − 0.05y

V (W ) = (0.03)2 y 2 + (0.09)2 (1000 − y)2

+ 2y(1000 − y)(0.03)(0.09)(−0.26)

= 0.010404y 2 − 17.604y + 8100

The investor’s utility function, u(x) = 1 − e−x/100 , is maximized when E [W ] − V (W ) /200 is maximized. E [W ] − V (W ) /200 = −0.00005202y 2 + 0.03802y + 1089.5 This quadratic expression is maximized when y ≈ 365.436. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

442

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(17) According to Theorem 12.7 αA = αB = αC = αD = αE =

1 0.24

+

1 0.41

+

1 0.24

+

1 0.41

+

1 0.24

+

1 0.41

+

1 0.24

+

1 0.41

+

1 0.24 1 0.27 1 0.41 1 0.27 1 0.27 1 0.27 1 0.16 1 0.27

+

1 0.33 1 0.27

1 0.24

+

1 0.41

+

1 0.16

+

1 0.33

+

1 0.16

+

1 0.33

+

1 0.16

+

1 0.33

+

1 0.16

+

1 0.33

+

1 0.16

+

1 0.33

= 0.212697 = 0.124505 = 0.189064 = 0.319045 = 0.154689

(18) If c is a real number then r(cw) = E [R(cw)] " n # X =E cwi (Ri − r)

(by Eq. (12.18))

i=1

= cE [R(w)]

(by Theorem 2.3)

= cr(w). Likewise σ 2 (cw) = V (R(cw)) ! n X =V cwi (Ri − r)

(by Eq. (12.18))

i=1

= c2 V (R(w))

(by exercise (22) of Chap. 2)

2 2

= c σ (w). (19) In this case, the variance in the rate of return is to be minimized subject to the constraint that the expected value of the rate of return must be one. We will again use Lagrange multipliers to find the minimum variance. We must solve the following system of two simultaneous equations.  ∇ σ 2 (w) = λ∇ (r(w)) r(w) = 1

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

443

The set of equations above are equivalent to the following set of n + 1 equations.

n X i=1

2wi σi2 = λ(ri − r)

for i = 1, 2, . . . , n,

wi (ri − r) = 1

If we solve the ith equation for wi for i = 1, 2, . . . , n and substitute into the last equation we have λ

n X (ri − r)2 i=1

2σi2

= 1.

This implies that λ = Pn

i=1

Therefore

wi =

(20) According to Lemma 12.3

1 (ri −r)2 2σi2

ri −r σi2 Pn (ri −r)2 i=1 σi2

.

.

αA = 4.44549 αB = 1.30112 αC = 7.9031 αD = 0.0 αE = 9.69925. (21) We must show that the set of equations below is satisfied by the expressions in Eqs. (12.29) and (12.30). This can be accomplished by direct substitution into the equations.   2 ri − rM = λ 2 (ρi,M σi − σM )σM + x(σi2 − 2ρi,m σi σM + σM ) 2 + 2x(1 − x)ρi,M σi σM σ 2 = x2 σi2 + (1 − x)2 σM

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

444

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(22) Substituting rf = 0.0475, rM = 0.0765, σM = 0.22, and Cov (i, M ) = 0.15 into Eq. (12.27) we obtain Cov (i, M ) (rM − rf ) 2 σM 0.15 ri − 0.0475 = (0.0765 − 0.0475) (0.22)2 ri = 0.0475 + (3.09917)(0.029) ri − rf =

= 0.137376. (23) Using the values S(0) = 83, µ = 0.13, σ = 0.25, r = 0.055, T = 1/4, t = 0, and K = 86 we can determine the value of the European call option as C = 3.33212. Suppose a portfolio consists of a position x in the security and a position y in the option. According to Eq. (12.34) the expected return of the portfolio is a linear function of x and y given by E [R] = 1.57093x + 0.808234y. The variance in the expected rate of return is a quadratic function of x and y described in detail by Eq. (12.36). In this example the quadratic function takes the form V (R) = 112.631x2 + 121.696xy + 43.0376y 2. We would like to find the minimum variance portfolio having E [R] = 10. If we set the expected rate of return to $10 and solve for y we obtain y = 12.3727 − 1.94366x. Substituting this expression into the variance of R produces the quadratic function of x given below. V (R) = 38.6829x2 − 564.248x + 6588.31 The minimum value occurs when x = 7.29325 and thus y = −1.80294. 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

B.13

BC8495/Chp. B

445

American Options

(1) If r = 0.06, S = 50, K = 51, T = 3/12, and Pa = 9.75 then according to inequality (13.3) 50 − 51 ≤ Ca − 9.75 ≤ 50 − 51e−0.06(3/12) 8.75 ≤ Ca ≤ 9.51.

(2) If r = 0.0574, S = 93, K = 90, T = 2/12, and Ca = 11.77 then according to inequality (13.3) 93 − 90 ≤ 11.77 − Pa ≤ 93 − 90e−0.0574(2/12) 0.56 ≤ Pa ≤ 8.77.

(3) If the dividend paid is 10% of S = 115 then the present value of the dividend due in three months assuming r = 0.0375 is PV(div) = (0.10)(115)e−(0.0375)(3/12) ≈ 11.3927. Substituting r, S, K = 110, and T = 6/12 into the inequality in Eq. (13.5) yields 115 − 110 − 11.3927 ≤ C a − P a ≤ 115 − 110e−(0.0375)(6/12) −6.3927 ≤ C a − P a ≤ 7.04328.

(4) The present value of the dividends paid is when the continuously compounded interest rate is r = 0.0295 is Z

4/12

12e

−0.0295t

0

4/12 12 −0.0295t dt = e ≈ 3.98. −0.0295 0

Substituting r, S = 98, K = 10, and T = 4/12 into the inequality in Eq. (13.5) yields 98 − 100 − 3.98 ≤ C a − P a ≤ 98 − 100e−(0.0295)(4/12) −5.98 ≤ C a − P a ≤ −1.02.

(5) The binomial model parameters will be assigned as follows: ∆t = 1/12. The proportional increase and decrease factors in the price of the security between time steps are respectively

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

446

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

√ u = e0.25 1/12 ≈ 1.07484 √ d = e−0.25 1/12 ≈ 0.930374. The probability of an increase in the price of the security occurring between time steps is

p ≈ 0.550973. The binomial lattice of security prices resembles the following. 64.6953

60.1909

56.

56.

52.1009

48.4733

Using the algorithm described in Eq. (13.10), the lattice of put values is shown in the following.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

447

Solutions to Chapter Exercises

0.

0.889118

3.10749

2.

5.89906

9.52665

(6) It is sufficient to show that u ≥ er∆t . To the contrary suppose that u < er∆t then an investor could sell short the security at time t = 0 and invest the income in a risk-free bond at rate r compounded continuously. At t = ∆t, the bond will be worth S(0)er∆t while the maximum value of the security will be uS(0) < er∆ S(0). Thus the investor may close out the short position in the security and earn an amount

(er∆t − u)S(0) > 0 without risk. Consequently arbitrage is present. (7) By assumption (K − uS(ti ))+ > 0 so (K − uS(ti ))+ = K − uS(ti ) and (K − dS(ti ))+ = K − dS(ti ). 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

448

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

e−r∆t E [Q(ti+1 )] = e−r∆t [p(K − uS(ti )) + (1 − p)(K − dS(ti ))] = e−r∆t [K − (pu + (1 − p)d)S(ti )]

= e−r∆t [K − (p(u − d) + d)S(ti )]   = e−r∆t K − er∆t S(ti ) (by Eq. (13.8)) = e−r∆t K − S(ti ) < K − S(ti )

= (K − S(ti ))+ = Q(ti ).

(since (K − uS(ti ))+ > 0)

(8) Assigning the values S = 80, K = 78, r = 0.05, ∆t = 1/12, n = 3, and σ = 0.25 we have from Eqs. (13.6), (13.7), and (13.8) u ≈ 1.07521,

d ≈ 0.93005,

and p ≈ 0.510642.

The value of the American put and the intrinsic value are listed in the table below. i 3

2

1 0

S(i∆t) 99.4424 86.0159 74.4040 64.3588 92.4864 80 69.1994 86.0169 74.4040 80

(K − S(i∆t))+ 0 0 3.59604 13.6412 0 0 8.80063 0 3.59604 0

Pa (i∆t) 0 0 3.59604 13.6412 0 1.75243 8.80063 0.854001 5.1799 2.95856

The earliest time that the value of the option equals its intrinsic value is i∗ = 2 corresponding to two months into the life of the option. We can verify that   2.95856 = e−0.05(2/12) p2 (0) + 2p(1 − p)(1.75243) + (1 − p)2 (8.80063) .

(9) Assigning the values S = 100, K = 10, r = 0.0325, ∆t = 1/12, n = 4, and σ = 0.29 we have from Eqs. (13.6), (13.7), and (13.8) u ≈ 1.08762, 14:58:13.

d ≈ 0.919436,

and p ≈ 0.495138.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

449

Solutions to Chapter Exercises

The value of the American put and the intrinsic value are listed in the table below. i 4

3

2

1 0

S(i∆t) 139.931 118.293 100 84.5362 71.4637 128.658 108.762 91.9436 77.7256 118.293 100 84.5362 108.762 91.9436 100

(K − S(i∆t))+ 0 0 0 15.4638 28.5363 0 0 8.05644 22.2744 0 0 15.4638 0 8.05644 0

Pa (i∆t) 0 0 0 15.4638 28.5363 0 0 8.05644 22.2744 0 4.05639 15.4638 2.04238 9.78902 5.93726

The earliest time that the value of the option equals its intrinsic value is i∗ = 2 corresponding to two months into the life of the option. We can verify that   5.93726 = e−0.0325(2/12) p2 (0) + 2p(1 − p)(4.05639) + (1 − p)2 (15.4638) .

(10) Starting with the inequality in Eq. (13.3) note that C a − P a ≤ S − Ke−rT

⇐⇒

C a + Ke−rT ≤ P a + S.

⇐⇒

P a + K ≤ C a + K.

Likewise S − K ≤ Ca − P a

Thus the two inequalities are equivalent. (11) It is readily understood that the price of the option, given the American style of exercise must satisfy C ≤ S, otherwise it would be cheaper to buy the stock than the option. Therefore the following system of

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

450

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

inequalities holds for all T ≥ 0. S − Ke−rT ≤ C ≤ S

lim (S − Ke−rT ) ≤ C ≤ S

T →∞

S≤C≤S

This implies C = S. (12) From Eq. (7.7) we know P e + S(0) − PV(div) = C e + Ke−rT

P e + S(0) − PV(div) − Ke−rT = C e

S(0) − PV(div) − Ke−rT ≤ C e .

(13) If the American call option is exercised immediately before the div− idend date the option is worth C(S(t− d ), td ) = S(td ) − K. If the option is exercised just after the dividend date the option satisfies the inequality + −r(T −td ) C(S(t+ d ), td ) ≥ S(td ) − D − Ke

according to the result of exercise (12). If S(td ) − D − Ke−r(T −td) ≥ S(td ) − K

D + Ke−r(T −td) ≤ K h i D ≤ K 1 − e−r(T −td )

then the option should not be exercised before the dividend is paid. (14) Suppose there are stock prices S1 < S2 for which C a (S1 ) > C a (S2 ). Let α1 = S1 /S(0) and α2 = S2 /S(0). Note that 0 < α1 < α2 . An investor can sell an American call on α1 shares of the stock and purchase an American call on α2 shares of the stock, both with strike price K. The initial cash flow is C a (α1 S(0)) − C a (α2 S(0)) = C a (S1 ) − C a (S2 ) > 0 which can be invested at the risk-free rate. If the sold call is exercised at time t ≤ T , the investor can also exercise the purchased call. At that time the cash flow is (α2 S(t) − K)+ − (α1 S(t) − K)+ > 0 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

451

since α2 > α1 . Thus arbitrage is present. The proof for put options is similar. (15) Using the Mean Value Theorem and the pricing formula for European call options we have C e (S2 ) − C e (S1 ) ∂C e = = Φ (S ∗ ) ≤ 1 S2 − S1 ∂S S ∗ C e (S2 ) − C e (S1 ) ≤ S2 − S1 where S1 ≤ S ∗ ≤ S2 . Since the values of European call options and American call options are the same for non-dividend-paying stocks then C a (S2 ) − C a (S1 ) ≤ S2 − S1 . For the case of American puts, suppose for some stock prices S1 < S2 that P a (S1 ) − P a (S2 ) > S2 − S1 . Let α1 = S1 /S(0) and α2 = S2 /S(0) where 0 < α1 < α2 . An investor will create a portfolio by selling a put option on α1 shares of stock, purchasing a put option on α2 shares of stock, purchasing α2 − α1 shares of stock. Both the sold and purchased options have a strike price of K. The initial cash flow is then P a (α1 S(0))−P a (α2 S(0))−(α2 −α1 )S(0) = P a (S1 )−P a (S2 )−(S2 −S1 ) > 0. This amount can be invested at the risk-free rate. If the sold put option is exercised at time t ≤ T , the purchased option can be exercised and the purchased stock can be sold at the same time. This produces a cash flow of (α2 − α1 )S(t) + (K − α2 S(t))+ − (K − α1 S(t))+ > 0 since α2 > α1 and thus arbitrage is present. If the sold put option expires unused, the investor may keep the principal and interest of the initial cash flow. (16) Suppose that γ is a fixed number in the interval (0, 1). Let α, α1 , and

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

452

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

α2 be values for which αS(0) = γS1 + (1 − γ)S2

α1 S(0) = S1

α2 S(0) = S2 . Note that α = γα1 + (1 − γ)α2 and 0 < α1 < α2 . If there are prices S1 and S2 for which the following inequality holds C a (γS1 + (1 − γ)S2 ) > γC a (S1 ) + (1 − γ)C(S2 ) then an investor may create a portfolio by selling an American call option on α shares of the stock, purchase γ American call options on α1 shares of the stock, and purchase 1 − γ American call options on α2 shares of the stock. All options will have the strike price K. The initial cash flow is C a (αS(0)) − γC a (α1 S(0)) − (1 − γ)C a (α2 S(0))

= C a (γS1 + (1 − γ)S2 ) − γC a (S1 ) − (1 − γ)C(S2 )

> 0.

The positive cash flow can be invested at the risk-free interest rate. If the sold option is exercised at time t ≤ T the investor can exercise the two purchased options. Note that (αS(t) − K)+ = ([γα1 + (1 − γ)α2 ]S(t) − [γ + (1 − γ)]K)+ = (γ[α1 S(t) − K] + (1 − γ)[α2 S(t) − K])+

≤ γ(α1 S(t) − K)+ + (1 − γ)(α2 S(t) − K)+

and hence an arbitrage opportunity exists since γ(α1 S(t) − K)+ + (1 − γ)(α2 S(t) − K)+ − (αS(t) − K)+ ≥ 0. If the sold option is not exercised, the investor can keep the principal and interest from the initial cash flow. The property for American put options is proved in a similar way. (17) Since K1 < K2 < K3 there exists a 0 < γ < 1 such that K2 = γK1 + (1 − γ)K3 . 14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Solutions to Chapter Exercises

BC8495/Chp. B

453

In this case K2 − K1 = (1 − γ)(K3 − K1 ) and K3 − K2 = γ(K3 − K1 ) therefore, C(K2 ) − C(K1 ) ≤ K2 − K1 C(K2 ) − C(K1 ) ≤ (1 − γ)(K3 − K1 ) C(K2 ) − C(K1 ) ≤ 1−γ C(K2 ) ≤

C(K3 ) − C(K2 ) K3 − K2 C(K3 ) − C(K2 ) γ(K3 − K1 ) C(K3 ) − C(K2 ) γ γC(K1 ) + (1 − γ)C(K3 ).

The property for put options is proved in a similar way. (18) Letting K1 = 100, K2 = 105, and K3 = 115 we see that   2 2 K2 = K1 + 1 − K3 3 3 and according to the result of exercise (17) C(K2 ) ≤ γC(K1 ) + (1 − γ)C(K3 ) 1 2 3.80 ≤ (4.57) + (2.13) 3 3 3.80 6≤ 3.7566. Thus an arbitrage opportunity exists. (19) Suppose T1 < T2 and C a (T1 ) > C a (T2 ). An investor can sell an American call with strike time T1 and purchase an American call with strike time T2 . The initial cash flow is C a (T1 ) − C a (T2 ) > 0 and can be invested at the risk free rate. If the sold option expires unused the investor may keep the principal and interest on the initial cash flow. If the sold option is exercised at t ≤ T1 then the investor may exercise the purchased option at the same time. Thus the investor’s portfolio at time t is worth (C a (T1 ) − C a (T2 )) ert + (S(t) − K)+ − (S(t) − K)+ = (C a (T1 ) − C a (T2 )) ert > 0.

Hence an arbitrage opportunity exists. The property for American put options is proved in a similar way.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

454

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

(20) Consider the case of the American call option. According to the results of exercise (14), C a is a non-decreasing function of S(t). Therefore C a (S(t)) − (S(t) − K)+ is a non-decreasing function of S(t) for all S(t) ≤ K. Likewise, according to the results of exercise (15), C a (S2 ) − C a (S1 ) ≤ S2 − S1 for all S1 < S2 . Therefore C a (S2 ) − C a (S1 ) ≤ (S2 − K) − (S1 − K)

a

C (S2 ) − (S2 − K) ≤ C a (S1 ) − (S1 − K)

C a (S2 ) − (S2 − K)+ ≤ C a (S1 ) − (S1 − K)+ if K ≤ S1 < S2 . Thus for S(t) ≥ K the function C a (S(t)) − (S(t) − K)+ is a non-increasing function of S(t). Consequently C a (S(t)) − (S(t) − K)+ is maximized when S(t) = K.

14:58:13.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

Index

absence of arbitrage, 175, 256 absorbing boundary condition, 118 Accenture, 81 Addition rule, 24 additivity, 52 American option, 173, 315 amount due, 2 annuity, 9 arbitrage, 81, 182 Type A, 81 Type B, 81 Arbitrage Theorem, 83, 104, 174 area as probability, 48 arithmetic mean, 312 ask price, 82, 156 asset pricing, 104 at the money, 252 average, see expected value backwards parabolic equation, 184 backwards substitution, 125 bear spread, 191 bell curve, 55 Bernoulli random variable, 28, 113 beta, 303 betting strategy, 103 bid price, 82, 156 bid/ask spread, 156 bid/offer spread, see bid/ask spread binomial model, 180, 199, 211, 315 binomial probability, 218 binomial random variable, 29, 56

14:31:48.

Black-Scholes equation, 174, 182 Black-Scholes Option Pricing formula, 222 boundary condition, 184 absorbing, 118 European call option, 186 Brownian motion, 111, 128, 305 bull spread, 191 butterfly spread, 193 call option, 173 canonical form, 86 canonical linear program, 86 cap, 188 capital, 2 Capital Asset Pricing Model, 300 capital market line, 299, 301 CAPM, see Capital Asset Pricing Model centered difference formula, 60 Central Limit Theorem, 64, 293 certainty equivalent, 289 chain rule, 143, 228 multivariable, 145 closing price, 148 Complementary Slackness, 97 complementary variable, 97 compound amount, 2 compound interest, 3 compounding period, 3 concavity, 231, 284 conditional exit time, 126

BC8495/Chp. B

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

460

Juliet

Undergrad Introd to... 3rd edn

An Undergraduate Introduction to Financial Mathematics

conditional expectation, see conditional expected value, see conditional expected value conditional expected value, 36, 53 conditional probability, 25 constrained optimization, 296 constraints, 90 equality, 86 inequality, 86 continuous distribution, 48 continuous income stream, 15 continuous probability, 48 continuous random variable, 47 continuous random walk, 127 continuously compounded interest, 5 convex set, 87 correlation, 271, 275 covariance, 271 covered put, 189 covered call, 188 covered position, 242 Cox-Ross-Rubinstein model, 211 cumulative distribution function, 63 currency forward, 171 decision variables, 86 Delta, 229, 245 Delta neutrality, 245, 249 density function, 48 dependent random variables, 52 derivative, 227 financial, 155, 182 deterministic process, 137 deviation, 54 difference equation, 122 differential equation, 12 stochastic, 150 differentiation chain rule, 143 Dirac delta function, 261 discounted value, 324 discrete dividends, 179 discrete outcome, 22 discrete random process, 113 discrete stochastic process, 113 distribution

14:31:48.

continuous, 48 hypergeometric, 357 distribution function, 48 dividend yield, 179 dividends, 158, 175, 315, 319 dot product, 85 Dow Jones Industrial average, 82 drift, 66, 138, 261, 278, 305 dual, 92 Duality Theorem, 91, 99 effective interest rate, 4 efficient markets, 81 equality constraints, 86 Euclidean inner product, 85 Euler Identity, 200 European call option boundary condition, 186 final condition, 185 European option, 173 events, 21, 22 certain, 22 compound, 23 discrete, 21 impossible, 22 independent, 27 mutually exclusive, 23 excess rate of return, 303 exercise time, 173 exit time, 126 exotic options, 199 expectation conditional, 38, 54 expected utility, 288, 332 expected value, 21, 30, 50 conditional, 36, 53 of normal random variable, 61 experiment, 21, 22 expiry, see expiry date expiry date, 173 exponential decay, 207 exponential growth, 5, 137 Extreme Value Theorem, 295 Farkas Alternative Lemma, 100 feasible solution, 87

BC8495/Chp. B

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

461

Index

feasible vector, see feasible solution final condition, 184 European call option, 185 financial derivative, 155, 182 finite differences, 60 First Fundamental Theorem of Asset Pricing, 104 floor, 187 forward, see forward contract currency, 171 forward contract, 155, 158, 160 prepaid, 157 Fourier Convolution, 202 Fourier Transform, 199 alternative definition, 203 definition, 200 existence, 200 of a convolution, 202 of a derivative, 201 free variables, 89 frequency, 200 Fubini’s Theorem, 202 fully leveraged purchase, 157 function concave, 284 utility, 282 Fundamental Theorem of Calculus, 228 future value, 6, 16 futures, 155, 166 Gamma, 231, 247, 250 Gamma neutrality, 250 Gaussian elimination, 125 general linear program, 89 generalized Wiener process, 138 geometric Brownian motion, 212, 321 geometric mean, 312 geometric series, 7 gradient operator, 296 Greeks, 227 harmonic mean, 312 heat equation, 184, 204 hedge and forget, 245 hedging, 231, 241, 279

14:31:48.

BC8495/Chp. B

hedging ratio, 281 hockey stick, 185 homogeneity spatial, 114 temporal, 114 hypergeometric distribution, 357 in the money, 185, 252 income stream, 15 increments, 113 independent events, 27 independent random variables, 51, 52 inequality constraints, 86 inflation, 10 rate of, 10 inflation-adjusted rate, 10 initial condition, 184 inner product, 85 insurance, 186 integrating factor, 13 interest, 1 compound, 3 continuously compounded, 5 effective rate, 4 real rate, 10 simple, 2 interest rate, 1 intrinsic value, 321 Inverse Fourier Transform, 203 investor risk-averse, 285 risk-loving, 285 risk-neutral, 285 Itˆ o process, 143 Itˆ o’s Lemma, 146, 182, 305 Jensen’s Inequality, 286, 287 Continuous Version, 287 Discrete Version, 286 joint probability function, 32, 51 Lagrange Multipliers, 296, 303 Langevin equation, 147 Laplacian equation, 122 lattice model, 212 linear program, 84, 85

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

462

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

canonical, 86 general, 89 normal, 86 standard, 86 symmetric, 86 linear programming, 91 linear systems tridiagonal, 125 linearity, 38, 236 loans, 7 lognormal random variable, 66, 278, 306 long position, 158, 174, 305 maintenance margin, 167 margin, 166 margin call, 167 marginal distribution, 51 marginal probability, 33 market, 156 market maker, 82, 156 marking-to-market, 166 Markov property, 117, 129 martingale, 116, 131, 334 martingale property, 116 Martingale Stopping Theorem, 140 matrix notation, 87 mean, see expected value, see expected value mean square, 131 minimum variance, 278, 280, 295 Monty Hall problem, 24 Multiplication rule, 26 naked position, 242 negative part, 87 New York Stock Exchange, 81 normal form, 86 normal linear program, 86 normal probability distribution, 56, 63 normal random variable, 47, 55 NYSE, see New York Stock Exchange objective function, 86 odds, 83

14:31:48.

offer price, 156 operations research, 91 optimal exercise, 331 optimal portfolio, 280 optimal solution, 88 optimal stopping time, 332 optimal time, 331 option, 173 American, 173, 315 call, 173 European, 173 perpetual call, 337 put, 173 option elasticity, 238 option pricing formula European call, 209 European call with dividends, 257 European put, 210 Optional Stopping Theorem, 140 Ornstein-Uhlenbeck mean reverting, 152 Ornstein-Uhlenbeck equation, 147 out of the money, 185, 252 outcomes, 21 discrete, 21, 22 outright purchase, 157 part negative, 87 positive, 86 partial differential equation, 183 Black-Scholes, 174, 235 linear, 184 parabolic type, 184 second order, 184 PDE, see partial differential equation perpetual call, 337 Poisson equation, 124 polar coordinates, 61 Polya’s urn, 150 portfolio, 182, 236, 277 Delta neutral, 245 Gamma neutral, 250 optimal, 280 rebalancing, 245 portfolio selection problem, 290, 292

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

463

Index

Portfolio Separation Theorem, 297 position covered, 242 long, 158, 174 naked, 242 short, 158, 174 positive part, 70, 86 prepaid forward contract, 157 present value, 1, 6, 16, 182, 319 pricing formula European call option, 209 European call option with dividends, 257 European put option, 210 primal, 92 principal, see principal amount principal amount, 2 probability, 22 classical, 23 conditional, 25 continuous, 48 empirical, 23 marginal, 33 probability density function, 48 probability distribution, 27 normal, 63 probability distribution function, 48 probability function, 27 probability measure risk-neutral, 104 Proctor and Gamble, 82 product rule, 153 profit, 161 purchase fully leveraged, 157 outright, 157 put option, 173 Put-Call Parity, 176, 230 continuous dividends, 179 discrete dividends, 179 quadratic variation, 116, 130 random experiment, 282 random process discrete, 113

14:31:48.

BC8495/Chp. B

random variable, 21 Bernoulli, 113 binomial, 21, 56 continuous, 47 discrete, 21 lognormal, 66, 278 normal, 47, 55 uncorrelated, 275, 295 uniform, 49 random walk, 111 continuous, 127 increments, 113 symmetric, 112 unbiased, 112 rate of inflation, 10 rate of return, 1, 14, 271, 278 excess, 303 real rate of interest, 10 rebalancing, 245 recombining trees, 216 retirement, 8 return rate of, see rate of return, 14 Rho, 233 Riemann sum, 16 risk, 271, 303 risk-averse investor, 285 risk-loving investor, 285 risk-neutral investor, 285 risk-neutral probability measure, 104 round trip cost, 168 saddle point, 108 sample means, 64 Schwarz Inequality, 276 secant lines, 284 sensitivity, 227 separation of variables, 184, 207 short position, 158, 174, 305 simple interest, 2 simplex method, 86 slack variables, 88 spatial homogeneity, 114, 120, 129 spot rate, 12 spread, see standard deviation bear, 191

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

464

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

bull, 191 butterfly, 193 spreads, 190 standard deviation, 21, 41, 55 standard form, 86 standard linear program, 86 standard normal random variable, 63 Stirling’s formula, 58 stochastic calculus, 111, 128 stochastic differential equation, 150, 261 stochastic integral, 131, 134 differential form, 134 stochastic process, 128, 150, 182 discrete, 113 stopped process, 140 stopping time, 121, 139, 331, 332 straddle, 192 strangle, 192 strike price, 173 strike time, 173 stub quotes, 82 surrogate, 279 symmetric form, 86 symmetric linear program, 86 symmetric random walk, 112

unbiased random walk, 112 uncorrelated random variable, 275 uniform random variable, 49 utility, 282, 332 decreasing, 284 utility function, 282, 284 value discounted, 324 intrinsic, 321 vanilla options, 199 variables complementary, 97 unrestricted in sign, 89 variance, 38, 54 alternative formula, 54 minimum, 278 of normal random variable, 62 vector inequalities, 86 Vega, 232 volatility, 66, 138, 148, 192, 233, 261, 278, 305 Weak Duality Theorem, 95 Wiener process, 128, 306 yield curve, 13

Taylor remainder, 145 Taylor series, 249 Taylor’s Theorem, 144 temporal homogeneity, 114 Theta, 227 transaction costs, 156 Treasury Bonds, 175 tridiagonal linear systems, 125 tridiagonal matrix, 125

14:31:48.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

Bibliography

Barrow, J. D. (2008). One Hundred Essential Things You Didn’t Know You Didn’t Know: Math Explains Your World, W. W. Norton and Company, Inc., New York, New York, USA. Bleecker, D. and Csordas, G. (1996). Basic Partial Differential Equations, International Press, Cambridge, Massachusetts, USA. Boyce, W. E. and DiPrima, R. C. (2001). Elementary Differential Equations and Boundary Value Problems, 7th edition, John Wiley and Sons, Inc., New York, USA. Bradley, S. P., Hax, A. C. and Magnanti, T. L. (1977). Applied Mathematical Programming, Addison-Wesley Publishing Company, Reading, MA, USA. Broverman, S. A. (2004). Mathematics of Investment and Credit, 3rd edition, ACTEX Publications, Winsted, CT, USA. Burden, R. L. and Faires, J. D. (2005). Numerical Analysis, 8th edition, Thomson Brooks/Cole, Belmont, CA, USA. Chawla, M. M. (2006), Accurate computation of the Greeks, International Journal of Applied Mathematics, 7, 4, pp. 379–388. Chawla, M. M. and Evans, D. J. (2005), High-accuracy finite difference methods for the valuation of options, International Journal of Computer Mathematics, 82, 9, pp. 1157–1165. Churchill, R. V., Brown, J. W. and Verhey, R. F. (1976). Complex Variables and Applications, 3rd edition, McGraw-Hill Book Company, New York, USA. Cornuejols, G. and T¨ ut¨ unc¨ u, R. (2007). Optimization Methods in Finance, Cambridge University Press, Cambridge, UK. Courant, R. and Robbins, H. (1969). What is Mathematics?, Oxford University Press, Oxford, UK. Cox, J.C., Ross, S. and Rubinstein, M. (1979), Option pricing: a simplified approach, Journal of Financial Economics, 7, pp. 229–264. Cuthbertson, K. and Nitzsche, D. (2004). Quantitative Financial Economics: Stocks, Bonds, and Foreign Exchange, 2nd edition, John Wiley & Sons Inc., West Sussex, UK.

14:31:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

456

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Dean, J., Corvin, J. and Ewell, D. (2001). List of Playboy’s playmates of the month with data sheet stats, Available at http://www3.sympatico.ca/ jimdean/pmstats.txt. DeGroot, M. H. (1975). Probability and Statistics, Behavioral Science: Quantitative Methods. Addison-Wesley Publishing Company, Reading, MA, USA. Durrett, R. (1996). Stochastic Calculus: A Practical Introduction, CRC Press, Boca Raton, FL, USA. Franklin, J. (1980). Methods of Mathematical Economics, Springer-Verlag, New York, NY, USA. Gale, D., Kuhn, H. W. and Tucker, A. W. (1951). Linear programming and the theory of games, in T.C. Koopmans, editor, Activity Analysis of Production and Allocation, Wiley, New York, NY, USA, pp. 317–329. Gale, D. (1960). The Theory of Linear Economic Models, McGraw-Hill, New York, NY, USA. Gard, T.C. (1988). Introduction to Stochastic Differential Equations, Marcel Dekker, New York, New York, USA. Gardner, M. (October 1959). “Mathematical Games” column, Scientific American, pp. 180–182. Goldstein, L. J., Lay, D. C. and Schneider, D. I. (1999). Brief Calculus and Its Applications, 8th edition, Prentice Hall, Upper Saddle River, NJ, USA. Golub, G. H. and Van Loan, C. F. (1989). Matrix Computations, 2nd edition, Johns Hopkins University Press, Baltimore, Maryland, USA. Gonz´ alez-D´ıaz, J., Garc´ıa-Jurado, I. and Fiestras-Janiero, M. G. (2010). An Introductory Course on Mathematical Game Theory, Vol. 115 of Graduate Studies in Mathematics, American Mathematical Society, Providence, Rhode Island, USA. Guo, W. and Su, T. (2006). “Option Put-Call Parity Relations When the Underlying Security Pays Dividends”, International Journal of Business and Economics, 5, pp. 225–230. Greenberg, M. D. (1998). Advanced Engineering Mathematics, 2nd edition, Prentice-Hall, Inc., Upper Saddle River, New Jersey, USA. Grimmett, G. R. and Stirzaker, D. R. (1982). Probability and Random Processes, Oxford University Press, London, UK. Hull, J. C. (2000). Options, Futures, and Other Derivatives, Prentice-Hall, Inc., Upper Saddle River, New Jersey, USA. Jeffrey, A. (2002). Advanced Engineering Mathematics, Harcourt Academic Press, Burlington, Massachusetts, USA. Karatzas, I. and Shreve, S. E. (1991). Brownian Motion and Stochastic Calculus, Springer-Verlag, New York, NY, USA. Kijima, M. (2003). Stochastic Processes with Applications to Finance, Chapman & Hall/CRC, Boca Raton, FL, USA. Lawler, G. F. (2006). Introduction to Stochastic Processes, 2nd edition, Chapman & Hall/CRC, Boca Raton, FL, USA. Luenberger, D. G. (1998). Investment Science, Oxford University Press, New York, USA.

14:31:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

Juliet

Bibliography

Undergrad Introd to... 3rd edn

BC8495/Chp. B

457

Marsden, J. E. and Hoffman, M. J. (1987). Basic Complex Analysis, W. H. Freeman and Company, New York, NY, USA. McDonald, R. L. (2006). Derivatives Markets, Addison-Wesley, Boston, MA, USA. Mehta, N., Thomasson, L. and Barret, P. M. (2010). The Machines that Ate the Market, Bloomberg Businessweek, 4180, pp. 48–55. Mikosch, T. (1998). Elementary Stochastic Calculus With Finance in View, Vol. 6 of Advanced Series on Statistical Science & Applied Probability, World Scientific Publishing, River Edge, New Jersey, USA. Neftci, S. N. (2000). An Introduction to the Mathematics of Financial Derivatives, 2nd edition, Academic Press, San Diego, CA, USA. Noble, D. and Daniel, J. W. (1988). Applied Linear Algebra, 3rd edition, Prentice Hall, Englewood Cliffs, NJ, USA. Øksendal, B. (2003). Stochastic Differential Equations: An Introduction with Applications, 6th editiion, Springer-Verlag, Berlin, Germany. Redner, S. (2001). A Guide to First Passage Processes, Cambridge University Press, Cambridge, UK. Ross, S. M. (1999). An Introduction to Mathematical Finance: Options and Other Topics, Cambridge University Press, Cambridge, UK. Ross, S. M. (2003). Introduction to Probability Models, 8th edition, Academic Press, San Diego, CA, USA. Ross, S. M. (2006). A First Course in Probability, 7th edition, Prentice Hall, Inc., Upper Saddle River, NJ, USA. Selvin, S. (1975). A problem in probability, American Statistician, 29, 1, p. 67. Seydel, R. (2002). Tools for Computational Finance, Springer-Verlag, New York, NY, USA. Shodor Education Foundation, Inc., “Graphing and Interpreting Bivariate Data”, 20 May 2008 hhttp://www.shodor.org/interactivate/discussions/ GraphingData/i. Shreve, S. E. (2004). Stochastic Calculus for Finance I, Springer-Verlag, New York, NY, USA. Shreve, S. E. (2004). Stochastic Calculus for Finance II, Springer-Verlag, New York, NY, USA. Smith, R. T. and Minton, R. B. (2002). Calculus, 2nd edition, McGraw-Hill, Boston, MA, USA. Steele, J. M. (2001). Stochastic Calculus and Financial Applications, Volume 45 in Applications of Mathematics, Springer-Verlag, New York, NY, USA. Stewart, J. (1999). Calculus, 4th edition, Brooks/Cole Publishing Company, Pacific Grove, CA, USA. Strang, G. (1986). Introduction to Applied Mathematics, Wellesley-Cambridge Press, Wellesley, MA, USA. Taylor, A. E. and Mann, W. R. (1983). Advanced Calculus, 3rd edition, John Wiley & Sons, Inc., New York, NY, USA. vos Savant, M. (February 1990). “Ask Marilyn” column, Parade Magazine, p. 12. Williams, A. C. (1970). Complementarity Theorems for Linear Programming, SIAM Review, 12, 1, pp. 135–137.

14:31:39.

May 25, 2012

14:36 WSPC/Book Trim Size for 9in x 6in

458

Juliet

Undergrad Introd to... 3rd edn

BC8495/Chp. B

An Undergraduate Introduction to Financial Mathematics

Wilmott, P. (2006). Paul Wilmott on Quantitative Finance, 2nd edition, John Wiley & Sons, Inc., Hoboken, NJ, USA. Wilmott, P., Howison, S. and Dewynne, J. (1995). The Mathematics of Financial Derivatives: A Student Introduction, Cambridge University Press, Cambridge, UK. Winston, W. L. (1994). Operations Research: Applications and Algorithms, International Thompson Publishing, Belmont, CA, USA.

14:31:39.