Ruin Probabilities: Smoothness, Bounds, Supermartingale Approach [1 ed.] 1785482181, 978-1-78548-218-2, 105-105-106-1

Table of contents :
Content: Part 1: Smoothness of the Survival Probabilities with Applications 1: Classical Results on the Ruin Probabilities 2: Classical Risk Model with Investments in a Risk-Free Asset 3: Risk Model with Stochastic Premiums Investments in a Risk-Free Asset 4: Classical Risk Model with a Franchise and a Liability Limit 5: Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model 6: Risk Models with Investments in Risk-Free and Risky Assets Part 2: Supermartingale Approach to the Estimation of Ruin Probabilities 7: Risk Model with Variable Premium Intensity and Investments in One Risky Asset 8: Risk Model with Variable Premium Intensity and Investments in One Risky Asset up to the Stopping Time of Investment Activity 9: Risk Model with Variable Premium Intensity and Investments in One Risk-Free and a Few Risky Assets

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Ruin Probabilities

Series Editor Nikolaos Limnios

Ruin Probabilities Smoothness, Bounds, Supermartingale Approach

Yuliya Mishura Olena Ragulina

First published 2016 in Great Britain and the United States by ISTE Press Ltd and Elsevier Ltd

Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms and licenses issued by the CLA. Enquiries concerning reproduction outside these terms should be sent to the publishers at the undermentioned address: ISTE Press Ltd 27-37 St George’s Road London SW19 4EU UK

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Notices Knowledge and best practice in this field are constantly changing. As new research and experience broaden our understanding, changes in research methods, professional practices, or medical treatment may become necessary. Practitioners and researchers must always rely on their own experience and knowledge in evaluating and using any information, methods, compounds, or experiments described herein. In using such information or methods they should be mindful of their own safety and the safety of others, including parties for whom they have a professional responsibility. To the fullest extent of the law, neither the Publisher nor the authors, contributors, or editors, assume any liability for any injury and/or damage to persons or property as a matter of products liability, negligence or otherwise, or from any use or operation of any methods, products, instructions, or ideas contained in the material herein. For information on all our publications visit our website at http://store.elsevier.com/ © ISTE Press Ltd 2016 The rights of Yuliya Mishura and Olena Ragulina to be identified as the authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. British Library Cataloguing-in-Publication Data A CIP record for this book is available from the British Library Library of Congress Cataloging in Publication Data A catalog record for this book is available from the Library of Congress ISBN 978-1-78548-218-2 Printed and bound in the UK and US

Contents

Preface ................. .

Xl

Part 1. Smoothness of the Survival Probabilities with Applications

..

Chapter 1. Classical Results on the Ruin Probabilities ............ .

3

1.1. Classical risk model .....

3

I .1.1. Description of the model

3

1.1.2. Net profit condition and the behavior of the infinite-horizon ruin probability . . . . . . . . .

5

1.1.3. Integro-differential equations for the survival probabilities ....................... .

6

1.1.4. Integral equation for the infinite-horizon ruin probability ............................ .

8

1.1.5. Laplace transform of the infinite-horizon survival probability ............... .

10

1.1.6. Analytic expressions for the infinite-horizon survival probability .......................... .

11

1.1.7. Cramer-Lundberg approximation for the infinite-horizon ruin probability................

13

1.1.8. Lundberg inequality for the infinite-horizon ruin probability ............. .

16

1.1.9. Bibliographical notes ...... .

18

1.2. Risk model with stochastic premiums

22

vi

Ruin Probabilities

1.2.1. Description of the model

22

1.2.2. Basic results . . . . .

23

1.2.3. Bibliographical notes ..

25

Chapter 2. Classical Risk Model with Investments ........... .

27

2. 1. Description of the model . . . . . . .

27

in a Risk-Free Asset

2.2. Continuity and differentiability of the infinite-horizon survival probability........ .

28

2.3. Continuity of the finite-horizon survival probability and existence of its partial derivatives .

32

. .

32

2.3.2. Examples . . . . .

40

2.3.1. Main theorem

2.4. Bibliographical notes

47

Chapter 3. Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset .

49

3.1. Description of the model .......

49

3.2. Continuity and differentiability of the infinite-horizon survival probability.

50

3.2.1. Main results ........... .

50

3.2.2. Zero interest rate ......... .

56

3.3. Continuity of the finite-horizon survival probability and existence of its partial derivatives .

59

Chapter 4. Classical Risk Model with a Franchise and a Liability Limit 4.1. Introduction ............ .

65 65

4.2. Survival probability in the classical risk model with a franchise ........... .

68

4.2.1. Analytic expression for the survival probability . . . . . . . . . . . . . . . . . .

68

4.2.2. Case of small and large enough initial surpluses .................... .

80

4.3. Survival probability in the classical risk model with a liability limit........................... .

84

4.3.1. Analytic expression for the survival probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85

Contents

vii

4.3.2. Case of small enough and large enough initial surpluses ................. .

91

4.4. Survival probability in the classical risk model with both a franchise and a liability limit.

93

4.4.1. Analytic expression for the survival probability.............. .

4.4.2. Case of small initial surpluses ...

94 102

Chapter 5. Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model 5.1. Introduction ............... .

105 105

5.2. Optimal control by the franchise amount

106

5.2.1. Problem statement .......... .

106

5.2.2. Hamilton-Jacobi-Bellman equation .

108

5.2.3. Existence theorem .......... .

110

5.2.4. Verification theorem......... .

113

5.2.5. Exponentially distributed claim sizes

116

5.3. Optimal control by the deductible amount

119

5.3.1. Problem statement ...........

119

5.3.2. Hamilton-Jacobi-Bellman equation .

121

5.3.3. Existence and verification theorems .

122

5.3.4. Exponentially distributed claim sizes

123

5.4. Bibliographical notes .......... .

124

Chapter 6. Risk Models with Investments in Risk-Free and Risky Assets ....... . 6. 1. Description of the models . . . . . . . . . . . . . .

127 127

6.2. Classical risk model with investments in risk-free and risky assets ..................... . . . . . . . . . 129

6.2.1. Upper and lower bounds for the infinite-horizon survival probability..

129

6.2.2. Continuity and differentiability of the infinite-horizon survival probability.....

136

6.2.3. Continuity of the finite-horizon survival probability and existence of its partial derivatives ......... . 140

6.3. Risk model with stochastic premiums and investments in risk-free and risky assets ............... . 147

Contents

ix

Chapter 9. Risk Model with Variable Premium Intensity and Investments in One Risk-Free and a Few Risky Assets

. .. . . . . . .

205

9. 1. Description of the model . . . . . .... . . .

205

9.2. Existence and uniqueness theorem...... .

209

9.3. Supermartingale property for the exponential process ........................ .

209

9.4. Upper exponential bound for the ruin probability

214

9.5. Case of one risky asset

223

9.6. Examples ..

224

Appendix ..... .

231

Bibliography

239

Abbreviations and Notation

255

Index

259

Preface

Risk theory is traditionally considered as a branch of insurance mathematics and its active area of research. Since Filip Lundberg [LUN 03] introduced the collective risk model in 1903, the estimation of ruin probabilities has been one of the central directions for investigations in risk theory. The infinite- and finite-horizon ruin probabilities are the probabilities that the surplus of an insurance company becomes negative on infinite and finite time intervals, respectively. The infinite-horizon ruin probability is usually considered as a function of the initial surplus, and the finite-horizon ruin probability is assumed to be a function of the initial surplus and time interval. Instead of ruin probabilities, one sometimes deals with the corresponding survival probabilities, which are the probabilities that the surplus of the insurance company does not become negative. So, the survival of the insurance company at least implies that it is able to reimburse claims and meet its obligations. Although other characteristics have also been proposed and investigated (e.g. distribution of the ruin time, surplus prior to ruin, deficit at ruin and Gerber–Shiu expected discounted penalty functions), the ruin or survival probabilities are still the main objects of interest in risk theory. In this book, we will consider continuous-time risk models. Due to the obvious relationship between ruin and survival probabilities, from the mathematical viewpoint it makes no difference which of these two characteristics is studied. Since sometimes it is more convenient to deal with the ruin probabilities and sometimes with the survival probabilities, in what

xii

Ruin Probabilities

follows we consider both of them. To be more precise, we investigate both survival probabilities in Chapters 2–6 and ruin probabilities in Chapters 7–9. This monograph mainly provides the results obtained by the authors recently in this field. It is a revised and extended version of the material published in [AND 07a, AND 07b, BON 12, MIS 15, RAG 10, RAG 11, RAG 12, RAG 14, RAG 15]. The book deals with several aspects of risk theory. The first of them is the smoothness of the survival probabilities. Particularly, we investigate the continuity and differentiability of the infinite- and finite-horizon survival probabilities in detail for different risk models. The initial motivation for this investigation is the reasonable derivation of integro-differential equations for these functions. These equations turn out to be very useful for calculation of the survival probabilities. In some cases, one can find their closed-form solutions. In other cases, the integro-differential equations provide some analytic information concerning the survival probabilities. In particular, they are used for obtaining bounds and approximations for the survival probabilities or investigation of their asymptotics. Note that it is not a problem to write down the integro-differential equations for the survival probabilities and most of the equations we derive have already been obtained and used for further investigation. Furthermore, one can easily write down all these equations in terms of infinitesimal generators for stopped surplus processes, which are Markov processes (see, e.g. [ASM 10, p. 248] or [SCH 08b, pp. 220, 225]). Nevertheless, usually one has to make the additional assumption concerning the differentiability of the survival probabilities or conditions when these functions that are really differentiable are not studied. Thus, investigation of the continuity and differentiability of the survival probabilities is of great interest. As we will see, these functions are not always differentiable in the classical sense. Moreover, sometimes points of discontinuity are also possible. So, the continuity and differentiability of the survival probabilities must be investigated separately for each risk model, and we study these properties for different models in this book. Next, we concentrate on some possible applications of the results concerning the smoothness of the survival probabilities. Firstly, we find analytic expressions for the infinite-horizon survival probabilities in a few cases when the c.d.f. of the insurance compensation is a sum of absolutely continuous and discrete components. Secondly, we solve the problems of

Preface

xiii

optimal control by franchise and deductible amounts in the classical risk model from the viewpoint of survival probability maximization. Thirdly, we obtain relationships connecting accuracy and reliability of uniform approximations of the survival probabilities from their statistical estimates. Finally, we introduce and apply the supermartingale approach to obtain upper exponential bounds for the infinite-horizon ruin probabilities in different generalizations of the classical risk model with risky investments. This approach generalizes the martingale one introduced by Gerber [GER 73] to obtain the famous exponential bound for the infinite-horizon ruin probability in the classical risk model, which is known as the Lundberg inequality. The book is not a textbook in risk theory and does not provide general results. Nevertheless, we give some basic results concerning ruin probabilities in the classical risk model and the risk model with stochastic premiums in Chapter 1. The reader can find the general results in risk theory in [ASM 10, BEA 84, BÜH 70, DIC 10, GER 79, GRA 91, ROL 99]. Furthermore, we refer the reader to [SCH 08b] for a systematic description of optimal control methods applied to insurance problems and [AZC 14] for the viscosity approach to optimal control problems in insurance. This work is intended for researchers in probability theory, actuarial sciences and financial mathematics, as well as for graduate and postgraduate students. Though the book is mainly theoretical, it is also accessible to practitioners who want to extend their knowledge in insurance mathematics. The book is self-contained and does not require any complementary material. All its main results are provided with proofs. It is assumed that the reader has a basic knowledge in probability theory, stochastic processes, stochastic calculus, ordinary and stochastic differential equations and calculus. We provide some well-known mathematical results which are often used in the main text in the Appendix. Moreover, we refer the reader to [DOO 53, ELL 82, JAC 06, KAR 91, ØKS 07, ROG 00, ROL 99, SHR 04] for basic notions of the theory of stochastic processes and stochastic calculus and to [CHE 05, IKE 89, KAR 91, ØKS 03] for basic facts concerning stochastic differential equations.

xiv

Ruin Probabilities

The main material of the book is divided into two parts and organized as follows. In Part 1, we investigate the continuity and differentiability of survival probabilities for different risk models and concentrate on some possible applications of these results. In Chapter 1, we give some basic results concerning ruin probabilities in the classical risk model and the risk model with stochastic premiums. In Chapter 2, we deal with a generalization of the classical risk model where an insurance company invests all surplus in a risk-free asset. We investigate the continuity and differentiability of infiniteand finite-horizon survival probabilities in detail. In Chapter 3, we obtain the corresponding results for the risk model with stochastic premiums and investments in a risk-free asset. In Chapter 4, we consider the classical risk model under the additional assumption that an insurance company applies a franchise and a liability limit. Assuming that claim sizes are exponentially distributed we find analytic expressions for the infinite-horizon survival probabilities and investigate how a franchise and a liability limit change the survival probability for small and large enough initial surpluses. In Chapter 5, we solve the problems of optimal control by franchise and deductible amounts in the classical risk model from the viewpoint of survival probability maximization. In Chapter 6, we extend the results of Chapters 2 and 3 to a case where an insurance company invests all surplus in risk-free and risky assets proportionally and the price of the risky asset follows a jump process. Moreover, we obtain relationships connecting accuracy and reliability of uniform approximations of the survival probabilities by their statistical estimates. In Part 2, we apply the supermartingale approach to obtain upper exponential bounds for infinite-horizon ruin probabilities in different generalizations of the classical risk model where a premium intensity depends on a current surplus of an insurance company. To this end, in Chapter 7 we assume that all surplus is invested in one risky asset, the price of which follows a geometric Brownian motion, and the premium intensity grows rapidly with increasing surplus. Furthermore, we investigate the question concerning the probability of explosion of the surplus process between claim arrivals in detail. In Chapter 8, we suppose that all surplus is also invested in one risky asset, the price of which follows a geometric Brownian motion, but an insurance company stops its investment activity when the price of the risky asset goes below some fixed level or outside some fixed interval. In Chapter 9, we assume that the surplus is invested in one risk-free and a few risky assets, the prices of which follow geometric Brownian motions, and borrowing is also possible.

Preface

xv

We conclude the preface by thanking all those without whose help and encouragement this book would never have been written. First of all, it is our pleasure to thank Hanspeter Schmidli, who is a great authority in risk theory, for his hospitality during our visits to the University of Cologne and numerous helpful discussions of the problems considered in the book. Moreover, we are grateful to the co-authors of our papers, in particular Maryna Lundgren (Androshchuk), for their contribution to our joint work.

Yuliya M ISHURA Olena R AGULINA Kyiv May, 2016

PART 1

SMOOTHNESS OF THE SURVIVAL PROBABILITIES WITH ApPLICATIONS

1 Classical Results on the Ruin Probabilities

In this chapter, we formulate some basic results concerning ruin probabilities in the classical risk model and the risk model with stochastic premiums. These results are used in Chapters 2–9. Let (Ω, F, P) be a probability space satisfying the usual conditions and let all the stochastic objects that we use in Chapters 1–9 be defined on it. 1.1. Classical risk model We now deal with the classical risk model, which is also called the Cramér– Lundberg model. For details and more information, see, for example [ASM 10, BEA 84, BÜH 70, DIC 10, GER 79, GRA 91, KAA 08, ROL 99, SCH 08b]. 1.1.1. Description of the model In the classical risk model, claim sizes form a sequence (Yi )i≥1 of nonnegative i.i.d. r.v.’s with c.d.f. F (y) = P[Yi ≤ y] and finite expectation μ > 0. Let τi be the time when the ith claim arrives. The number of claims on the time interval [0, t] is a Poisson process (Nt )t≥0 with constant intensity λ > 0. The r.v.’s (Yi )i≥1 and the process independent. Thus, the total  t (Nt )t≥0 are mutually 0 claims on [0, t] equal N Y . We set Y = 0 if Nt = 0. Let (Ft )t≥0 be i=1 i i=1 i the filtration generated by (Yi )i≥1 and (Nt )t≥0 . The insurance company has a non-negative initial surplus x and receives premiums with constant intensity c > 0.

4

Ruin Probabilities

Let Xt (x) be the surplus of the insurance company at   time t provided that its initial surplus is x. Then, the surplus process Xt (x) t≥0 follows: Xt (x) = x + ct −

Nt 

Yi ,

t ≥ 0.

[1.1]

i=1

The ruin time is defined by τ (x) = inf{t ≥ 0 : Xt (x) < 0}. We suppose that τ (x) = ∞ if Xt (x) ≥ 0 for all t ≥ 0. The infinite-horizon ruin probability is given by   ψ(x) = P inf s≥0 Xs (x) < 0 , which is equivalent to ψ(x) = P[τ (x) < ∞]. The finite-horizon ruin probability is given by   ψ(x, t) = P inf 0≤s≤t Xs (x) < 0 , which is equivalent to ψ(x, t) = P[τ (x) ≤ t]. The corresponding infinite-horizon and finite-horizon survival probabilities equal ϕ(x) = 1 − ψ(x) and ϕ(x, t) = 1 − ψ(x, t). The notions of ruin time as well as infinite- and finite-horizon ruin and survival probabilities introduced above are applicable to all risk models considered in this book, unless otherwise stated.

Classical Results on the Ruin Probabilities

5

It is easily seen that ϕ(x) and ϕ(x, t) are non-decreasing functions w.r.t. x. Furthermore, ϕ(x, t) is a non-increasing function w.r.t. t. But it is not clear whether these functions are continuous or differentiable. As we will see later, survival probabilities are not always differentiable. Moreover, sometimes points of discontinuity of survival probabilities are also possible (see Chapter 3). So, the continuity and differentiability of these functions must be investigated separately for each risk model. Since there is elementary dependence between ruin probabilities and the corresponding survival probabilities, in what follows, we formulate results for either of them. Sometimes it is more convenient to deal with the ruin probabilities and sometimes with the survival probabilities. 1.1.2. Net profit condition and the behavior of the infinite-horizon ruin probability It turns out that the behavior of ψ(x) depends substantially on whether the condition c > λμ is true or not.   L EMMA 1.1.– Let the surplus process Xt (x) t≥0 follow [1.1]. i) If c ≤ λμ, then ψ(x) = 1 for all x ≥ 0. ii) If c > λμ, then limx→+∞ ψ(x) = 0. To prove lemma 1.1, we will need the following theorem from the theory of random walks. T HEOREM 1.1.– ([ROL 99, p. 233], theorem 6.3.1) Let (Zi )i≥1 be a sequence of i.i.d. r.v.’s, such that P[Zi > 0] > 0, P[Zi < 0]> 0 and E[Zi ] < ∞. Moreover, let the r.v.’s Z˜n , n ≥ 1, be given by Z˜n = ni=1 Zi .   i) If E[Zi ] > 0, then P limn→∞ Z˜n = +∞ = 1.   ii) If E[Zi ] < 0, then P limn→∞ Z˜n = −∞ = 1.   iii) If E[Zi ] = 0, then P lim supn→∞ Z˜n = +∞ = 1 and   ˜ P lim inf n→∞ Zn = −∞ = 1. Note that the proof of theorem 1.1 is straightforward in the first two cases. Indeed, by the strong law of large numbers, we get

6

Ruin Probabilities

    P limn→∞ Z˜n /n = E[Zi ] = 1. Therefore, P limn→∞ Z˜n = +∞ = 1 if   E[Zi ] > 0 and P limn→∞ Z˜n = −∞ = 1 if E[Zi ] < 0. The case E[Zi ] = 0 is much more difficult (see, for example [ROL 99]). P ROOF (Proof of lemma 1.1).– Let Zi = Yi − c(τi − τi−1 ), i ≥ 1. It is easily seen that the r.v.’s Zi satisfy the requirements of theorem 1.1 and E[Zi ] = μ − c/λ.   Nt Next, let Z˜n = ni=1 Zi , n ≥ 1, and M = supt≥0 Y − ct . It is i=1 i evident that M = supn≥1 Z˜n , which gives ψ(x) = P[M > x],

x ≥ 0.

[1.2]

  If c ≤ λμ, then E[Zi ] ≥ 0 and P lim supn→∞ Z˜n = +∞ = 1 by theorem 1.1. Hence, from [1.2] we conclude that ψ(x) = 1 for all x ≥ 0.   If c > λμ, then E[Zi ] < 0 and P limn→∞ Z˜n = −∞ = 1 by theorem 1.1, which yields P[M < ∞] = 1. Therefore, limx→+∞ ψ(x) = 0 by [1.2].  The condition c > λμ iscalled the net  profit condition. This name can be explained by the fact that E Xt (x) − x = (c − λμ)t. In what follows, we concentrate on that case. Moreover, we often assume that the insurance company uses the expected value principle for premium calculation, which means that c = λμ(1 + θ), where θ > 0 is a safety loading. 1.1.3. Integro-differential equations for the survival probabilities T HEOREM  1.2.–  ([ROL 99, pp. 162–163], theorem 5.3.1) Let the surplus process Xt (x) t≥0 follow [1.1]. Then, ϕ(x) is continuous on R+ with right and left derivatives ϕ+ (x) and ϕ− (x), respectively. Moreover, we have

x ϕ(x − y) dF (y), x ≥ 0, [1.3] cϕ+ (x) = λϕ(x) − λ 0

and cϕ− (x)

= λϕ(x) − λ

x− 0

ϕ(x − y) dF (y),

x > 0.

[1.4]

Classical Results on the Ruin Probabilities

7

P ROOF.– It is easily seen that ruin does not occur up to the time τ1 and it will not take place after τ1 if and only if Y1 ≤ x + cτ1 and ruin does not happen on the time interval (τ1 , +∞) with the initial surplus x + cτ1 − Y1 . Moreover, note that τ1 is exponentially distributed with mean 1/λ. So, we have P[τ1 > t] = e−λt for all t ≥ 0. Thus, by the law of total probability, we get

x+cs

t −λt −λs ϕ(x) = e ϕ(x + ct) + λe ϕ(x + cs − y) dF (y) ds [1.5] 0

0

for all x ≥ 0 and t ≥ 0. Letting t ↓ 0 in [1.5] we conclude that ϕ(x) is right-continuous on R+ . Furthermore, we can rewrite [1.5] as c

1 − e−λt ϕ(x + ct) − ϕ(x) = ϕ(x + ct)

tct

x+cs t 1 − λe−λs ϕ(x + cs − y) dF (y) ds. t 0 0

[1.6]

This shows that ϕ(x) is differentiable from the right on R+ . Letting t ↓ 0 in [1.6] yields [1.3]. Next, for x > 0 and t ≤ x/c, we can rewrite [1.5] in the form ϕ(x − ct) = e−λt ϕ(x)

x−c(t−s)

t −λs + λe ϕ(x − c(t − s) − y) dF (y) ds. 0

[1.7]

0

Letting t ↓ 0 in [1.7] we deduce that ϕ(x) is left-continuous on (0, +∞). Rewriting [1.7] in the form c

1 − e−λt ϕ(x) − ϕ(x − ct) = ϕ(x) ct t



1 t −λs x−c(t−s) − λe ϕ(x − c(t − s) − y) dF (y) ds t 0 0

[1.8]

shows that ϕ(x) is differentiable from the left on (0, +∞). Letting t ↓ 0 in [1.8] yields [1.4]. 

8

Ruin Probabilities

R EMARK 1.1.– It follows from [1.3] and [1.4] that if c > λμ, then ϕ(x) is differentiable on R+ , except at positive points y where F (y) is not continuous. Equation [1.3] involves both the derivative and an integral of ϕ(x), so it is not easily solved in the general case. Furthermore, we can easily show that ϕ(x, t) satisfies the following partial integro-differential equation:

x ∂ϕ(x, t) ∂ϕ(x, t) −c + λϕ(x, t) − λ ϕ(x − y, t) dF (y) = 0. ∂t ∂x 0 The question regarding the existence of partial derivatives of ϕ(x, t) is investigated in section 2.3. 1.1.4. Integral equation for the infinite-horizon ruin probability   T HEOREM 1.3.– Let the surplus process Xt (x) t≥0 follow [1.1] and c > λμ. Then, ψ(x) satisfies the integral equation

+∞

x     cψ(x) = λ 1−F (y) dy+λ ψ(x−y) 1−F (y) dy, x ≥ 0. [1.9] x

0

P ROOF.– Integrating [1.3] over [0, x] yields

x   ϕ(z) dz − λ c ϕ(x) − ϕ(0) = λ

=λ

=λ

=λ

=λ

0

x 0 x 0 x 0

=λ

0

ϕ(z) dz − λ

ϕ(z) dz − λ

ϕ(z) dz − λ

0

0

0

0

0

x x y

ϕ(z − y) dF (y) dz ϕ(z − y) dz dF (y)

x x−y 0

x x−z 0

ϕ(z) dz dF (y) dF (y) ϕ(z) dz

x

  ϕ(z) 1 − F (x − z) dz

x

  ϕ(x − y) 1 − F (y) dy.

0





x z

Classical Results on the Ruin Probabilities

9

Thus, we get   c ϕ(x) − ϕ(0) = λ



x 0

  ϕ(x − y) 1 − F (y) dy,

x ≥ 0.

[1.10]

By monotone convergence, letting x → +∞ in [1.10] gives  c lim ϕ(x) − ϕ(0) = λμ lim ϕ(x). x→+∞

x→+∞

Hence, we have  λμ ϕ(0) = 1 − lim ϕ(x). x→+∞ c

[1.11]

Since c > λμ, limx→+∞ ϕ(x) = 1 by lemma 1.1. Consequently, from [1.11] we get ϕ(0) = 1 −

λμ . c

[1.12]

Finally, substituting ϕ(x) = 1 − ψ(x) into [1.10] and using [1.12] we obtain

x

x     cψ(x) = λμ−λ 1−F (y) dy +λ ψ(x−y) 1−F (y) dy, x ≥ 0, 0

0

which is equivalent to [1.9].



Equation [1.9] is called a defective renewal equation. R EMARK 1.2.– Note that there is linear dependence [1.11] between ϕ(0) and limx→+∞ ϕ(x) in this risk model. In particular, [1.12] holds if and only if limx→+∞ ϕ(x) = 1. If the insurance company uses the expected value principle for premium calculation, then [1.12] can be rewritten as ϕ(0) = θ/(1 + θ).

10

Ruin Probabilities

1.1.5. Laplace transform of the infinite-horizon survival probability ˆ ϕ (z) be the Laplace–Stieltjes transform of F (y) and the Let ˆlF (z) and L Laplace transform of ϕ(x), respectively, i.e.

+∞ ˆlF (z) = e−zy dF (y), z ≥ 0, 0

and

ˆ ϕ (z) = L

+∞ 0

e−zx ϕ(x) dx,

z > 0.

  T HEOREM 1.4.– Let the surplus process Xt (x) t≥0 follow [1.1] and c > λμ. Then c − λμ ˆ ϕ (z) = [1.13] L   , z > 0. cz − λ 1 − ˆlF (z) P ROOF.– Multiplying [1.3] by e−zx and integrating over [0, +∞) gives

+∞

+∞ c ϕ+ (x) e−zx dx = λ ϕ(x) e−zx dx 0

+∞ x



−λ

0

0

0

[1.14]

ϕ(x − y) dF (y) e−zx dx,

It is easily seen that

+∞

ϕ+ (x) e−zx dx = −ϕ(0) + z 0

+∞ 0

z > 0.

ϕ(x) e−zx dx

[1.15]

ˆ ϕ (z) = −ϕ(0) + z L and



+∞ x

0

0

+∞ +∞

=

=

=

ϕ(x − y) dF (y) e−zx dx

0

y

+∞ +∞

0

0 +∞

0

ϕ(x − y) e−zx dx dF (y) [1.16] ϕ(x) e

ϕ(x) e

−zx

−z(x+y)

dx

+∞ 0

dx dF (y) ˆ ϕ (z) ˆlF (z). e−zy dF (y) = L

Classical Results on the Ruin Probabilities

Substituting [1.15] and [1.16] into [1.14] yields     ˆ ϕ (z) 1 − ˆlF (z) , z > 0. ˆ ϕ (z) − ϕ(0) = λL c zL

11

[1.17]

Since c > λμ, we have ϕ(0) = 1 − λμ/c. Hence, [1.17] can be rewritten as   ˆ ϕ (z) − c + λμ = λL ˆ ϕ (z) 1 − ˆlF (z) , z > 0, cz L which is equivalent to [1.13].



R EMARK 1.3.– Since there is a one-to-one correspondence between a function and its Laplace transform, theorem 1.4 implies the existence and uniqueness of the solution to [1.3], such that limx→+∞ ϕ(x) = 1, provided that c > λμ. 1.1.6. Analytic expressions for the infinite-horizon survival probability There are only a few cases where we can find analytic expressions for the infinite-horizon survival probability. We now consider two such cases.   T HEOREM 1.5.– Let the surplus process Xt (x) t≥0 follow [1.1], c > λμ and the claim sizes be exponentially distributed with mean μ. Then  (c − λμ) x λμ exp − ϕ(x) = 1 − , x ≥ 0. [1.18] c cμ P ROOF.– Since F (y) is continuous, by theorem 1.2, ϕ(x) is differentiable on R+ and satisfies the equation

x λ ϕ(y) ey/μ dy. [1.19] cϕ (x) = λϕ(x) − e−x/μ μ 0 From [1.19] we see that ϕ (x) is differentiable on R+ and

λ λ −x/μ x   cϕ (x) = λϕ (x) + 2 e ϕ(y) ey/μ dy − ϕ(x) μ μ 0  λ 1  = λϕ (x) − cϕ (x) − λϕ(x) − ϕ(x) μ μ  c = λ− ϕ (x). μ

12

Ruin Probabilities

Thus, we get the differential equation  c  − λ ϕ (x) = 0. cϕ (x) + μ

[1.20]

The general solution to [1.20] is  (c − λμ)x ϕ(x) = C1 − C2 exp − , cμ where C1 ∈ R and C2 ∈ R. Since limx→+∞ ϕ(x) = 1, we conclude that C1 = 1. Since ϕ(0) = 1 − λμ/c, we get C2 = λμ/c, which completes the proof.  If the insurance company uses the expected value principle for premium calculation, then [1.18] can be rewritten as  θx 1 exp − , x ≥ 0. ϕ(x) = 1 − 1+θ μ(1 + θ) Using theorem 1.4, we can prove the following theorem for the case where the claim sizes have a degenerate distribution.   T HEOREM 1.6.– Let the surplus process Xt (x) t≥0 follow [1.1], c > λμ and P[Yi = μ] = 1, i ≥ 1. Then ∞   k λ k λμ  1 max{x − kμ, 0} − ϕ(x) = 1 − c k! c k=0 [1.21]  λ max{x − kμ, 0} , x ≥ 0. × exp c R EMARK 1.4.– It is easy to check that ϕ(x) given by [1.21] is not differentiable at the point x = μ and  λ λμ   ϕ− (μ) − ϕ+ (μ) = 1− > 0. [1.22] c c Indeed, from [1.21], we see that   λx λμ ϕ(x) = 1 − exp , c c

0 ≤ x ≤ μ,

Classical Results on the Ruin Probabilities

λx exp ϕ(x) = c  λ(x − μ) λ − (x − μ) exp , c c 

λμ 1− c



13



μ ≤ x ≤ 2μ.

Hence, we have   λx λ λμ  ϕ (x) = 1− exp , 0 ≤ x ≤ μ, c c c     λ(x − μ) λx λ λμ λ  1− exp − (x − μ) exp ϕ (x) = c c c c c  λ(x − μ) , μ ≤ x ≤ 2μ. − exp c Consequently, ϕ− (μ) =

λ c

and ϕ+ (μ)

λ = c

 1− 

λμ c

λμ 1− c



 exp

λμ c



  λμ exp −1 , c

which gives [1.22]. Note that the same conclusion follows from theorem 1.2.



1.1.7. Cramér–Lundberg approximation for the infinite-horizon ruin probability ˆ > 0, such that We now assume that there is R

 λ +∞ Ry ˆ  1 − F (y) dy = 1. e c 0 Let λ μ ˆ= c



+∞ 0

 ˆ  yeRy 1 − F (y) dy.

[1.23]

14

Ruin Probabilities

Using the integral equation [1.9] and the key renewal theorem, we can prove the following theorem, which is called the Cramér–Lundberg theorem.   T HEOREM 1.7.– Let the surplus process Xt (x) t≥0 follow [1.1], c > λμ and ˆ > 0, such that [1.23] holds. there be R i) If μ ˆ < +∞, then ˆ

lim ψ(x) eRx =

x→+∞

c − λμ . ˆμ cRˆ

[1.24]

ii) If μ ˆ = +∞, then ˆ

lim ψ(x) eRx = 0.

x→+∞

From [1.24] we conclude that ψ(x) ∼

c − λμ −Rx ˆ e ˆ cRˆ μ

as

x → +∞.

[1.25]

This approximation for ψ(x) is called the Cramér–Lundberg approximation. Particularly, if the insurance company uses the expected value principle for premium calculation, then [1.25] can be rewritten as ψ(x) ∼

θ ˆμ (1 + θ)Rˆ

ˆ

e−Rx

as x → +∞.

ˆ > 0, such that [1.23] We now investigate the conditions when there is R holds. Let h : R+ → R+ be the shifted moment generating function of Yi , such that h(0) = 0, i.e.   h(R) = E eRYi − 1. We make the following classical assumption concerning h(R): there is R∞ ∈ (0, +∞], such that h(R) < +∞ for all R ∈ [0, R∞ ) and limR↑R∞ h(R) = +∞ (see [GRA 91, p. 2] and [ROL 99, p. 170]). It is easily seen that

Classical Results on the Ruin Probabilities

15

h(R) is increasing, convex and continuous on [0, R∞ ). Moreover, we can show that h(R) is infinitely differentiable on [0, R∞ ). Integrating by parts gives

+∞

+∞     ˆ  1 ˆ  Ry 1 − F (y) dy = 1 − F (y) d eRy e ˆ 0 R 0  +∞ ˆ h(R) 1 ˆ Ry . e dF (y) − 1 = = ˆ ˆ R R 0 Hence, we can rewrite [1.23] in the form ˆ = cR. ˆ λh(R)

[1.26]

L EMMA 1.2.– Let c > λμ and there be R∞ ∈ (0, +∞], such that h(R) < +∞ for all R ∈ [0, R∞ ) and limR↑R∞ h(R) = +∞. Then, there is a unique ˆ ∈ (0, R∞ ) to [1.26]. solution R The proof of lemma 1.2 follows easily from the properties of h(R). ˆ ∈ (0, R∞ ) to [1.26], if it exists, is called the The unique solution R adjustment coefficient or the Lundberg exponent. If the conditions of lemma 1.2 hold, then

 λ +∞ Ry ˆ  μ ˆ= 1 − F (y) dy ye c 0 

 y λ +∞  1 ˆ = 1 − F (y) d − eRy ˆ R ˆ2 c 0 R

+∞  y λ 1 λ ˆ = − + eRy dF (y) ˆ R ˆ2 ˆ2 c 0 R cR 

+∞

+∞ λ 1 1 1 ˆ ˆ Ry Ry = e dF (y) + ye dF (y) − ˆ2 R ˆ 0 ˆ2 0 c R R  ˆ ˆ + 1 λ 1 h(R) h (R) = − + ˆ2 ˆ ˆ2 c R R R  ˆ ˆ −c ˆ λh (R) λ h (R) cR/λ = − < +∞. = ˆ ˆ2 ˆ c R R cR

16

Ruin Probabilities

Therefore, in this case from [1.25] we get ψ(x) ∼

c − λμ ˆ e−Rx ˆ −c λh (R)

as

x → +∞,

ˆ ∈ (0, R∞ ) is the unique solution to [1.26]. where R 1.1.8. Lundberg inequality for the infinite-horizon ruin probability We now give an upper exponential bound for the infinite-horizon ruin probability. To this end, we use the martingale approach introduced by Gerber [GER 73].   For all R ≥ 0, we define the exponential process Vt (R) t≥0 by 



Vt (R) = exp −R ct −

Nt 

Yi .

i=1

  We now establish the martingale property for Vt (R) t≥0 .

  ˆ> L EMMA 1.3.– Let the surplus process Xt (x) t≥0 follow [1.1]. If there is R   ˆ 0, such that [1.26] holds, then Vt (R) is an (Ft )-martingale. t≥0

  P ROOF.– For all R > 0, such that E eRYi < +∞, if any, we have E[Vt (R)] = e

−cRt



E exp R

Nt 

 Yi

i=1

= e−cRt

∞  j=0

e−λt

(λt)j   RYi j E e j!

     = exp t λ E eRYi − λ − cR = exp(t (λh(R) − cR)).

[1.27]

Classical Results on the Ruin Probabilities

17

 ˆ  ˆ > 0, such that [1.26] holds, then E eRY i < +∞, so for all If there is R t2 ≥ t1 ≥ 0, we have 

ˆ / Ft E Vt2 (R) 1



Nt 2

     ˆ ct2 − = E exp −R Yi Ft1 i=1





ˆ ct1 − = E exp −R



Nt 1



Yi

i=1





ˆ c(t2 − t1 ) − × E exp −R

Nt 2 

Yi



 Ft1

i=Nt1 Nt2 −t1

        ˆ ˆ ˆ , = E Vt1 (R) E exp −R c(t2 − t1 ) − Yi = E Vt1 (R) i=1

  ˆ which shows that Vt (R) is an (Ft )-martingale. t≥0 Here we used the fact that Nt2 −t1      ˆ ˆ =1 ˆ Yi R) = exp (t2 −t1 )(λh(R)−c E exp −R c(t2 −t1 )−



i=1

by [1.26] and [1.27].



  T HEOREM 1.8.– Let the surplus process Xt (x) t≥0 follow [1.1]. If there is ˆ > 0, such that [1.26] holds, then for all x ≥ 0, we have R ˆ

ψ(x) ≤ e−Rx .

[1.28]

P ROOF.– It is easily seen that the ruin time τ (x) is an (Ft )-stopping time. Hence,  τ (x) ∧ t is a bounded (Ft )-stopping time for any t ≥ 0.  The process  ˆ ˆ Vt (R) is an (F )-martingale by lemma 1.3. Moreover, Vt (R) is t t≥0 t≥0 positive a.s. by its definition. Consequently, applying the optional stopping theorem yields   ˆ = E Vτ (x)∧t (R) ˆ 1 = V0 (R)     ˆ {τ (x) 0. The number of claims on the time interval [0, t] is a Poisson process (Nt )t≥0 with constant intensity λ > 0. Next, premium sizes form a sequence (Y¯i )i≥1 of non-negative i.i.d. r.v.’s with c.d.f. F¯ (y) = P[Y¯i ≤ y] and finite expectation μ ¯ > 0. The number of premiums on the time interval [0, t] ¯ > 0. The r.v.’s (Yi )i≥1 ¯t )t≥0 with constant intensity λ is a Poisson process (N ¯t )t≥0 are mutually independent. and (Y¯i )i≥1 and the processes (Nt )t≥0 and (N  Nt N¯t ¯ Thus, the total claims premiums on [0, t] equal i=1 Yi and i=1 Yi , and  0 0 ¯ ¯ respectively. We set i=1 Yi = 0 if Nt = 0 and i=1 Yi = 0 if Nt = 0. ¯t )t≥0 . Let (Ft )t≥0 be the filtration generated by (Yi )i≥1 , (Y¯i )i≥1 , (Nt )t≥0 , (N Moreover, the insurance company has a non-negative initial surplus x. Let Xt (x) be the surplus of the insurance company  at time  t provided that its initial surplus is x. Then, the surplus process Xt (x) t≥0 follows the equation Xt (x) = x +

¯t N  i=1

Y¯i −

Nt 

Yi ,

t ≥ 0.

[1.30]

i=1

The ruin time τ (x), the infinite-horizon ruin probability ψ(x), the finite-horizon ruin probability ψ(x, t), the infinite-horizon survival

Classical Results on the Ruin Probabilities

23

probability ϕ(x) and the finite-horizon survival probability ϕ(x, t) are defined as in section 1.1.1 for the classical risk model. As in section 1.1, we formulate some results for the ruin probabilities and other results for the survival probabilities. 1.2.2. Basic results Applying theorem 1.1 and arguments similar to those in the proof of lemma 1.1 one can prove the following lemma.   L EMMA 1.4.– Let the surplus process Xt (x) t≥0 follow [1.30]. ¯ μ ≤ λμ, then ψ(x) = 1 for all x ≥ 0. i) If λ¯ ¯ μ > λμ, then limx→+∞ ψ(x) = 0. ii) If λ¯ ¯ μ > λμ is the analogue of the net profit In this model, the condition λ¯ condition in the classical risk model.   T HEOREM 1.9.– Let the surplus process Xt (x) t≥0 follow [1.30]. Then, ϕ(x) satisfies the integral equation

+∞

x ¯ ¯ ¯ (λ + λ)ϕ(x) = λ ϕ(x + y) dF (y) + λ ϕ(x − y) dF (y). [1.31] 0

0

Equation [1.31] can be derived in a standard way applying the law of total probability. Note that this equation is the analogue of the integro-differential equation [1.3] in the classical risk model, but it does not involve the derivative of ϕ(x). That is why one does not need to investigate the differentiability of this function. ¯ μ > λμ, then ϕ(0) cannot be found explicitly, but it is easy to show If λ¯ that ϕ(0) > 0 in this case. Indeed, from [1.31], we get ¯ +∞ λ ϕ(0) = ¯ ϕ(y) dF¯ (y). [1.32] λ+λ 0 Moreover, limx→+∞ ϕ(x) = 1 by lemma 1.4 and the function ϕ(x) is nondecreasing by its definition. Assume that the equality ϕ(0) = 0 is true. Then by [1.32], there is a point x0 > 0, such that ϕ(x) = 0 for all x ∈ [0, x0 ]. Substituting x = x0 into [1.31] and applying the same arguments give that

24

Ruin Probabilities

ϕ(x) = 0 for all x ∈ [0, 2x0 ]. Thus, by induction, one can show that ϕ(x) = 0 for all x ≥ 0, which contradicts the fact that limx→+∞ ϕ(x) = 1. Furthermore, the function ϕ(x, t) satisfies the partial integro-differential equation ∂ϕ(x, t) ¯ + λ) ϕ(x, t) + (λ ∂t

+∞

¯ −λ ϕ(x + y, t) dF¯ (y) − λ 0

x 0

ϕ(x − y, t) dF (y) = 0.

The question concerning the existence of a partial derivative of ϕ(x, t) w.r.t. t is investigated in section 3.3. Theorems 1.10 and 1.11 below give analytic expressions for the infinitehorizon survival probability in two particular cases.   ¯ μ > λμ, T HEOREM 1.10.– Let the surplus process Xt (x) t≥0 follow [1.30], λ¯ the premium sizes be exponentially distributed with mean μ ¯ and the claim sizes be exponentially distributed with mean μ. Then  ¯ λ(μ + μ ¯) (λ¯ μ − λμ) x ϕ(x) = 1 − , x ≥ 0. ¯ exp − μ¯ ¯ μ ¯(λ + λ) μ(λ + λ)   ¯ > λ and T HEOREM 1.11.– Let the surplus process Xt (x) t≥0 follow [1.1], λ P[Y¯i = 1] = P[Yi = 1] = 1, i ≥ 1. Then  x +1 λ ϕ(x) = 1 − ¯ , x ≥ 0. λ The proof of theorem 1.10 is similar to the proof of theorem 1.5. To prove theorem 1.11, it is sufficient to notice that ϕ(x) = ϕ( x) for all x ≥ 0, and for integer x ≥ 1, integral equation [1.31] becomes the difference equation ¯ ¯ λϕ(x + 1) − (λ + λ)ϕ(x) + λϕ(x − 1) = 0, which can be solved in a standard way. In particular, it can be easily seen from theorem 1.11 that ϕ(x) may not be continuous on R+ in this risk model. In theorem 1.11, ϕ(x) is not continuous at all integer points x ≥ 1.

Classical Results on the Ruin Probabilities

25

Theorem 1.12 below gives an upper exponential bound for the infinite-horizon ruin probability.   T HEOREM 1.12.– Let the surplus process Xt (x) t≥0 follow [1.30]. If there is ˆ > 0, such that R     ˆ     i ¯ E e−Rˆ Y¯i − 1 + λ E eRY − 1 = 0, λ

[1.33]

then for all x ≥ 0, we have ˆ

ψ(x) ≤ e−Rx . The proof of theorem 1.12 is similar to the proof of theorem 1.8 and based on the martingale approach. In particular,   to prove theorem 1.12, we need to ˆ show that the exponential process Vt (R) t≥0 given by ¯t   N Nt  ˆ = exp −R ˆ Y¯i − Yi Vt (R) i=1

i=1

ˆ > 0 is such that [1.33] holds. is an (Ft )-martingale, where R 1.2.3. Bibliographical notes The risk model with stochastic premiums was introduced independently by Boikov [BOI 03] and Livshits [LIV 99] as a natural generalization of the classical risk model. All the basic results given in section 1.2.2 can be found in [BOI 03]. Some similar results were obtained in [LIV 99] by other techniques. Zinchenko and Andrusiv [ZIN 08] suggested an analogue of the De Vylder approximation for ψ(x), got diffusion approximations for ψ(x) and ψ(x, t) and considered the strong invariance principle for the risk process with stochastic premiums. Bening and Korolev [BEN 03] dealt with a generalized risk model where the premium rate and the claim intensity are stochastic. They constructed a statistical estimate for the infinite-horizon ruin probability and investigated its asymptotic properties.

26

Ruin Probabilities

Norkin [NOR 06] derived integral equations for ϕ(x) in the risk model where the premium intensity depends on the current surplus of the insurance company, and stochastic premium and claim arrivals are not necessarily Poisson. The method of successive approximations was proposed for numerical solution of these equations. In particular, necessary and sufficient conditions for the existence and uniqueness of solutions of the equations were obtained. Some distributions of functionals for the risk processes with stochastic premiums were studied by Gusak [GUS 05a, HUS 07] and Gusak and Karnaukh [GUS 05b].

2 Classical Risk Model with Investments in a Risk-Free Asset

In this chapter, we deal with a generalization of the classical risk model where an insurance company invests all surplus in a risk-free asset. We investigate the continuity and differentiability of the infinite- and finitehorizon survival probabilities in detail. Moreover, we derive integrodifferential equations for these functions and get bounds for their derivatives w.r.t. the initial surplus. 2.1. Description of the model We consider a generalization of the classical risk model described in section 1.1.1: an insurance company invests all surplus in a risk-free asset, the price of which equals Bt = B0 ert at time t, where B0 is the price of the asset at the time t = 0 and r > 0 is a risk-free interest rate.   Then, the surplus process Xt (x) t≥0 follows the equation  Xt (x) = x +

Nt   rXs (x) + c ds − Yi ,

t 0

t ≥ 0.

[2.1]

i=1

The rest of the chapter is organized as follows. In section 2.2, we investigate the continuity and differentiability of the infinite-horizon survival probability. Section 2.3 is devoted to investigation of these properties for the finite-horizon survival probability. Some bibliographical notes concerning the classical risk model with investments in a risk-free asset are given in section 2.4.

28

Ruin Probabilities

2.2. Continuity and differentiability of the infinite-horizon survival probability T HEOREM 2.1.–

  1) Let the surplus process Xt (x) t≥0 follow [2.1] under the above assumptions. Then, the function ϕ(x) is continuous on R+ . 2) Moreover, let c > λμ. Then, the following assertions hold: i) The function ϕ(x) is continuously differentiable on R+ , except at positive points of discontinuity of F (y). ii) If x > 0 is a point of discontinuity of F (y) and F (x) − F (x− ) = p, then ϕ(x) has the left and right derivatives ϕ− (x) and ϕ+ (x), respectively, and ϕ− (x) − ϕ+ (x) =

λpϕ(0) > 0. rx + c

iii) The function ϕ(x) satisfies the integro-differential equation  x  (rx + c)ϕ+ (x) = λϕ(x) − λ ϕ(x − y) dF (y)

[2.2]

[2.3]

0

on R+ with the boundary condition limx→+∞ ϕ(x) = 1. iv) We have  λ  sup ϕ+ (x) ≤ . c x∈[0,+∞)

[2.4]

P ROOF.– It follows immediately from [2.1] that Xt (x) = (x + c/r) ert − c/r up to the time τ1 . Let τ1 = s and Y1 = y. It is easily seen that ruin does not occur up to the time τ1 and it will not take place after τ1 if and only if y ≤ (x + c/r) ers − c/r and ruin does not occur on the time interval (s, +∞) with the initial surplus (x + c/r) ers − c/r − y. Note that τ1 is exponentially distributed with mean 1/λ. Therefore, by the law of total probability, for all x ≥ 0, we get  +∞ ϕ(x) = λe−λs 0



(x+c/r) ers −c/r

× 0

  ϕ (x + c/r) ers − c/r − y dF (y) ds.

[2.5]

Classical Risk Model with Investments in a Risk-Free Asset

29

Here and subsequently, all integrals w.r.t. a c.d.f. are assumed to be the Lebesgue–Stieltjes integrals. Changing the variable (x + c/r) ers − c/r = u in the outer integral in the right-hand side of [2.5] we have λ/r

+∞ 



ϕ(x) = λ(rx + c)

x

1 ru + c

λ/r+1 

u

ϕ(u − y) dF (y) du. [2.6]

0

the integral u Since ϕ(x) is non-decreasing and bounded, 0 ϕ(u − y) dF (y) is also a non-decreasing and bounded function of u. λ/r+1

u 1 Hence, the function ru+c 0 ϕ(u − y) dF (y) is Riemann integrable on any finite interval in [x, +∞) and the improper integral in the right-hand side of [2.6] is a continuous function of x by theorem A.1 and remark A.1. Thus, ϕ(x) is continuous in R+ , which proves the first assertion of the theorem. By Lebesgue’s decomposition theorem, we can write F (y) = F c (y) + F d (y), where F c (y) and F d (y) are continuous and discrete components of F (y), respectively. Let (yk )k≥1 be a sequence of all points of discontinuity of F (y), if any. Set pk = F (yk ) − limy↑yk F (y). Then, we can rewrite [2.6] as ϕ(x) = ϕ1 (x) + ϕ2 (x),

[2.7]

where ϕ1 (x) = λ(rx + c)

λ/r

+∞ 

 x

1 ru + c

λ/r+1

I(u) du,

ϕ2 (x) = λ(rx + c)λ/r  λ/r+1    +∞  1 × pk ϕ(u − yk ) du, ru + c x k : yk ≤u

[2.8]

[2.9]

30

Ruin Probabilities



u

I(u) =

ϕ(u − y) dF c (y).

0

We now prove that I(u) is left-continuous on (0, +∞). Fix any u0 > 0. For u ≤ u0 , we have  u   I(u0 ) − I(u) = ϕ(u0 − y) − ϕ(u − y) dF c (y) 0



+

u0 u

[2.10] ϕ(u0 − y) dF (y). c

It is obvious that  u0   ϕ(u0 − y) dF c (y) ≤ lim F c (u0 ) − F c (u) = 0. 0 ≤ lim u↑u0

u↑u0

u

[2.11]

Since ϕ(x) is continuous on [0, u0 ], it is uniformly continuous on this interval by the Heine-Cantor theorem.  Hence, for any ε > 0, there is δ > 0, (u0 − y) − (u − y) < δ, u ∈ [0, u0 ], y ∈ [0, u], we have such that for   ϕ(u0 − y) − ϕ(u − y) < ε. Therefore, for |u0 − u| < δ, we get  u      c  ϕ(u0 − y) − ϕ(u − y) dF (y) < εF (u0 ) ≤ ε. [2.12]  0

Thus, [2.10]–[2.12] imply the left continuity of I(u) at any u0 > 0. The right continuity of I(u) on R+ is proved similarly. We only note that for u ≥ u0 , we have  u0   I(u) − I(u0 ) = ϕ(u − y) − ϕ(u0 − y) dF c (y) 0



+

u u0

ϕ(u − y) dF c (y).

Thus, the integrand in the right-hand side of [2.8] is continuous in R+ w.r.t. u. Hence, by [2.8], theorem A.1 and remark A.1, ϕ1 (x) is differentiable on R+ and ϕ1 (x) =

λ λ ϕ1 (x) − I(x). rx + c rx + c

[2.13]

Classical Risk Model with Investments in a Risk-Free Asset

31

Since the integrand in the right-hand side of [2.9] is right-continuous on R+ w.r.t. u, by theorem A.1 and remarks A.1   and A.2, there is a right derivative of ϕ2 (x) on R+ (we denote it by ϕ2 (x) + ) and   ϕ2 (x) + =



λ λ ϕ2 (x) − rx + c rx + c

pk ϕ(x − yk ).

[2.14]

k : yk ≤x

Since the Riemann integral does not depend on values of the integrand at a countable number of points, we can rewrite [2.9] as ϕ2 (x) = λ(rx + c)λ/r  λ/r+1    +∞  [2.15] 1 × pk ϕ(u − yk ) du. ru + c x k : yk λμ, then limx→+∞ ϕ(x) = 1 and ϕ(0) = 1 − λμ/c > 0 in the classical risk model without any investments (see sections 1.1.2 and 1.1.4). Moreover, it is easily seen that at any time the surplus in the risk model with investments in the risk-free asset is not less than the surplus in the corresponding risk model without investments. Therefore, investments in the risk-free asset do not decrease the survival probability. From this we get limx→+∞ ϕ(x) = 1 and ϕ(0) > 0 for the model with investments. Hence, from [2.17] it follows that ϕ2 (x) is differentiable on R+ , except at positive points yk , k ≥ 1. Consequently, taking into account [2.7], we deduce

32

Ruin Probabilities

that ϕ(x) is also differentiable on R+ , except at positive points of discontinuity of F (y), and [2.2] holds. Moreover, from [2.7], [2.13] and [2.14], we obtain    λ  ϕ1 (x) + ϕ2 (x) − I(x) − pk ϕ(x − yk ) ϕ+ (x) = rx + c λ = rx + c



k : yk ≤x



x

ϕ(x) −



ϕ(x − y) dF (y) ,

0

which yields [2.3]. Note that the right-hand side of [2.3] is continuous on R+ , except at positive points of discontinuity of F (y). Therefore, ϕ+ (x) is also continuous on R+ , except at these points. Bound [2.4] follows immediately from [2.3], which completes the proof.  R EMARK 2.1.– Slight changes in the proof show that theorem 2.1 is also true for r = 0. Another proof of this theorem for the case r = 0 is given in [ROL 99, pp. 162–163] (see also section 1.1.3). R EMARK 2.2.– Bound [2.4], as well as all similar bounds that will be obtained later, can be used for derivation of a relation connecting accuracy and reliability of a uniform approximation of the survival probability by its statistical estimate (see Chapter 6). 2.3. Continuity of the finite-horizon survival probability and existence of its partial derivatives 2.3.1. Main theorem   T HEOREM 2.2.– Let the surplus process Xt (x) t≥0 follow [2.1] under the above assumptions. 1) If Yi , i ≥ 1, have a p.d.f. f (y), which is continuous on R+ , then ϕ(x, t) is continuous on R2+ as a function of two variables. 2) If the p.d.f. f (y) of Yi , i ≥ 1, has the derivative f  (y) on R+ , such that is integrable and bounded on R+ , then:

|f  (y)|

i) ϕ(x, t) has partial derivatives w.r.t. x and t on R2+ , which are continuous as functions of two variables;

Classical Risk Model with Investments in a Risk-Free Asset

33

ii) ϕ(x, t) satisfies the partial integro-differential equation ∂ϕ(x, t) ∂ϕ(x, t) − (rx + c) + λϕ(x, t) ∂t ∂x  x −λ ϕ(x − y, t) dF (y) = 0

[2.18]

0

on

R2+

with the boundary conditions ϕ(x, 0) = 1 and limx→+∞ ϕ(x, t) = 1;

iii) for all T > 0, we have    ∂ϕ(x, t)   ≤ C1 C 2 , sup  ∂x  x∈[0,+∞),

[2.19]

t∈[0,T ]

where

⎧ ⎨ 1 − e(r−λ)T if λ = r, C1 = λ−r ⎩ T if λ = r,    +∞  C2 = λ f (0) + |f (y)| dy . 0

rt P ROOF.– As in the proof of theorem 2.1, note that Xt (x) = (x+c/r)  e −c/r up to the time τ1 and let τ1 = s and Y1 = y. Note also that P τ1 > t = e−λt for all t ≥ 0. Moreover, if τ1 ≤ t, ruin does not occur up to the time τ1 and it will not take place after τ1 if and only if y ≤ (x + c/r) ers − c/r and ruin does not occur on the time interval (s, t] with the initial surplus (x + c/r) ers − c/r − y. Hence, by the law of total probability, for all x ≥ 0 and t ≥ 0, we obtain  t  (x+c/r) ers −c/r  −λs ϕ(x, t) = λe ϕ (x + c/r) ers [2.20] 0 0  − c/r − y, t − s dF (y) ds + e−λt .

Changing the variable t − s = v in the outer integral in the right-hand side of [2.20] we get  t  (x+c/r) er(t−v) −c/r  ϕ(x, t) = λe−λt eλv ϕ (x + c/r) er(t−v) [2.21] 0 0  −λt − c/r − y, v dF (y) dv + e .

34

Ruin Probabilities

Note that ϕ(x, t) is bounded, non-decreasing w.r.t. x and non-increasing w.r.t. t. Hence, the inner integral in the right-hand side of [2.21] is a nonincreasing function w.r.t. v. Consequently, it is Riemann integrable on [0, t] and the right-hand sides of [2.20] and [2.21] are well-defined. We now use that Yi , i ≥ 1, have a p.d.f. f (y). Changing the variable (x + c/r) er(t−v) − c/r − y = u in the inner integral in the right-hand side of [2.21] we have  t  (x+c/r) er(t−v) −c/r ϕ(x, t) = λe−λt eλv ϕ(u, v) [2.22] 0 0   r(t−v) −λt × f (x + c/r) e − c/r − u du dv + e . Let



t

I(x, t) =

e

λv

0



(x+c/r) er(t−v) −c/r

ϕ(u, v)

0

  × f (x + c/r) er(t−v) − c/r − u du dv.

We now use that f (y) is continuous on R+ . By [2.22], in order to prove that ϕ(x, t) is continuous on R2+ as a function of two variables, it is sufficient to show that the function I(x, t) is so, that is, for any fixed (x0 , t0 ) ∈ R2+ and any ε > 0, there is a δ > 0, such that   I(x, t) − I(x0 , t0 ) < ε [2.23] whenever |x − x0 | < δ, |t − t0 | < δ and (x, t) ∈ R2+ . We now fix (x0 , t0 ) ∈ (0, +∞) × (0, +∞) and assume that 0 ≤ x ≤ x0 and 0 ≤ t ≤ t0 (similar arguments can be applied to other cases). We have   I(x, t) − I(x0 , t0 )  (x+c/r) er(t−v) −c/r  t    f (x0 + c/r) er(t0 −v) − c/r − u ≤ eλv 0

0

  − f (x + c/r) er(t−v) − c/r − u  du dv +  ×



t

eλv

0 (x0 +c/r) er(t0 −v) −c/r (x+c/r) er(t−v) −c/r

  f (x0 + c/r) er(t0 −v) − c/r − u du dv

Classical Risk Model with Investments in a Risk-Free Asset

 + 

t0 t

eλv

(x0 +c/r) er(t0 −v) −c/r

×

35

0



f (x0 + c/r) e

r(t0 −v)

 − c/r − u du dv.

[2.24]

Consider first summand in The  now the r(t−v)  the right-hand side of [2.24]. 2 function f (x + c/r) e − c/r − u is continuous w.r.t. (x, t) ∈ R+ , so it is uniformly continuous on any compact set in R2+ , that is, for any ε1 > 0, there is a δ1 > 0, such that for all v ∈ [0, t0 ] and u ∈ [0, (x0 + c/r) er(t0 −v) − c/r], we have      f (x0 + c/r) er(t0 −v) − c/r − u − f (x + c/r) er(t−v) − c/r − u  < ε1 whenever 0 ≤ x0 − x < δ1 , 0 ≤ t0 − t < δ1 and (x, t) ∈ [0, x0 ] × [0, t0 ]. Therefore, for 0 ≤ x0 −x < δ1 , 0 ≤ t0 −t < δ1 and (x, t) ∈ [0, x0 ]×[0, t0 ], we get 

t

e 0

λv



(x+c/r) er(t−v) −c/r 0

   f (x0 + c/r) er(t0 −v) − c/r − u

  − f (x + c/r) er(t−v) − c/r − u  du dv  t0   eλv (x0 + c/r) er(t0 −v) − c/r dv ≤ ε1

[2.25]

0

 ≤ ε1

 (x0 + c/r)(eλt0 − ert0 ) c(eλt0 − 1) − . λ−r λr

Consider now the second summand in the right-hand side of [2.24]. Since f (y) is continuous, it is bounded and attains its maximum on any finite interval in R+ . Set M = maxy∈[0, (x0 +c/r) ert0 −c/r] f (y). The function (x + c/r) er(t−v) is continuous as a function of two variables x and t. Hence, it is continuous on any compact set in R2+ , that is, for any ε2 > 0, there is a δ2 > 0, such that for all v ∈ [0, t0 ], we have   (x0 + c/r) er(t0 −v) − (x + c/r) er(t−v)  < ε2 whenever 0 ≤ x0 − x < δ2 , 0 ≤ t0 − t < δ2 and (x, t) ∈ [0, x0 ] × [0, t0 ].

36

Ruin Probabilities

Therefore, we get 

t

e

λv



(x0 +c/r) er(t0 −v) −c/r (x+c/r) er(t−v) −c/r

0



≤ ε2 M

t0

eλv dv =

0

  f (x0 + c/r) er(t0 −v) − c/r − u du dv

ε2 M (eλt0 − 1) . λ

[2.26]

Consider now the third summand in the right-hand side of [2.24]. Since 

t0 t

eλv

(x0 +c/r) er(t0 −v) −c/r 0

 ≤

 t0

t

  f (x0 + c/r) er(t0 −v) − c/r − u du dv

eλv dv ≤ eλt0 (t0 − t),

for any ε3 > 0, there is a δ3 > 0, such that 

t0 t

eλv



(x0 +c/r) er(t0 −v) −c/r 0

  f (x0 + c/r) er(t0 −v) − c/r − u du dv

< ε3

[2.27]

whenever 0 ≤ t0 − t < δ3 and t ∈ [0, t0 ]. We now fix any ε > 0. Let ε ε1 =  , λt 0 (x0 + c/r)(e − ert0 ) c(eλt0 − 1) − 3 λ−r λr ελ , 3M (eλt0 − 1) ε ε3 = . 3 ε2 =

By [2.24]–[2.27], we conclude that [2.23] holds whenever 0 ≤ x0 − x < δ, 0 ≤ t0 − t < δ and (x, t) ∈ [0, x0 ] × [0, t0 ], where δ = δ1 ∧ δ2 ∧ δ3 . We now use that the p.d.f. f (y) of Yi , i ≥ 1, has the derivative f  (y) on R+ , such that |f  (y)| is integrable and bounded on R+ . By [2.22], in order to

Classical Risk Model with Investments in a Risk-Free Asset

37

prove the existence of partial derivatives of ϕ(x, t), it is sufficient to show the existence of partial derivatives of I(x, t). Let 

(x+c/r) er(t−v) −c/r

I(x, t, v) =

  ϕ(u, v) f (x+c/r) er(t−v) −c/r −u du.

0

By theorem A.2, to prove the existence of a partial derivative of I(x, t) w.r.t. x for any fixed t, it suffices to show that I(x, t, v) is continuous w.r.t. v on [0, t] for any fixed (x, t) ∈ R2+ and has a partial derivative w.r.t. x, which is continuous as a function of two variables x and v, where (x, v) ∈ [0, +∞) × [0, t]. The continuity of I(x, t, v) w.r.t. v on [0, t] is evident. By theorems A.3 and A.5, to prove the existence of a partial derivative of I(x, t, v) w.r.t. x, it suffices to show that for any fixed t ∈ [0, +∞) and v ∈ [0, t], the following conditions hold: – the function (x + c/r) er(t−v) − c/r has a derivative w.r.t. x;   – the function ϕ(u, v) f (x + c/r) er(t−v) − c/r − u is continuous as a function of two variables x and u, where (x, u) ∈ [0, +∞) × [0, (x + c/r) er(t−v) − c/r];   – the function f (x + c/r) er(t−v) − c/r − u has a derivative w.r.t. x, which is integrable w.r.t. u on [0, (x + c/r) er(t−v) − c/r] and bounded for all x ∈ [0, +∞) and u ∈ [0, (x + c/r) er(t−v) − c/r]. It is easily seen that all these conditions hold. Hence, there is a partial derivative of I(x, t, v) w.r.t. x and    ∂I(x, t, v) r(t−v) =e f (0) ϕ (x + c/r) er(t−v) − c/r, v ∂x   (x+c/r) er(t−v) −c/r    r(t−v) ϕ(u, v) f (x + c/r) e − c/r − u du . + 0

Arguments similar to those in the proof of the continuity of I(x, t) show that the partial derivative of I(x, t, v) w.r.t. x is continuous as a function of

38

Ruin Probabilities

two variables x and v, where (x, v) ∈ [0, +∞) × [0, t], for all fixed t ∈ R+ . We only note that to this end, we use the equality  (x+c/r) er(t−v) −c/r   ϕ(u, v) f  (x + c/r) er(t−v) − c/r − u du 0



(x+c/r) er(t−v) −c/r

=

  ϕ (x + c/r) er(t−v) − c/r − u, v f  (u) du.

0

Thus, there is a partial derivative of ϕ(x, t) w.r.t. x and   t   ∂ϕ(x, t) (r−λ)t (λ−r)v = λe e f (0) ϕ (x + c/r) er(t−v) − c/r, v ∂x 0  (x+c/r) er(t−v) −c/r [2.28] ϕ(u, v) + 0

   r(t−v) × f (x + c/r) e − c/r − u du dv. 

We conclude from [2.28] that      +∞  t  ∂ϕ(x, t)   (r−λ)t   |f (y)| dy e e(λ−r)v dv  ∂x  ≤ λ f (0) + 0 0 for all (x, t) ∈ R2+ . Hence, we get [2.19]. Next, by theorem A.2, to prove the existence of a partial derivative of I(x, t) w.r.t. t for any fixed x, it suffices to show that I(x, t, v) is continuous as a function of two variables t and v, where t ∈ [0, +∞) and v ∈ [0, t], for all fixed x ∈ R+ and I(x, t, v) has a partial derivative w.r.t. t, which is continuous as a function of two variables t and v, where t ∈ [0, +∞) and v ∈ [0, t], for all fixed x ∈ R+ . Arguments similar to those in the proof of the continuity of I(x, t) show that I(x, t, v) is continuous. By theorems A.3 and A.5, to prove the existence of a partial derivative of I(x, t, v) w.r.t. t, it suffices to show that for any fixed (x, v) ∈ [0, +∞) × [0, t], the following conditions hold: – the function (x + c/r) er(t−v) − c/r has a derivative w.r.t. t;   – the function ϕ(u, v) f (x + c/r) er(t−v) − c/r − u is continuous as a function of two variables t and u, where (t, u) ∈ [0, +∞) × [0, (x + c/r) er(t−v) − c/r];

Classical Risk Model with Investments in a Risk-Free Asset

39

  – the function f (x + c/r) er(t−v) − c/r − u has a derivative w.r.t. t, which is integrable w.r.t. u on [0, (x + c/r) er(t−v) − c/r] and bounded for all t ∈ [0, +∞) and u ∈ [0, (x + c/r) er(t−v) − c/r]. It is easily seen that the first two conditionshold. In the third condition, the derivative of f (x + c/r) er(t−v) − c/r − u w.r.t. t may not be bounded for all t ∈ [0, +∞) and u ∈ [0, (x + c/r) er(t−v) − c/r], but it is bounded if t ∈ [0, T ] for any T > 0. So, we must prove the existence of a partial derivative of I(x, t, v) w.r.t. t when t ∈ [0, T ] for any T > 0. This implies the existence of the partial derivative for all t ∈ [0, +∞). Hence, there is a partial derivative of I(x, t, v) w.r.t. t and    ∂I(x, t, v) r(t−v) = (rx + c)e f (0) ϕ (x + c/r) er(t−v) − c/r, v ∂t   (x+c/r) er(t−v) −c/r    r(t−v) ϕ(u, v) f (x + c/r) e − c/r − u du . + 0

Arguments similar to those in the proof of the continuity of I(x, t) show that the partial derivative of I(x, t, v) w.r.t. t is continuous as a function of two variables t and v, where t ∈ [0, +∞) and v ∈ [0, t], for all fixed x ∈ R+ . Thus, there exists a partial derivative of ϕ(x, t) w.r.t. t and  (x+c/r) er(t−v) −c/r  t ∂ϕ(x, t) λv 2 −λt = −λ e e ϕ(u, v) ∂t 0 0   × f (x + c/r) er(t−v) − c/r − u du dv  x +λ ϕ(u, t) f (x − u) du + λ(rx + c) e(r−λ)t 0



t

×

e 

 (λ−r)v

  f (0) ϕ (x + c/r) er(t−v) − c/r, v

0 (x+c/r) er(t−v) −c/r

+ 0 



× f (x + c/r) e

r(t−v)

ϕ(u, v)   − c/r − u du dv − λe−λt .

[2.29]

40

Ruin Probabilities

Combining [2.22], [2.28] and [2.29], we have ∂ϕ(x, t) ∂ϕ(x, t) = (rx + c) − λϕ(x, t) + λ ∂t ∂x



x

ϕ(x − y, t) dF (y),

0

which yields [2.18]. Arguments similar to those in the proof of the continuity of I(x, t) show that the partial derivatives of ϕ(x, t) are continuous. The boundary condition ϕ(x, 0) = 1 is evident. The condition limx→+∞ ϕ(x, t) = 1 follows immediately from the fact that there can be only a finite number of claims on any finite time interval a.s. This completes the proof.  R EMARK 2.3.– Slight changes in the proof show that theorem 2.2 is also true for r = 0. R EMARK 2.4.– The requirements that we impose on the distribution of the claim sizes in theorem 2.2 are not necessary for the continuity of ϕ(x, t) or the existence of its partial derivatives. This is verified by example 2.1 below. Furthermore, it is clear that the second assertion of the theorem is true if f (y) is continuous and has a derivative everywhere except at a finite number of points, where it has finite one-sided derivatives, and |f  (y)| is integrable and bounded on R+ . On the other hand, ϕ(x, t) may not have partial derivatives in some cases. This can be seen from example 2.2 below. 2.3.2. Examples E XAMPLE 2.1.– Let Yi , i ≥ 1, be uniformly distributed on [0, 1]. It is evident that the p.d.f. of Yi is not continuous at the points y = 0 and y = 1. Hence, it is not differentiable at these points. Nevertheless, we now show that ϕ(x, t) has partial derivatives w.r.t. x and t on R2+ , which are continuous as functions of two variables. We can rewrite [2.22] as ϕ(x, t) = λe

−λt



t

λv

e 0



(x+c/r) er(t−v) −c/r 0 ∨ ((x+c/r) er(t−v) −c/r−1)

ϕ(u, v) du dv + e−λt . [2.30]

Classical Risk Model with Investments in a Risk-Free Asset

41

Consider now three cases.   r+c 1) If (x, t) ∈ [0, 1] × 0, 1r ln rx+c , then for all v ∈ [0, t], we have (x + c/r) er(t−v) − c/r − 1 ≤ (x + c/r)(ert − 1) ≤ 0. Hence, we can rewrite [2.30] as  t  (x+c/r) er(t−v) −c/r −λt λv ϕ(x, t) = λe e ϕ(u, v) du dv + e−λt . [2.31] 0

0

2) If (x, t) ∈ [0, 1] × r(t−v)



1 r

 r+c ln rx+c , +∞ , then

− c/r − 1 ≥ 0

for

(x + c/r) er(t−v) − c/r − 1 ≤ 0

for

(x + c/r) e

Hence, we can rewrite [2.30] as  1 r+c  t− r ln rx+c −λt λv ϕ(x, t) = λe e 0

 +

t t− r1 ln

r+c rx+c

eλv



 r+c 1 , v ∈ 0, t − ln r rx + c   r+c 1 ,t . v ∈ t − ln r rx + c 

(x+c/r) er(t−v) −c/r (x+c/r) er(t−v) −c/r−1

(x+c/r) er(t−v) −c/r

ϕ(u, v) du dv



ϕ(u, v) du dv

+ e−λt .

[2.32]

0

3) If (x, t) ∈ (1, +∞) × [0, +∞), then for all v ∈ [0, t], we have (x + c/r) er(t−v) − c/r − 1 ≥ x − 1 > 0. Hence, we can rewrite [2.30] as  t  (x+c/r) er(t−v) −c/r ϕ(x, t) = λe−λt eλv ϕ(u, v) du dv +e−λt . [2.33] 0

(x+c/r) er(t−v) −c/r−1

Thus, we can divide the domain of ϕ(x, t) into three sets   r+c 1 S1 = [0, 1] × 0, ln , r rx + c

42

Ruin Probabilities

 S2 = [0, 1] ×

 1 r+c ln , +∞ , r rx + c

S3 = (1, +∞) × [0, +∞), where it satisfies equations [2.31], [2.32] and [2.33], respectively. Applying arguments similar to those in the proof of theorem 2.2 and using [2.31], [2.32] and [2.33], we can show that ϕ(x, t) is continuous as a function of two variables on S1 , S2 , S3 . Moreover, we can show that ϕ(x, t) is continuous as a function of two variables on R2+ . Then, it satisfies equations [2.31], [2.32] and [2.33] on   r+c 1 ¯ , S1 = [0, 1] × 0, ln r rx + c   1 r + c S¯2 = [0, 1] × ln , +∞ , r rx + c S¯3 = [1, +∞) × [0, +∞), respectively. Applying arguments similar to those in the proof of theorem 2.2 and using [2.31], [2.32] and [2.33], we can also show that there are continuous partial derivatives of ϕ(x, t) w.r.t. x and t on S¯1 , S¯2 , S¯3 . Note that we imply one-sided derivatives on the boundary of these sets. If (x, t) ∈ S¯1 , then ∂ϕ(x, t) = λe(r−λ)t ∂x



t

  e(λ−r)v ϕ (x + c/r) er(t−v) − c/r, v dv, [2.34]

0

 t  (x+c/r) er(t−v) −c/r ∂ϕ(x, t) = −λ2 e−λt eλv ϕ(u, v) du dv ∂t 0 0  x ϕ(u, t) du + λ(rx + c) e(r−λ)t +λ 0



t

× 0

  e(λ−r)v ϕ (x + c/r) er(t−v) − c/r, v dv − λe−λt .

[2.35]

Classical Risk Model with Investments in a Risk-Free Asset

43

If (x, t) ∈ S¯2 , then  1 r+c t− r ln rx+c ∂ϕ(x, t) = λe(r−λ)t e(λ−r)v ∂x 0    × ϕ (x + c/r) er(t−v) − c/r, v   − ϕ (x + c/r) er(t−v) − c/r − 1, v dv   t   (λ−r)v r(t−v) + e ϕ (x + c/r) e − c/r, v dv , t− 1r ln

r+c rx+c

[2.36]

∂ϕ(x, t) = −λ2 e−λt ∂t  ×

r+c rx+c

t− r1 ln

(x+c/r) er(t−v) −c/r−1 t t− 1r ln



x

+λ

eλv

0

(x+c/r) er(t−v) −c/r

 +



r+c rx+c

eλv



ϕ(u, v) du dv

(x+c/r) er(t−v) −c/r

 ϕ(u, v) du dv

0

ϕ(u, t) du

0



+ λ(rx + c) e

t− r1 ln

(r−λ)t

r+c rx+c

e(λ−r)v

0

   × ϕ (x + c/r) er(t−v) − c/r, v   − ϕ (x + c/r) er(t−v) − c/r − 1, v dv   t   (λ−r)v r(t−v) + e ϕ (x + c/r) e − c/r, v dv t− 1r ln

r+c rx+c

− λe−λt . [2.37]

44

Ruin Probabilities

If (x, t) ∈ S¯3 , then

 t    ∂ϕ(x, t) = λe(r−λ)t e(λ−r)v ϕ (x + c/r) er(t−v) − c/r, v ∂x 0 [2.38]   r(t−v) − c/r − 1, v dv, − ϕ (x + c/r) e  t  (x+c/r) er(t−v) −c/r ∂ϕ(x, t) 2 −λt λv = −λ e e ϕ(u, v) du dv ∂t 0 (x+c/r) er(t−v) −c/r−1  t  x (r−λ)t ϕ(u, t) du + λ(rx + c) e e(λ−r)v [2.39] +λ x−1

0

   × ϕ (x + c/r) er(t−v) − c/r, v   − ϕ (x + c/r) er(t−v) − c/r − 1, v dv − λe−λt . By [2.34]–[2.37], we conclude that there are continuous partial derivatives r+c of ϕ(x, t) on the boundary of S¯1 and S¯2 , i.e. for x ∈ [0, 1] and t = 1r ln rx+c . Furthermore, by [2.36]–[2.39], we see that there are continuous partial derivatives of ϕ(x, t) on the boundary of S¯2 and S¯3 , i.e. for x = 1 and t ∈ R+ . Thus, ϕ(x, t) has continuous partial derivatives on R2+ and we can show that it satisfies [2.18]. E XAMPLE 2.2.– Let P[Yi = 1] = 1, i ≥ 1. We now show that ϕ(x, t) does not have partial derivatives on some sets on R2+ . Consider the following three cases:   r+c 1) If (x, t) ∈ [0, 1] × 0, 1r ln rx+c , then by [2.21], we have ϕ(x, t) = e−λt .

[2.40]

 r+c ln rx+c , +∞ , then   r+c 1 r(t−v) − c/r > 1 for v ∈ 0, t − ln (x + c/r) e . r rx + c

2) If (x, t) ∈ [0, 1] ×



1 r

Hence, we can rewrite [2.21] as  t− 1 ln r+c   r rx+c eλv ϕ (x + c/r) er(t−v) − c/r − 1, v dv ϕ(x, t) = λe−λt 0

+e

−λt

.

[2.41]

Classical Risk Model with Investments in a Risk-Free Asset

45

3) If (x, t) ∈ (1, +∞) × [0, +∞), then for all v ∈ [0, t], we have (x + c/r) er(t−v) − c/r − 1 ≥ x > 1. Hence, we can rewrite [2.21] as  t   −λt ϕ(x, t) = λe eλv ϕ (x+c/r) er(t−v) −c/r −1, v dv +e−λt . [2.42] 0

Thus, as in example 2.1, we can divide the domain of ϕ(x, t) into the same three sets S1 , S2 , S3 , where it is defined by [2.40], [2.41] and [2.42], respectively. The sets S¯1 , S¯2 and S¯3 are also defined as in example 2.1. If (x, t) ∈ S¯1 , then it can easily be seen from [2.40] that ϕ(x, t) is continuous as a function of two variables and has continuous partial derivatives on S¯1 . Moreover, ∂ϕ(x, t) =0 ∂x

and

∂ϕ(x, t) = −λe−λt . ∂t

[2.43]

On the contrary, [2.41] and [2.42] do not imply these properties of ϕ(x, t) on S¯2 and S¯3 . Nevertheless, we make the additional assumption, which is intuitively natural, that ϕ(x, t) is continuous as a function of two variables on R2+ and has continuous partial derivatives on S¯2 and S¯3 . Let us denote by ϕ1 (·, ·) the partial derivative of ϕ(x, t) w.r.t. the first argument. If (x, t) ∈ S¯2 , then by [2.41], we get     ∂ϕ(x, t) λ r+c rx + c λ/r 1 = ϕ 0, t − ln + λe(r−λ)t ∂x rx + c r + c r rx + c  t− 1 ln r+c   r rx+c × e(λ−r)v ϕ1 (x + c/r) er(t−v) − c/r − 1, v dv, [2.44] 0

46

Ruin Probabilities

 t− 1 ln r+c r rx+c ∂ϕ(x, t) 2 −λt =−λ e eλv ∂t 0   × ϕ (x + c/r) er(t−v) − c/r − 1, v dv     rx + c λ/r r+c 1 +λ ϕ 0, t − ln r+c r rx + c  t− 1 ln r+c r rx+c (r−λ)t + λ(rx + c) e e(λ−r)v

[2.45]

0

 

 × ϕ1 (x + c/r) er(t−v) − c/r − 1, v dv − λe−λt . If (x, t) ∈ S¯3 , then by [2.42], we get ∂ϕ(x, t) = λe(r−λ)t ∂x



t 0

  e(λ−r)v ϕ1 (x+c/r) er(t−v) −c/r−1, v dv, [2.46]

∂ϕ(x, t) = − λ2 e−λt ∂t



t

  eλv ϕ (x + c/r) er(t−v) − c/r − 1, v dv

0



+ λϕ(x − 1, t) + λ(rx + c) e

t

(r−λ)t

e(λ−r)v

0

  × ϕ1 (x + c/r) er(t−v) − c/r − 1, v dv − λe−λt . However, the partial derivatives of ϕ(x, t) w.r.t. x and t do not exist on the boundary of S¯1 and S¯2 even under these additional assumptions. Moreover, the partial derivative of ϕ(x, t) w.r.t. x does not exist on the boundary of S¯2 and S¯3 . Indeed, if x ∈ [0, 1] and t = ∂ϕ(x, t) =0 ∂x

1 r

r+c ln rx+c , then by [2.43], we have

∂ϕ(x, t) = −λe−λt ∂t

and

on S¯1 and by [2.44] and [2.45], we have ∂ϕ(x, t) λ = ∂x rx + c



rx + c r+c

λ/r

Classical Risk Model with Investments in a Risk-Free Asset

47

and   rx + c λ/r ∂ϕ(x, t) =λ − λe−λt ∂t r+c on S¯2 . Furthermore, if x = 1 and t ∈ R+ , then by [2.44], we have ∂ϕ(x, t) λ = ϕ(0, t) ∂x r+c  t   (r−λ)t + λe e(λ−r)v ϕ1 (1 + c/r) er(t−v) − c/r − 1, v dv 0

on S¯2 and by [2.46], we have  t   ∂ϕ(x, t) = λe(r−λ)t e(λ−r)v ϕ1 (1 + c/r) er(t−v) − c/r − 1, v dv ∂x 0 on S¯3 . It only remains to note that ϕ(0, t) > 0 for all t ∈ R+ by [2.40] and [2.41].  2.4. Bibliographical notes For some early results concerning the classical risk model with investments in a risk-free asset, we refer the reader to [BOO 87, BOO 88, DEL 87, EMA 75, GER 71, HAR 77, VIT 87]. In particular, Delbaen and Haezendonck [DEL 87] studied the influence of macro-economic factors such as interest and inflation on the classical surplus process. Boogaert and Crijns [BOO 87] investigated the influence of a negative safety loading on the upper bounds for the ruin probabilities when macro-economic factors are taken into account. Later, Embrechts and Schmidli [EMB 94] and Schmidli [SCH 94] studied risk models with borrowing, investment and inflation. In the classical risk model with investments in a risk-free asset, Sundt and Teugels [SUN 95, SUN 97] derived integro-differential and integral equations

48

Ruin Probabilities

for ϕ(x) and obtained some approximations and upper and lower bounds for it. Moreover, they found the closed-form solution to the integro-differential equation when the claim sizes are exponentially distributed. Pervozvansky [PER 98] derived an integro-differential equation for ϕ(x, t) and found the closed-form solution to it when the claim sizes are exponentially distributed and the risk-free interest rate is zero. Asmussen [ASM 98], Kalashnikov and Konstantinides [KAL 00] and Klüppelberg and Stadtmüller [KLÜ 98] obtained asymptotic approximations for ψ(x) when the claim sizes are heavy-tailed. These approximations differ considerably from those in the classical risk model without any investments. Thus, investments in a risk-free asset change the asymptotics of the infinite-horizon ruin probability. Tang [TAN 05] established an asymptotic formula for ψ(x, t) in the case of subexponential claims. Two-sided bounds for the infinite-horizon ruin probability in the case of heavy-tailed claim sizes were obtained by Kalashnikov [KAL 96] and Konstantinides, Tang and Tsitsiashvili [KON 02, KON 04]. Cai and Dickson [CAI 03] derived upper exponential bounds for ψ(x) by martingale and recursive techniques in the more general Sparre Andersen model and gave applications of the results to the classical risk model. Kasozi and Paulsen [KAS 05] proposed a numerical procedure for estimation of ψ(x). Albrecher, Teugels and Tichy [ALB 01] got explicit formulas for the finite-horizon ruin probability in the case of exponentially distributed claim sizes in terms of finite gamma series. Knessl and Peter [KNE 96] studied detailed asymptotics of ψ(x, t) for small interest rate and exponentially distributed claim sizes. Brekelmans and De Waegenaere [BRE 01] presented an algorithm that allows us to determine lower and upper bounds for the finite-horizon ruin probability. Cardoso and Waters [CAR 03, CAR 05] suggested numerical algorithms for calculation of ψ(x, t) based on discrete time Markov chains and obtained upper and lower bounds for ψ(x, t). For bibliographical notes concerning the ruin or survival probabilities in the classical risk model without investments, see section 1.1.9.

3 Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset

In this chapter, we consider a generalization of the risk model with stochastic premiums where an insurance company invests all surplus in a risk-free asset. We investigate the continuity and differentiability of the infinite- and finite-horizon survival probabilities and compare these results with those in the case of zero interest rate. Furthermore, we derive integro-differential equations for the survival probabilities and get bounds for their derivatives w.r.t. an initial surplus. 3.1. Description of the model We consider a generalization of the risk model with stochastic premiums described in section 1.2.1: an insurance company invests all surplus in a riskfree asset, the price of which equals Bt = B0 ert at time t, where B0 is the price of the asset at the time t = 0 and r is a risk-free interest rate. In what follows, we assume that r > 0, but deal with the case r = 0 in section 3.2.2.   Then, the surplus process Xt (x) t≥0 follows the equation  Xt (x) = x + r

t 0

Xs (x) ds +

¯t N  i=1

Y¯i −

Nt 

Yi ,

t ≥ 0.

[3.1]

i=1

In this chapter, we discuss the continuity and differentiability of the survival probabilities in this risk model. Moreover, we compare these

50

Ruin Probabilities

properties of the survival probabilities with those in the risk model with stochastic premiums and without any investments. They turn out to be quite different in contrast to the classical risk model where investments in a risk-free asset do not have an influence on the continuity and differentiability of these functions. In particular, we will show that the infinite-horizon survival probability may not be differentiable at some points. Moreover, points of discontinuity of this function are also possible. The rest of the chapter is organized as follows. Section 3.2 is devoted to investigation of the continuity and differentiability of the infinite-horizon survival probability. In section 3.3, we investigate these properties for the finite-horizon survival probability. The bibliographical notes regarding the risk model with stochastic premiums are given in section 1.2.3. 3.2. Continuity and differentiability of the infinite-horizon survival probability 3.2.1. Main results T HEOREM 3.1.–

  1) Let the surplus process Xt (x) t≥0 follow [3.1] under the above assumptions. Then, the function ϕ(x) is continuous on (0, +∞). ¯ μ > λμ. Then the following assertions hold: 2) Moreover, let λ¯ i) The function ϕ(x) is continuous on R+ . ii) The function ϕ(x) is continuously differentiable on (0, +∞), except at positive points of discontinuity of F (y). iii) If x > 0 is a point of discontinuity of F (y) and F (x) − F (x− ) = p, then ϕ(x) has the left and right derivatives ϕ− (x) and ϕ+ (x), respectively, and ϕ− (x) − ϕ+ (x) =

λpϕ(0) > 0. rx

[3.2]

iv) The function ϕ(x) satisfies the integro-differential equation ¯ + λ)ϕ(x) rxϕ+ (x) = (λ   +∞ ¯ ¯ ϕ(x + y) dF (y) − λ −λ 0

x 0

ϕ(x − y) dF (y)

[3.3]

Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset

51

on (0, +∞) with the boundary condition limx→+∞ ϕ(x) = 1 and its value at the point x = 0 is given by ¯  +∞ λ ϕ(0) = ¯ ϕ(y) dF¯ (y). [3.4] λ+λ 0 v) For all x∗ > 0, we have ¯+λ  λ  . sup ϕ+ (x) ≤ rx∗ x∈[x∗ ,+∞)

[3.5]

P ROOF.– It follows easily from [3.1] that Xt (x) = xert up to the time τ¯1 ∧ τ1 . Let τ¯1 ∧ τ1 = s. Furthermore, let Y¯1 = y if τ¯1 ∧ τ1 = τ¯1 and Y1 = y if τ¯1 ∧ τ1 = τ1 . It is evident that ruin does not occur up to the time τ¯1 ∧ τ1 and it will not take place after that time if and only if one of the following two conditions holds: – τ¯1 ∧ τ1 = τ¯1 and ruin does not occur on the time interval (s, +∞) with the initial surplus xers + y; – τ¯1 ∧ τ1 = τ1 , y ≤ xers and ruin does not occur on the time interval (s, +∞) with the initial surplus xers − y.     ¯ λ ¯ + λ), P τ¯1 ∧ τ1 = τ1 = λ/(λ ¯ + λ) and Next, P τ¯1 ∧ τ1 = τ¯1 = λ/( ¯ τ¯1 ∧ τ1 is exponentially distributed with mean 1/(λ + λ). Hence, by the law of total probability, for all x ≥ 0, we get  +∞  +∞   ¯ −(λ+λ)s ¯ ϕ(x) = e ϕ xers + y dF¯ (y) λ 0

0



xers

+λ







[3.6]

ϕ xers − y dF (y) ds.

0

We see at once that [3.4] follows immediately from [3.6] for x = 0. Changing the variable xers = u in the outer integral in the right-hand side of [3.6] for x > 0 we obtain  +∞ (λ+λ)/r+1 ¯ ¯ 1 x(λ+λ)/r ϕ(x) = r u x [3.7]  +∞

 u ¯ × λ ϕ(u + y) dF¯ (y) + λ ϕ(u − y) dF (y) du. 0

0

52

Ruin Probabilities

Since the inner integrals in the right-hand side of [3.7] are non-decreasing and bounded functions of u, the integrand in the outer integral is integrable on any [x, +∞), where x > 0. Therefore, ϕ(x) is continuous on (0, +∞), which proves the first assertion of the theorem. By Lebesgue’s decomposition theorem, we have F (y) = F c (y) + F d (y), where F c (y) and F d (y) are continuous and discrete components of F (y), respectively. Denote by (yk )k≥1 a sequence of all points of discontinuity of F (y), if any. Set pk = F (yk ) − limy↑yk F (y). Then, we can rewrite [3.7] as ϕ(x) = ϕ1 (x) + ϕ2 (x), where

[3.8]

+∞

¯  1 (λ+λ)/r+1  ¯ ¯ λI(u) + λI(u) du, u x  +∞ (λ+λ)/r+1 ¯ ¯ 1 λx(λ+λ)/r ϕ2 (x) = r u x 

× pk ϕ(u − yk ) du, ¯

x(λ+λ)/r ϕ1 (x) = r

 ¯ I(u) =



k : yk ≤u +∞

ϕ(u + y) dF¯ (y),



u

I(u) =

0

[3.9]

[3.10]

ϕ(u − y) dF c (y).

0

The continuity of I(u) is established in the proof of theorem 2.1. Next, for all u > 0, u0 > 0 and M > 0, we have  +∞     ¯ − I(u ¯ 0 ) ≤ I(u) ϕ(u + y) − ϕ(u0 + y) dF¯ (y) 0  [3.11]    ≤ 1 − F¯ (M ) + sup ϕ(u + y) − ϕ(u0 + y). y∈[0, M ]

Choosing a large enough value of M , we can first make 1 − F¯ (M ) small enough and then can make the second summand in [3.11] small enough when ¯ on (0, +∞). u is close to u0 . This proves the continuity of I(u)

Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset

53

Therefore, the integrand in the right-hand side of [3.9] is continuous on (0, +∞) w.r.t. u. By [3.9], theorem A.1 and remark A.1, ϕ1 (x) is differentiable on (0, +∞) and ϕ1 (x) =

¯+λ ¯ I(x) ¯ + λI(x) λ λ ϕ1 (x) − . rx rx

[3.12]

Since the integrand in the right-hand side of [3.10] is right-continuous on (0, +∞) w.r.t.  u, there is a right derivative of ϕ2 (x) on (0, +∞) (we denote it by ϕ2 (x) + ) and ¯+λ   λ λ ϕ2 (x) + = ϕ2 (x) − rx rx



pk ϕ(x − yk ).

[3.13]

k : yk ≤x

Since the Riemann integral does not depend on values of the integrand at a countable number of points, we can rewrite [3.10] as  +∞ (λ+λ)/r+1 ¯ ¯ 1 λx(λ+λ)/r ϕ2 (x) = r u x 

× pk ϕ(u − yk ) du.

[3.14]

k : yk 0 in the for x > 0. If λ¯ risk model with stochastic premiums and without any investments (see section 1.2.2). Moreover, it is easily seen that at any time the surplus in the

54

Ruin Probabilities

risk model with investments in the risk-free asset is not less than the surplus in the corresponding risk model without investments. Therefore, investments in the risk-free asset do not decrease the survival probability. From this we get limx→+∞ ϕ(x) = 1 and ϕ(0) > 0 for the model with investments. Therefore, from [3.16] we see that ϕ2 (x) is differentiable on (0, +∞), except at positive points yk , k ≥ 1. Hence, taking into account [3.8] we conclude that ϕ(x) is also differentiable on (0, +∞), except at positive points of discontinuity of F (y), and [3.2] holds. Furthermore, from [3.8], [3.12] and [3.13] we get ϕ+ (x) =

¯+λ ¯  λ λ ¯ ϕ1 (x) + ϕ2 (x) − I(x) rx rx

 λ − I(x) + pk ϕ(x − yk ) rx k : yk ≤x

 x ¯+λ ¯  +∞ λ λ λ ¯ = ϕ(x) − ϕ(x + y) dF (y) − ϕ(x − y) dF (y), rx rx 0 rx 0 which gives [3.3]. Since the right-hand side of [3.3] is continuous on (0, +∞), except at positive points of discontinuity of F (y), we deduce that ϕ+ (x) is also continuous on (0, +∞), except at these points. Bound [3.5] follows easily from [3.3]. What is left is to show that ϕ(x) is right-continuous at x = 0. Indeed, taking into account [3.4] and [3.7], the differentiability of ϕ(x) in the right semi-neighborhood of the point x = 0 and L’Hopital’s rule we have  +∞ (λ+λ)/r+1 ¯ 1 1 lim ϕ(x) = lim ¯ x↓0 x↓0 rx−(λ+λ)/r u x  +∞

 u ¯ × λ ϕ(u + y) dF¯ (y) + λ ϕ(u − y) dF (y) du 0

0

(λ+λ)/r+1 ¯ 1 1 = lim ¯ ¯ + λ) x−(λ+λ)/r−1 x↓0 (λ x  +∞

 x ¯ ¯ × λ ϕ(x + y) dF (y) + λ ϕ(x − y) dF (y) 0

¯ λ = ¯ λ+λ

0



+∞

ϕ(y) dF¯ (y) = ϕ(0),

0

which completes the proof.



Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset

55

R EMARK 3.1.– In contrast to the classical risk model, theorem 3.1 is not true for r = 0. The continuity and differentiability of the infinite-horizon survival probability in the risk model with stochastic premiums and without investments is considered in section 3.2.2. R EMARK 3.2.– Theorem 3.1 does not indicate whether the right derivative of ϕ(x) exists at x = 0 and ϕ (x) is bounded in the right semi-neighborhood of this point. Proposition 3.1 below yields sufficient conditions for existence of the right derivative of ϕ(x) at x = 0 and boundedness of ϕ (x) for all x ∈ R+ .   P ROPOSITION 3.1.– Let the surplus process Xt (x) t≥0 follow [3.1] under ¯ + λ. If Y¯i and Yi , i ≥ 1, have p.d.f.’s the above assumptions with r < λ ¯ f (y) and f (y), respectively, which have the derivatives f¯ (y) and f  (y) on R+ , such that |f¯ (y)| and |f  (y)| are integrable and bounded on R+ , then ϕ(x) is continuously differentiable on R+ and       ¯ f¯ (y)| + λ|f  (y)| dy ¯(0) − λf (0) + +∞ λ|    ¯ λ f 0 sup ϕ (x) ≤ . [3.17] ¯+λ−r λ x∈[0,+∞) P ROOF.– If Y¯i and Yi , i ≥ 1, have p.d.f.’s f¯(y) and f (y), respectively, then for all x ≥ 0, we can rewrite [3.6] as  +∞  +∞   ¯ −(λ+λ)s ¯ ϕ(x) = e ϕ(y) f¯ y − xers dy λ 0

xers



xers

+λ



rs





[3.18]

ϕ(y) f xe − y dy ds.

0

If the assumptions of the proposition hold, then by [3.18] and theorems A.3, A.4 and A.5, it follows that ϕ(x) is continuously differentiable on R+ and 



+∞

ϕ (x) =

¯

e−(λ+λ−r)s

0



  rs  ¯ ¯ × −λ f (0) ϕ xe +   rs  +λ f (0) ϕ xe +

+∞ xers

xers 0

  ϕ(y) f¯ y − xers dy 



rs



ϕ(y) f xe − y dy





ds.

[3.19]

56

Ruin Probabilities

From [3.19] we conclude that  +∞    ¯ ϕ (x) ≤ e−(λ+λ−r)s 0



  λf¯(0) − λf (0) + × ¯



    ¯ f¯ (y) + λf  (y) dy λ

+∞ 



0

for all x ≥ 0, which yields [3.17]. This proves the proposition.



3.2.2. Zero interest rate   Let the surplus process Xt (x) t≥0 be given by Xt (x) = x +

¯t N  i=1

Y¯i −

Nt 

Yi ,

t ≥ 0.

[3.20]

i=1

In [BOI 03], it was shown that ϕ(x) satisfies the following integral equation:  ¯ + λ)ϕ(x) = λ ¯ (λ 0

+∞



x

ϕ(x + y) dF¯ (y) + λ

ϕ(x − y) dF (y)

[3.21]

0

on R+ , which is an analogue of integro-differential equation [3.3] (see also section 1.2.2). The analytic properties of the infinite-horizon survival probability in this risk model are quite different from those established in section 3.2.1. Firstly, ϕ(x) may have points of discontinuity if the distributions of premium and claim sizes are discrete (see section 1.2.2 and [BOI 03]). Secondly, ϕ(x) may not be differentiable at some points if the distributions of premium and claim sizes are absolutely continuous, which can be seen from example 3.1 below. Note that if Y¯i and Yi , i ≥ 1, have p.d.f.’s f¯(y) and f (y), respectively, then we can rewrite [3.21] as  x ¯  +∞ λ λ ¯ ϕ(x) = ¯ ϕ(y) f (y−x) dy+ ¯ ϕ(y) f (x−y) dy. [3.22] λ+λ x λ+λ 0

Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset

57

E XAMPLE 3.1.– Let the premium and claim sizes be uniformly distributed on [¯ y∗ , y¯∗ ] and [y∗ , y ∗ ], respectively, where 0 < y¯∗ < y¯∗ < y∗ < y ∗ and ¯ y∗ + y¯∗ ) > λ(y∗ + y ∗ ). λ(¯ Then from [3.22] we have

 x+¯y∗ ¯ λ ϕ(x) = ¯ ϕ(y) dy, 0 ≤ x ≤ y∗ , (λ + λ)(¯ y ∗ − y¯∗ ) x+¯y∗  x+¯y∗ ¯ λ ϕ(x) = ¯ ϕ(y) dy (λ + λ)(¯ y ∗ − y¯∗ ) x+¯y∗  x−y∗ λ + ¯ ϕ(y) dy, y∗ ≤ x ≤ y ∗ , (λ + λ)(y ∗ − y∗ ) 0  x+¯y∗ ¯ λ ϕ(x) = ¯ ϕ(y) dy (λ + λ)(¯ y ∗ − y¯∗ ) x+¯y∗  x−y∗ λ + ¯ ϕ(y) dy, x ≥ y ∗ . (λ + λ)(y ∗ − y∗ ) x−y∗

From this we conclude that ϕ(x) is continuous on R+ and differentiable on [0, y∗ ], [y∗ , y ∗ ] and [y ∗ , +∞]. Moreover, we have ¯   λ ∗ ϕ(x + y ¯ ϕ (x) = ¯ ) − ϕ(x + y ¯ ) , ∗ (λ + λ)(¯ y ∗ − y¯∗ )

0 ≤ x ≤ y∗ ,

¯   λ ∗ ϕ(x + y ¯ ϕ (x) = ¯ ) − ϕ(x + y ¯ ) ∗ (λ + λ)(¯ y ∗ − y¯∗ ) λ ϕ(x − y∗ ), y∗ ≤ x ≤ y ∗ , + ¯ (λ + λ)(y ∗ − y∗ ) ¯   λ ϕ(x + y¯∗ ) − ϕ(x + y¯∗ ) ϕ (x) = ¯ ∗ (λ + λ)(¯ y − y¯∗ )   λ ϕ(x − y∗ ) − ϕ(x − y ∗ ) , + ¯ ∗ (λ + λ)(y − y∗ )

x ≥ y∗.

Nevertheless, ϕ(x) is not differentiable at the points x = y∗ and x = y ∗ . Indeed, from the above, we have λϕ(0) ϕ+ (y∗ ) − ϕ− (y∗ ) = ¯ (λ + λ)(y ∗ − y∗ )

58

Ruin Probabilities

and λϕ(0) ϕ− (y ∗ ) − ϕ+ (y ∗ ) = ¯ . (λ + λ)(y ∗ − y∗ ) ¯ y∗ + y¯∗ ) > λ(y∗ + y ∗ ) (see section 1.2.2).  Furthermore, ϕ(0) > 0 since λ(¯ The following proposition gives sufficient conditions for the continuity and differentiability of ϕ(x) in the risk model with stochastic premiums and without any investments.   P ROPOSITION 3.2.– Let the surplus process Xt (x) t≥0 follow [3.20] under the above assumptions. 1) If Y¯i and Yi , i ≥ 1, have p.d.f.’s f¯(y) and f (y), respectively, which are continuous on R+ , then ϕ(x) is continuous on R+ . 2) If the p.d.f.’s f¯(y) and f (y) of Y¯i and Yi , i ≥ 1, respectively, have the derivatives f¯ (y) and f  (y) on R+ , such that |f¯ (y)| and |f  (y)| are integrable and bounded on R+ , then ϕ(x) is continuously differentiable on R+ and       ¯ f¯ (y)| + λ|f  (y)| dy ¯(0) − λf (0) + +∞ λ|    ¯ λ f 0 sup ϕ (x) ≤ . [3.23] ¯+λ λ x∈[0,+∞) P ROOF.– Fix any x0 > 0. We now prove that ϕ(x) is left-continuous at this point. By [3.22], for all x < x0 we have ¯  M     ϕ(x0 ) − ϕ(x) ≤ λ f¯(y − x0 ) − f¯(y − x) dy ¯+λ λ x0

  + 2 1 − F¯ (M − x0 ) + F¯ (x0 − x) λ +¯ λ+λ



 f (x0 − y) − f (x − y) dy + F (x0 − x) ,

x 0

where M > x0 is  arbitrary. Hence, choosing a large enough value of M , we can first make 2 1 − F¯ (M − x0 ) small enough, and then by the uniform continuity of f¯(y) and f (y) on [0, M ] and [0, x0 ], respectively, we can make      M f¯(y − x0 ) − f¯(y − x) dy and x f (x0 − y) − f (x − y) dy small enough 0 x0 when x is close to x0 . Note that we can also make F¯ (x0 − x) and F (x0 − x)

Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset

59

small enough when x is close to x0 . Consequently, ϕ(x) is left-continuous at x = x0 . Similarly, we can show that ϕ(x) is right-continuous at any point x0 ≥ 0. Thus, ϕ(x) is continuous on R+ . If f¯(y) and f (y) have derivatives, which satisfy the conditions of the proposition, then by [3.22], the continuity of ϕ(x) and theorems A.3 and A.4, we conclude that ϕ(x) is continuously differentiable on R+ and

 +∞ ¯ λ   ϕ(y) f¯ (y − x) dy ϕ (x) = − ¯ f¯(0) ϕ(x) + λ+λ x [3.24]

 x λ +¯ ϕ(y) f  (x − y) dy . f (0) ϕ(x) + λ+λ 0 Bound [3.23] follows immediately from [3.24]. This completes the proof.  3.3. Continuity of the finite-horizon survival probability and existence of its partial derivatives   T HEOREM 3.2.– Let the surplus process Xt (x) t≥0 follow [3.1] under the above assumptions. 1) If Y¯i and Yi , i ≥ 1, have p.d.f.’s f¯(y) and f (y), respectively, which are continuous on R+ , then ϕ(x, t) is continuous on R2+ as a function of two variables. 2) If the p.d.f.’s f¯(y) and f (y) of Y¯i and Yi , i ≥ 1, respectively, have the derivatives f¯ (y) and f  (y) on R+ , such that |f¯ (y)| and |f  (y)| are integrable and bounded on R+ , then: i) ϕ(x, t) has partial derivatives w.r.t. x and t on R2+ , which are continuous as functions of two variables; ii) ϕ(x, t) satisfies the partial integro-differential equation ∂ϕ(x, t) ∂ϕ(x, t) ¯ + λ) ϕ(x, t) − rx + (λ ∂t ∂x [3.25]  +∞  x ¯ ¯ −λ ϕ(x + y, t) dF (y) − λ ϕ(x − y, t) dF (y) = 0 0

0

60

Ruin Probabilities

on R2+ with the boundary conditions ϕ(x, 0) = 1 and limx→+∞ ϕ(x, t) = 1; iii) for all T > 0, we have    ∂ϕ(x, t)   ≤ C1 C 2 , sup  ∂x  x∈[0,+∞),

[3.26]

t∈[0,T ]

where ⎧ ¯ ⎨ 1 − e(r−λ−λ)T ¯ C1 = ⎩ λ+λ−r T

if

¯ + λ = r, λ

if

¯ + λ = r, λ

  ¯ λf¯(0) − λf (0) + λ C2 = ¯



+∞

|f¯ (y)| dy + λ

0



+∞

|f  (y)| dy.

0

P ROOF.– Applying arguments similar to those in the proofs of theorems 2.2 and 3.1, for all x ≥ 0 and t ≥ 0, we get  +∞  t   ¯ ¯ ϕ(x, t) = e−(λ+λ)s λ ϕ xers + y, t − s dF¯ (y) 0

0



xers

+λ





rs



ϕ xe − y, t − s dF (y) ds + e

[3.27] ¯ −(λ+λ)t

.

0

Changing the variable t − s = v in the outer integral in the right-hand side of [3.27], we have ¯ −(λ+λ)t



t

ϕ(x, t) = e

e 0



xer(t−v)

+λ

¯ (λ+λ)v

 ¯ λ

+∞

  ϕ xer(t−v) + y, v dF¯ (y)

0

  ¯ ϕ xer(t−v) − y, v dF (y) dv + e−(λ+λ)t .

[3.28]

0

We now use that Y¯i and Yi , i ≥ 1, have p.d.f.’s f¯(y) and f (y), respectively. Changing the variables xer(t−v) + y = u and xer(t−v) − y = u in the first and

Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset

61

second inner integrals, respectively, in the right-hand side of [3.28] we get ϕ(x, t) = e

¯ −(λ+λ)t



t

e

¯ (λ+λ)v

 ¯ λ

0



xer(t−v)

+λ

+∞ xer(t−v)

  ϕ(u, v) f¯ u − xer(t−v) du

  ¯ ϕ(u, v) f xer(t−v) − u du dv + e−(λ+λ)t .

[3.29]

0

Let



t

I(x, t) =

e 0

¯ (λ+λ)v



 ¯ λ

xer(t−v)

+λ

+∞ xer(t−v)

  ϕ(u, v) f¯ u − xer(t−v) du 

ϕ(u, v) f xe

r(t−v)





− u du dv.

0

If f¯(y) and f (y) are continuous on R+ , then we can show the continuity of ϕ(x, t) as a function of two variables applying arguments similar to those in the proof of theorem 2.2. Indeed, for any fixed (x0 , t0 ) ∈ (0, +∞) × (0, +∞) and all 0 ≤ x ≤ x0 and 0 ≤ t ≤ t0 , we have  M  t      ¯ ( λ+λ)v ¯ f¯ u − x0 er(t0 −v) I(x, t) − I(x0 , t0 ) ≤ λ e x0 er(t0 −v)

0

     − f¯ u − xer(t−v)  du + 2 1 − F¯ M − x0 ert0 dv 

t

¯ +λ

e 

0 t

+λ

e 

¯ (λ+λ)v

¯ (λ+λ)v

0 t

+λ

e

¯ (λ+λ)v

  

x0 er(t0 −v) xer(t−v) x0 er(t0 −v) xer(t−v) xer(t−v) 0

0

  − f xer(t−v) − u  du dv  +∞  t0 ¯ (λ+λ)v ¯ e λ +

  f¯ u − xer(t−v) du dv   f x0 er(t0 −v) − u du dv

   f x0 er(t0 −v) − u

x0 er(t0 −v)

t



x0 er(t0 −v)

+λ 0



f x0 e

  f¯ u − x0 er(t0 −v) du

r(t0 −v)−u



du dv,

[3.30]

62

Ruin Probabilities

where M > x0 ert0 is arbitrary.  Consequently,choosing a large enough value of M we can first make 2 1 − F¯ M − x0 ert0 small enough and then by the continuity of f¯(y) and f (y), we can make the right-hand side of [3.30] small enough when x is close to x0 and t is close to t0 . Thus, ϕ(x, t) is continuous at (x0 , t0 ) for x ≤ x0 and t ≤ t0 . Similarly, we can consider other cases. If f¯(y) and f (y) have the derivatives f¯ (y) and f  (y) on R+ , such that  ¯ |f (y)| and |f  (y)| are integrable and bounded on R+ , then we can show the existence of partial derivatives of ϕ(x, t) as in the proof of theorem 2.2. Moreover, from [3.29] we conclude that  t   ∂ϕ(x, t) ¯ ¯ (r−λ−λ)t (λ+λ−r)v ¯ f¯(0) ϕ xer(t−v) , v =e e −λ ∂x 0

 +∞    r(t−v) ¯ du ϕ(u, v) f u − xe + xer(t−v)

  + λ f (0) ϕ xer(t−v) , v 

xer(t−v)

+



r(t−v) ϕ(u, v) f xe − u du dv  



[3.31]

0

and  +∞ ∂ϕ(x, t) ¯ + λ) ϕ(x, t) + λ ¯ = − (λ ϕ(u, t) f¯(u − x) du ∂t x  x ϕ(u, t) f (x − u) du +λ 0

+ rx e



t

e

¯ (λ+λ−r)v



  ¯ −λ f¯(0) ϕ xer(t−v) , v

0

 +

¯ (r−λ−λ)t

+∞ xer(t−v)



[3.32]

  ϕ(u, v) f¯ u − xer(t−v) du



  + λ f (0) ϕ xer(t−v) , v 

xer(t−v)

+ 0



  ϕ(u, v) f¯ xer(t−v) − u du

dv.

Risk Model with Stochastic Premiums and Investments in a Risk-Free Asset

63

Combining [3.31] and [3.32], we have  +∞ ∂ϕ(x, t) ¯ ¯ = −(λ + λ) ϕ(x, t) + λ ϕ(u, t) f¯(u − x) du ∂t x  x ∂ϕ(x, t) +λ ϕ(u, t) f (x − u) du + rx ∂x 0 ∂ϕ(x, t) ¯ + λ) ϕ(x, t) − (λ ∂x  +∞  ¯ ¯ +λ ϕ(x + y, t) dF (y) + λ

= rx

0

x

ϕ(x − y, t) dF (y),

0

which yields [3.25]. From [3.31], it follows that  

 +∞  +∞    ∂ϕ(x, t)    ¯ ¯ ¯   ≤ ¯  λf (0) − λf (0) + λ |f (y)| dy + λ |f (y)| dy  ∂x  0 0  t ¯ ¯ (r−λ−λ)t ×e e(λ+λ−r)v dv 0

for all x ≥ 0 and t ≥ 0, which gives [3.26]. As in theorem 2.2, the boundary condition ϕ(x, 0) = 1 is evident and the condition limx→+∞ ϕ(x, t) = 1 follows immediately from the fact that there can be only a finite number of claims on any finite time interval a.s. This completes the proof.  R EMARK 3.3.– Let r = 0. If Y¯i and Yi , i ≥ 1, have p.d.f.’s f¯(y) and f (y), respectively, then by the law of total probability, for all x ≥ 0 and t ≥ 0, we have  +∞  t ¯ ¯ (λ+λ)v −(λ+λ)t ¯ ϕ(x, t) = e e ϕ(x + y, v) dF¯ (y) λ 0



x

+λ

0



ϕ(x − y, v) dF (y) dv + e

[3.33] ¯ −(λ+λ)t

.

0

By [3.33], we can show that if f¯(y) and f (y) are continuous on R+ , then ϕ(x, t) is continuous on R2+ as a function of two variables, has a continuous

64

Ruin Probabilities

partial derivative w.r.t. t and satisfies the partial integro-differential equation ∂ϕ(x, t) ¯ + λ) ϕ(x, t) + (λ ∂t  +∞  ¯ −λ ϕ(x + y, t) dF¯ (y) − λ 0

x

ϕ(x − y, t) dF (y) = 0

0

on R2+ with the boundary conditions ϕ(x, 0) = 1 and limx→+∞ ϕ(x, t) = 1. Nevertheless, we can guarantee the existence of a partial derivative w.r.t. x provided that the conditions of theorem 3.2 hold. In this case, bound [3.26] is also true with r = 0 in the definition of C1 .  R EMARK 3.4.– As in theorem 2.2, the requirements that we impose on the distribution of the premium and claim sizes in theorem 3.2 are not necessary for the continuity of ϕ(x, t) or existence of its partial derivatives (see also remark 2.4). On the other hand, ϕ(x, t) may not have partial derivatives in some cases.

4 Classical Risk Model with a Franchise and a Liability Limit

In this chapter, we deal with the classical risk model under the additional assumption that an insurance company applies a franchise and a liability limit. To be more precise, we consider three cases: the insurance company establishes a franchise only, a liability limit only and both a franchise and a liability limit. Assuming that claim sizes are exponentially distributed we find analytic expressions for the infinite-horizon survival probabilities in these cases. The expressions turn out different on certain intervals. Moreover, we investigate how a franchise and a liability limit change the survival probability for small and large enough initial surpluses. 4.1. Introduction We consider the classical risk  model described in section 1.1.1. Recall that in this case the surplus process Xt (x) t≥0 follows Xt (x) = x + ct −

Nt 

Yi ,

t ≥ 0.

[4.1]

i=1

Let the net profit condition hold, i.e. c > λμ, throughout this chapter. Moreover, we assume that the insurance company uses the expected value principle for premium calculation, which means that c = λμ(1 + θ), where θ > 0 is a safety loading.

66

Ruin Probabilities

It is well known that ϕ(x) is a solution to the integro-differential equation  x ϕ(x − y) dF (y), [4.2] cϕ+ (x) = λϕ(x) − λ 0

where ϕ+ (x) is the right derivative of ϕ(x). The question concerning the differentiability of ϕ(x) is investigated in [ROL 99, pp. 162–163] and section 2.3 in different ways (see also section 1.1.3). Equation [4.2] has a unique solution, such that limx→+∞ ϕ(x) = 1, provided that the net profit condition holds, otherwise ruin on an infinite horizon occurs with probability 1. Moreover, it is known that the condition limx→+∞ ϕ(x) = 1 is true for ϕ(x) if and only if the condition ϕ(0) = 1 − λμ/c, which is equivalent to ϕ(0) = θ/(1 + θ), holds (see sections 1.1.2 and 1.1.4). So, if we find any solution to [4.2] with ϕ(0) = θ/(1 + θ), then we can be sure that this solution is the survival probability. If the claim sizes are exponentially distributed, then the closed form solution to [4.2] can be found and   θx 1 exp − ϕ(x) = 1 − . [4.3] 1+θ μ(1 + θ) In this chapter, we will deal with the classical risk model under the additional assumption that the insurance company uses a franchise and a liability limit. A franchise is a provision in an insurance policy whereby an insurer does not pay unless damage exceeds the franchise amount. It is applied to prevent a large number of trivial claims. A liability limit determines the maximum amount that is paid by an insurer. It is used to restrict the insurer’s liability to the insured. Let d and L be franchise and liability limit amounts, respectively. We choose them at the initial time and do not change them later. We make the following natural assumption concerning these amounts: 0 ≤ d < L ≤ +∞. In particular, if d = 0, then a franchise is not used and if L = +∞, then a (d, L) liability limit is not used. Let Yi , i ≥ 1, denote an insurance compensation for the ith claim provided that the franchise and liability limit (d, L) amounts are d and L. We let F (d, L) (y) stand for the c.d.f. of Yi .

Classical Risk Model with a Franchise and a Liability Limit

67

Normally, a franchise and a liability limit also imply reduction of insurance premiums. We suppose that the safety loading θ > 0 is constant. Thus, the premium intensity is given by  (d, L) c(d, L) = λ(1 + θ) E Yi provided that the insurance company uses the expected value principle for premium calculation. (d, L)

(x) be the surplus of the insurance company at time t provided Let Xt that its initial surplus is x and the franchise and liability limit amounts are d  (d, L)  and L, respectively. Then, [4.1] for the surplus process Xt (x) t≥0 can be rewritten as follows: (d, L)

Xt

(x) = x + c(d, L) t −

Nt 

(d, L)

Yi

,

t ≥ 0.

[4.4]

i=1

Let

ϕ(d, L) (x)

denote the corresponding infinite-horizon survival probability.

In what follows, we deal with exponentially distributed claim sizes only. This is one of the few cases where we succeed in finding an analytic expression for the survival probability. It is easily seen that the c.d.f. of the insurance compensation is a sum of absolutely continuous and discrete components when the claim sizes are exponentially distributed and the insurance company uses the franchise and liability limit. That is why analytic expressions for the survival probability turn out different on certain intervals as we will show later. To get them, we apply the results of Chapter 2, namely theorem 2.1 and remark 2.1 (see also section 1.1.3 and [ROL 99, pp. 162–163]). Furthermore, we investigate how a franchise and a liability limit change the survival probability for small and large enough initial surpluses. The rest of the chapter is organized as follows. In section 4.2, we consider the case where the insurance company establishes a franchise only. In section 4.3, we suppose that the insurance company applies a liability limit only. Section 4.4 deals with the case when both a franchise and a liability limit are established. The bibliographical notes concerning approaches to calculation of the ruin or survival probabilities in the classical risk model are given in section 1.1.9. For details concerning various benefits that can be provided by insurance policies, see, for example [PIT 14] and references therein.

68

Ruin Probabilities

4.2. Survival probability in the classical risk model with a franchise If the insurance company establishes a franchise only and the claim sizes are exponentially distributed, then equation [4.2] for ϕ(d, +∞) (x) can be written as   c(d, +∞) ϕ(d, +∞) (x) +  x [4.5] (d, +∞) = λϕ (x) − λ ϕ(d, +∞) (x − y) dF (d, +∞) (y), 0

where ⎧ ⎪ ⎨0 (d, +∞) (y) = 1 − e−d/μ F ⎪ ⎩ 1 − e−y/μ

if y < 0, if 0 ≤ y < d, if y ≥ d,

and  (d, +∞) c(d, +∞) = λ(1 + θ) E Yi  +∞ −y/μ ye = λ(1 + θ) dy = λ(1 + θ)(μ + d) e−d/μ . μ d In section 4.2.1, we derive analytic expressions for ϕ(d, +∞) (x) in the case of exponentially distributed claim sizes. In section 4.2.2, we investigate how a franchise changes the survival probability for small and large enough initial surpluses. Note that throughout this chapter all sums equal 0 provided that their lower summation indices are greater than the upper ones. 4.2.1. Analytic expression for the survival probability To formulate the next theorem, introduce the constants γ = (1 + θ)(μ + d), C1, 1 =

θ , 1+θ

Classical Risk Model with a Franchise and a Liability Limit

69

θ e−d/γ , (1 + θ)(γ + μ)   θ γμ + d(γ + μ) −d/γ = e , 1+ 1+θ (γ + μ)2

A2, 0 = − C2, 1

C2, 2 = −

θγμ ed/μ . (1 + θ)(γ + μ)2

Moreover, let the constants An+1, j , 0 ≤ j ≤ n − 1, be given in a recurrent way by formulas An, n−2 −d/γ e , n ≥ 2, n(γ + μ)  1 1 (j + 2)γμAn+1, j+1 + =− γ+μ j+1     n−2  i+1 e−d/γ , An, i (−d)i−j+1 × j

An+1, n−1 = − An+1, j

[4.6]

[4.7]

i=j−1

An+1, 0

1 ≤ j ≤ n − 2, n ≥ 3,  1 =− 2γμAn+1, 1 γ+μ    n−2  i+1 −d/γ + Cn, 1 + An, i (−d) e ,

[4.8] n ≥ 2.

i=0

Next, let the constants Bn+1, j , 0 ≤ j ≤ n − 2, be given in a recurrent way by formulas B3, 0 =

C2, 2 d/μ e , γ+μ

Bn+1, n−2 =

Bn, n−3 ed/μ , (n − 1)(γ + μ)

n ≥ 3,

[4.9]

70

Ruin Probabilities

Bn+1, j

 1 1 (j + 2)γμBn+1, j+1 + = γ+μ j+1     n−3  i−j+1 i + 1 d/μ e Bn, i (−d) , × j

[4.10]

i=j−1

Bn+1, 0

1 ≤ j ≤ n − 3, n ≥ 4,  1 = 2γμBn+1, 1 γ+μ    n−3  i+1 d/μ + Cn, 2 + Bn, i (−d) e ,

[4.11] n ≥ 3.

i=0

Finally, let the constants Cn+1, 1 and Cn+1, 2 be given by formulas Cn+1, 1

n−3 γμ(An, 0 − An+1, 0 )  + = Cn, 1 + An, i − An+1, i γ+μ i=0  (i + 2)γμ(An, i+1 − An+1, i+1 ) (nd)i+1 + γ+μ   nγμAn+1, n−1 + An, n−2 − An+1, n−2 − (nd)n−1 γ+μ

− An+1, n−1 (nd)n n−3  γμ + (i + 1)(Bn, i − Bn+1, i )(nd)i γ+μ i=0    γ+μ n−2 , − (n − 1)Bn+1, n−2 (nd) exp −nd γμ n ≥ 2, C3, 2 = C2, 2 +

γμB3, 0 − 2dB3, 0 γ+μ

  γμ(A3, 0 − A2, 0 + 4dA3, 1 ) γ+μ exp 2d + , γ+μ γμ

[4.12]

Classical Risk Model with a Franchise and a Liability Limit

Cn+1, 2

71

n−4 γμ(Bn+1, 0 − Bn, 0 )  + = Cn, 2 + Bn, i − Bn+1, i γ+μ i=0  (i + 2)γμ(Bn+1, i+1 − Bn, i+1 ) (nd)i+1 + γ+μ   (n − 1)γμBn+1, n−2 + Bn, n−3 − Bn+1, n−3 + γ+μ

× (nd)n−2 − Bn+1, n−2 (nd)n−1 n−2  γμ + (i + 1)(An+1, i − An, i )(nd)i γ+μ i=0    γ+μ n−1 , + nAn+1, n−1 (nd) exp nd γμ

[4.13]

n ≥ 3. Note that to compute the constants by the formulas above for any n ≥ 2, we have to know all the constants for n − 1. Moreover, for any fixed n ≥ 2, we start from the computation of An+1, j for j from n − 1 to 0, Bn+1, j for j from n − 2 to 0, and then we can compute Cn+1, 1 and Cn+1, 2 . We introduced all the constants only to formulate the next theorem. We will get them in the proof of this theorem.  (d, +∞)  (x) t≥0 follow [4.4] under T HEOREM 4.1.– Let the surplus process Xt the above assumptions with 0 < d < +∞ and L = +∞, and the claim sizes be exponentially distributed with mean μ. Then, (d, +∞)

ϕ(d, +∞) (x) = ϕn+1

(x) for all x ∈ [nd, (n + 1)d),

n ≥ 0,

where (d, +∞)

(x) = C1, 1 ex/γ ,   (d, +∞) ϕ2 (x) = C2, 1 + A2, 0 x ex/γ + C2, 2 e−x/μ ,   n−1  (d, +∞) An+1, j xj+1 ex/γ ϕn+1 (x) = Cn+1, 1 + ϕ1

[4.14] [4.15]

j=0

 + Cn+1, 2 +

n−2  j=0

[4.16]



Bn+1, j xj+1 e−x/μ ,

n ≥ 2.

72

Ruin Probabilities

P ROOF.– Substituting c(d, +∞) and F (d, +∞) (y) into [4.5] yields   (1 + θ)(μ + d) ϕ(d, +∞) (x) + = ϕ(d, +∞) (x), x ∈ [0, d),

[4.17]

and   μ(1 + θ)(μ + d) ϕ(d, +∞) (x) + = μϕ(d, +∞) (x)  x d/μ −e ϕ(d, +∞) (x − y) e−y/μ dy, x ∈ [d, +∞).

[4.18]

d

By theorem 2.1 and remark 2.1, ϕ(d, +∞) (x) is continuously differentiable on R+ (see also section 1.1.3 and [ROL 99, pp. 162–163]). Let us introduce (d, +∞) (d, +∞) the functions ϕn+1 (x), n ≥ 0, in the following way: ϕn+1 (x) is defined on [nd, (n + 1)d) and coincides with ϕ(d, +∞) (x) on this interval. Set γ = (1 + θ)(μ + d). Solving [4.17] gives [4.14], where the constant (d, +∞) C1, 1 can be found from ϕ1 (0) = θ/(1 + θ), which guarantees that the solution is the survival probability. Thus, C1, 1 = θ/(1 + θ). By [4.18], the function ϕ(d, +∞) (x) is easily seen to have the second derivative, which is continuous. Moreover, integro-differential equation [4.18] can be reduced to the differential one     γμ ϕ(d, +∞) (x) + (γ − μ) ϕ(d, +∞) (x) − ϕ(d, +∞) (x) [4.19] = −ϕ(d, +∞) (x − d), x ∈ [d, +∞), in a standard way (see section 1.1.6 and [ASM 10, BEA 84, BÜH 70, DIC 10, GER 79, GRA 91, ROL 99]). If x ∈ [d, 2d), then 0 ≤ x − d < d and the right-hand side of [4.19] has already been found. So, we only need to solve the linear differential equation  (d, +∞)   (d, +∞)  (d, +∞) (x) (x) − ϕ2 (x) + (γ − μ) ϕ2 γμ ϕ2 =−

θ e(x−d)/γ 1+θ

[4.20]

on [d, 2d) by standard techniques (see, for example [AGA 08, BIR 78, ROB 04]). Furthermore, since ϕ(d, +∞) (x) and its first derivative are

Classical Risk Model with a Franchise and a Liability Limit

73

continuous at the point x = d, the following conditions must hold to guarantee that the solution is the survival probability: ⎧ (d, +∞) (d, +∞) ⎪ (x) = ϕ2 (d), ⎪ ⎨ lim ϕ1 x↑d

[4.21]

 (d, +∞)   (d, +∞)  ⎪ ⎪ (x) = ϕ2 (d) . ⎩ lim ϕ1 x↑d

The solution to [4.20] can be found by writing (d, +∞)

ϕ2

(d, +∞)

(d, +∞)

(x) = ϕ2, gen (x) + ϕ2, part1 (x),

(d, +∞)

where ϕ2, gen (x) is a general solution to the linear homogeneous equation (d, +∞)

corresponding to [4.20] and ϕ2, part1 (x) is a particular solution to linear heterogeneous equation [4.20]. (d, +∞)

Note that ϕ2, part1 (x) can be written as (d, +∞)

ϕ2, part1 (x) = A2, 0 x ex/γ , (d, +∞)

where A2, 0 is a constant. Substituting ϕ2, part1 (x) into [4.20] and applying the method of undetermined coefficients yield A2, 0 . Moreover, we can write (d, +∞) ϕ2, gen (x) as ϕ2, gen (x) = C2, 1 ex/γ + C2, 2 e−x/μ , (d, +∞)

(d, +∞)

where the constants C2, 1 and C2, 2 are such that ϕ2 conditions [4.21].

(x) satisfies the

(d, +∞)

Thus, ϕ2 (x) is defined by [4.15] and the constants A2, 0 , C2, 1 and C2, 2 are given before the assertion of the theorem. It is easy to check that the second classical derivative of ϕ(d, +∞) (x) does not exist at the point x = d. If x ∈ [2d, 3d), then d ≤ x − d < 2d and the right-hand side of [4.19] has (d, +∞) already been found. Therefore, ϕ3 (x) is a solution to  (d, +∞)   (d, +∞)  (d, +∞) (x) (x) + (γ − μ) ϕ3 (x) − ϕ3 γμ ϕ3   γμ + d(γ + μ) −d/γ 1 θ −d/γ 1+ e e − (x − d) =− 1+θ (γ + μ)2 γ+μ ×e(x−d)/γ +

θμ(μ + d) 2d/μ −x/μ e e . (γ + μ)2

[4.22]

74

Ruin Probabilities

Furthermore, the continuity of ϕ(d, +∞) (x) and its derivative at the point x = 2d implies ⎧ (d, +∞) (d, +∞) ⎪ (x) = ϕ3 (2d), lim ϕ2 ⎪ ⎨ x↑2d [4.23]  (d, +∞)   (d, +∞)  ⎪ ⎪ (x) = ϕ3 (2d) . ⎩ lim ϕ2 x↑2d

(d, +∞)

We can write ϕ3 (d, +∞)

ϕ3

(x) as

(d, +∞)

(d, +∞)

(d, +∞)

(x) = ϕ3, gen (x) + ϕ3, part1 (x) + ϕ3, part2 (x),

(d, +∞)

where ϕ3, gen (x) is a general solution to the linear homogeneous equation (d, +∞)

(d, +∞)

corresponding to [4.22], and ϕ3, part1 (x) and ϕ3, part2 (x) are particular solutions to the equations  (d, +∞)   (d, +∞)  (d, +∞) γμ ϕ3 (x) + (γ − μ) ϕ3 (x) − ϕ3 (x)   γμ + d(γ + μ) −d/γ 1 θ −d/γ 1+ e e − (x − d) [4.24] =− 1+θ (γ + μ)2 γ+μ × e(x−d)/γ and

 (d, +∞)   (d, +∞)  (d, +∞) (x) + (γ − μ) ϕ3 (x) − ϕ3 (x) γμ ϕ3 =

[4.25]

θμ(μ + d) 2d/μ −x/μ e e , (γ + μ)2

respectively. (d, +∞)

(d, +∞)

Next, ϕ3, part1 (x) and ϕ3, part2 (x) can be written as (d, +∞)

ϕ3, part1 (x) = (A3, 0 + A3, 1 x) x ex/γ and ϕ3, part2 (x) = B3, 0 x e−x/μ , (d, +∞)

(d, +∞)

where A3, 0 , A3, 1 and B3, 0 are constants. Substituting ϕ3, part1 (x) and (d, +∞)

ϕ3, part2 (x) into [4.24] and [4.25], respectively, and applying the method of

Classical Risk Model with a Franchise and a Liability Limit

75

undetermined coefficients yield A3, 0 , A3, 1 and B3, 0 . Moreover, we can write (d, +∞) ϕ3, gen (x) as ϕ3, gen (x) = C3, 1 ex/γ + C3, 2 e−x/μ , (d, +∞)

(d, +∞)

(x) satisfies the where the constants C3, 1 and C3, 2 are such that ϕ3 conditions [4.23], which guarantees that the solution is the survival probability. All the constants are given before the assertion of the theorem. In particular, A3, 1 , A3, 0 and C3, 1 are given by [4.6], [4.8] and [4.12], respectively, with n = 2. (d, +∞)

(d, +∞)

(x), n ≥ 2, we can find ϕn+1 In general, if we know ϕn solving the equation  (d, +∞)   (d, +∞)  (d, +∞) γμ ϕn+1 (x) + (γ − μ) ϕn+1 (x) − ϕn+1 (x) +∞) = −ϕ(d, (x − d). n

(x) by

[4.26]

By the above, applying an induction argument yields (d, +∞)

(d, +∞)

+∞) +∞) (x) = ϕ(d, ϕ(d, n n, gen (x) + ϕn, part1 (x) + ϕn, part2 (x),

[4.27]

where +∞) x/γ ϕ(d, + Cn, 2 e−x/μ , n, gen (x) = Cn, 1 e (d, +∞) ϕn, part1 (x)

=

n−2  i=0

(d, +∞)

ϕn, part2 (x) =

n−3 

 An, i x

i

x ex/γ ,

 Bn, i xi x e−x/μ .

i=0

Thus, [4.27] can be rewritten as   n−2  (d, +∞) i+1 ϕn (x) = Cn, 1 + An, i x ex/γ i=0

 + Cn, 2 +

n−3  i=0



Bn, i xi+1 e−x/μ .

[4.28]

76

Ruin Probabilities

Therefore, we get (d, +∞)

ϕn+1

(d, +∞)

(d, +∞)

(d, +∞)

(x) = ϕn+1, gen (x) + ϕn+1, part1 (x) + ϕn+1, part2 (x),

where ϕn+1, gen (x) = Cn+1, 1 ex/γ + Cn+1, 2 e−x/μ , (d, +∞)

(d, +∞) ϕn+1, part1 (x)

=

n−1 

 An+1, i x

i

i=0

(d, +∞) ϕn+1, part2 (x)

=

n−2 

 Bn+1, i x

i

x ex/γ ,

[4.29]

x e−x/μ ,

[4.30]

i=0

which gives [4.16]. (d, +∞)

(x), n ≥ 2, which We now assume that we have already found ϕn means that we know all the constants in [4.28]. We will now derive formulas to find the constants in [4.16]. If x ∈ [nd, (n + 1)d), then (n − 1)d ≤ x − d < nd and  +∞) −ϕ(d, (x n

− d) = − Cn, 1 +

n−2 

 An, i (x − d)

i+1

e(x−d)/γ

i=0

 − Cn, 2 +

n−3 

 Bn, i (x − d)

i+1

e−(x−d)/μ .

[4.31]

i=0

Applying the binomial theorem in [4.31] yields +∞) (x − d) −ϕ(d, n     n−2 i+1   i+1 j i−j+1 x (−d) An, i = − Cn, 1 + e(x−d)/γ j i=0

 − Cn, 2 +

n−3  i=0

Bn, i

j=0

 i+1  j=0

  i+1 j i−j+1 x (−d) e−(x−d)/μ . j

[4.32]

Classical Risk Model with a Franchise and a Liability Limit

77

Interchanging the order of summation in [4.32] gives +∞) (x − d) − ϕ(d, n   n−2  i+1 An, i (−d) = − Cn, 1 + i=0

+

n−1  n−2  j=1

i+1 j

i=j−1

 −

 An, i (−d)i−j+1

Cn, 2 +

n−3 



 xj e(x−d)/γ

[4.33]

 Bn, i (−d)

i+1

i=0

+

n−2  n−3  j=1

 Bn, i (−d)

i−j+1

i=j−1

i+1 j



 x

j

e−(x−d)/μ .

To derive formulas for An+1, j , 0 ≤ j ≤ n − 1, we calculate  (d, +∞)   (d, +∞)  ϕn+1, part1 (x) and ϕn+1, part1 (x) using [4.29] and substitute them, as well as [4.33], into [4.26]. Thus, we get  μ (2An+1, 0 + 2γAn+1, 1 ) +

n−2  j=1

 An+1, j−1 + 2(j + 1)An+1, j + (j + 2)(j + 1)γAn+1, j+1 xj γ



  An+1, n−2 An+1, n−1 n n−1 + 2nAn+1, n−1 x x + (γ − μ) + + γ γ    n−1  An+1, j−1 An+1, n−1 n j + (j + 1)An+1, j x + x × An+1, 0 + γ γ j=1

−

n 

 An+1, j−1 x = − j

Cn, 1 +

j=1

+

 An, i (−d)

i+1

i=0

n−1  n−2  j=1

n−2 

i=j−1

 An, i (−d)i−j+1

i+1 j



 xj e−d/γ .

[4.34]

78

Ruin Probabilities

To find An+1, j , 0 ≤ j ≤ n − 1, from [4.34], we apply the method of undetermined coefficients. To be more precise, we equate the expressions of xn−1 , xj , 1 ≤ j ≤ n − 2, and x0 in [4.34] and obtain [4.6], [4.7] and [4.8], respectively. Thus, to find An+1, n−1 , it suffices to know An, n−2 . The constants An+1, j , 1 ≤ j ≤ n − 2, are expressed in a recurrent way in terms of An+1, j+1 and An, i , j − 1 ≤ i ≤ n − 2. Finally, An+1, 0 can be found in terms of An+1, 1 , Cn, 1 and An, i , 0 ≤ i ≤ n − 2. Similarly, to derive formulas for Bn+1, j , 0 ≤ j ≤ n − 2, we calculate  (d, +∞)   (d, +∞)  ϕn+1, part2 (x) and ϕn+1, part2 (x) using [4.30] and substitute them, as well as [4.34], into [4.26]. Thus, we obtain  γ (−2Bn+1, 0 + 2μBn+1, 1 ) +

n−3  j=1

Bn+1, j−1 μ 

− 2(j + 1)Bn+1, j + (j + 2)(j + 1)μBn+1, j+1 xj 

 Bn+1, n−3 − 2(n − 1)Bn+1, n−2 xn−2 + μ   Bn+1, n−2 n−1 x + + (γ − μ) Bn+1, 0 μ [4.35]   n−2  Bn+1, j−1 Bn+1, n−2 n−1 j + (j + 1)Bn+1, j x − x + − μ μ j=1

−

n−1 

 Bn+1, j−1 xj = −

Cn, 2 +

j=1

+

 Bn, i (−d)i+1

i=0

n−2  n−3  j=1

n−3 

i=j−1

 Bn, i (−d)

i−j+1

i+1 j



 x

j

ed/μ .

Equating the expressions of xn−2 , xj , 1 ≤ j ≤ n − 3, and x0 in [4.35] yields [4.9], [4.10] and [4.11], respectively.

Classical Risk Model with a Franchise and a Liability Limit

79

Since the continuity of ϕ(d, +∞) (x) and its derivative at the point x = nd implies ⎧ (d, +∞) (d, +∞) ⎪ (x) = ϕn+1 (nd), ⎪ ⎨ lim ϕn x↑nd

   (d, +∞)  ⎪ +∞) ⎪ (x) = ϕn+1 (nd) , ⎩ lim ϕ(d, n x↑nd

the constants Cn+1, 1 and Cn+1, 2 can be found by solving the system of linear equations  Cn+1, 1 +

n−1 

 An+1, i (nd)i+1

 end/γ +

Cn+1, 2 +

i=0

 =

Cn, 1 +

n−2  i=0

n−2 

 Bn+1, i (nd)i+1

e−nd/μ

i=0

 An, i (nd)i+1

 end/γ +

Cn, 2 +

n−3 

 Bn, i (nd)i+1

e−nd/μ

i=0

and   n−2  An+1, i Cn+1, 1 + An+1, 0 + + (i + 2)An+1, i+1 (nd)i+1 γ γ i=0   An+1, n−1 Cn+1, 2 n nd/γ (nd) e + Bn+1, 0 + + − γ μ  n−3  Bn+1, i + (i + 2)Bn+1, i+1 (nd)i+1 + − μ i=0  Bn+1, n−2 n−1 (nd) − e−nd/μ μ   n−3  An, i Cn, 1 + An, 0 + + (i + 2)An, i+1 (nd)i+1 = γ γ i=0   An, n−2 Cn, 2 n−1 nd/γ (nd) + Bn, 0 + + − e γ μ   n−4  Bn, i Bn, n−3 i+1 n−2 + (i + 2)Bn, i+1 (nd) (nd) + − e−nd/μ , − μ μ i=0

80

Ruin Probabilities

which guarantees that the solution is the survival probability. Thus, Cn+1, 1 and Cn+1, 2 are given by [4.12] and [4.13], which completes the proof.  R EMARK 4.1.– Theorem 4.1 gives the analytic expression for ϕ(d, +∞) (x) for all x ≥ 0. However, the computations become too tedious for large initial surpluses. So, this theorem is useful when the initial surpluses are not too large, otherwise it is reasonable to use the Cramér–Lundberg approximation (see section 1.1.7 and [ASM 10, BEA 84, BÜH 70, DIC 10, GER 79, GRA 91, ROL 99, SCH 08b]).    (2, +∞)  E XAMPLE 4.1.– Let the surplus processes Xt (x) t≥0 and Xt (x) t≥0 follow [4.1] and [4.4], respectively, under the above assumptions with d = 2 and L = +∞. Furthermore, let ϕ(x) and ϕ(2, +∞) (x) be the corresponding survival probabilities, the claim sizes be exponentially distributed with mean μ and θ = 0.1. To compute ϕ(x) and ϕ(2, +∞) (x) , we apply formula [4.3] and theorem 4.1, respectively. Table 4.1 gives a comparison of ϕ(x) and ϕ(2, +∞) (x) for small initial surpluses and different μ. The computational results show that a franchise can considerably decrease the survival probability when d is too large for given μ (d = μ = 2 in the example). If d is considerably less than μ, then a franchise decreases the survival probability  very slightly. 4.2.2. Case of small and large enough initial surpluses We now investigate how a franchise changes the survival probability for small and large enough initial surpluses.    (d, +∞)  (x) t≥0 T HEOREM 4.2.– Let the surplus processes Xt (x) t≥0 and Xt follow [4.1] and [4.4], respectively, under the above assumptions with 0 < d < +∞ and L = +∞. Moreover, let ϕ(x) and ϕ(d, +∞) (x) be the corresponding survival probabilities and the claim sizes be exponentially distributed with mean μ. i) If 

    μ(1 + θ) θd x ∈ 0, min ln 1 + , d , θ μ(1 + θ) then ϕ(d, +∞) (x) < ϕ(x) for any 0 < d < +∞.

[4.36]

Classical Risk Model with a Franchise and a Liability Limit

81

ii) For  d∈

μ(1 + θ) ln(1 + θ) 0, θ

 [4.37]

and large enough initial surpluses, we have ϕ(d, +∞) (x) > ϕ(x). x

μ

ϕ(x)

ϕ(2, +∞) (x)

0.01 0.20 0.50 1.00 2.00 4.00 8.00 0.01 0.20 0.50 1.00 2.00 4.00 8.00 0.01 0.20 0.50 1.00 2.00 4.00 8.00

2 2 2 2 2 2 2 10 10 10 10 10 10 10 50 50 50 50 50 50 50

0.09132222 0.09913610 0.11133723 0.13130633 0.16990844 0.24204280 0.36805097 0.09099173 0.09256048 0.09503195 0.09913610 0.10728866 0.12337311 0.15467783 0.09092562 0.09123961 0.09173516 0.09256048 0.09420887 0.09749668 0.10403653

0.09111594 0.09513668 0.10184951 0.11410656 0.14322337 0.20447048 0.31354244 0.09097799 0.09229699 0.09441867 0.09806373 0.10578145 0.12165650 0.15255999 0.09092499 0.09122751 0.09170723 0.09251238 0.09414395 0.09742971 0.10396548



ϕ(2, +∞) (x) ϕ(x)

 − 1 × 100%

-0.2259% -4.0343% -8.5216% -13.0990% -15.7056% -15.5230% -14.8101% -0.0151% -0.2847% -0.6453% -1.0817% -1.4048% -1.3914% -1.3692% -0.0007% -0.0133% -0.0304% -0.0520% -0.0689% -0.0687% -0.0683%

Table 4.1. Comparison of ϕ(x) and ϕ(2, +∞) (x) for different μ

P ROOF.– We now prove assertion i) of the theorem. Introduce the function g1 (x) =

ϕ(x) ϕ(d, +∞) (x)

for x ∈ [0, d]. Substituting [4.3] and [4.14] into [4.38] yields   θx   1 + θ − exp − μ(1+θ) x exp − . g1 (x) = θ (1 + θ)(μ + d)

[4.38]

82

Ruin Probabilities

Taking the derivative gives   x 1  g1 (x) = exp − θ(μ + d) (1 + θ)(μ + d)     μ(1 + θ) + θd θx exp − −1 . × μ(1 + θ) μ(1 + θ) Since

  μ(1 + θ) + θd θx exp − −1>0 μ(1 + θ) μ(1 + θ)

for all x given by [4.36], we have g1 (x) > 0 for these x. Furthermore, g1 (0) = 1, which gives g1 (x) > 1 for these x. Thus, assertion i) of the theorem follows and we now prove assertion ii). By the Cramér–Lundberg approximation (see section 1.1.7 and [ASM 10, BEA 84, BÜH 70, DIC 10, GER 79, GRA 91, ROL 99, SCH 08b]), we have ϕ(d, +∞) (x) ∼ 1 −

θ (1 +

θ) R(d, +∞) μ(d, +∞)

e−R

(d, +∞)

x

,

[4.39]

where R(d, +∞) is a unique positive solution (if it exists) to  +∞  (d, +∞) y  1 − F (d, +∞) (y) dy = (1 + θ)(μ + d) e−d/μ eR

[4.40]

0

and μ(d, +∞) =

1 (1 + θ)(μ + d) e−d/μ  +∞  (d, +∞) y  1 − F (d, +∞) (y) dy y eR ×

[4.41]

0

if the improper integral in the right-hand side of [4.41] is finite. First, we show that there is a unique positive solution R(d, +∞) to [4.40], such that the integral in the right-hand side of [4.41] is finite. If R(d, +∞) < 1/μ, then substituting F (d, +∞) (y) into [4.40] and doing elementary computations yield edR

(d, +∞)

= (1 + θ)(μ + d) R(d, +∞) + 2 +

1 μR(d, +∞)

−1

.

[4.42]

Classical Risk Model with a Franchise and a Liability Limit

83

If R(d, +∞) ≥ 1/μ, then the integral in the left-hand side of [4.40] is infinite. So, in what follows, we consider the case R(d, +∞) < 1/μ only. For R ∈ [0, 1/μ), introduce the functions g2 (R) = edR and g3 (R) = (1 + θ)(μ + d) R + 2 +

1 . μR − 1

Note that g2 (0) = 1 and g3 (0) = 1. Moreover, we have g2 (R) ≈ 1 + dR and g3 (R) ≈ 1 + g3 (0) R ≈ 1 + (d + θd + θμ) R for small enough R > 0. Consequently, g3 (R) > g2 (R) in some right semineighborhood of the point R = 0. The function g2 (R) is increasing on [0, 1/μ). The function g3 (R) is increasing on [0, R∗ ) and decreasing on (R∗ , 1/μ), where    μ 1 ∗ 1− . R = μ (1 + θ)(μ + d) Thus, [4.42] has the unique solution R(d, +∞) on (0, 1/μ). Note that is the unique positive solution to [4.40]. It is evident that μ(d, +∞) is finite in this case.

R(d, +∞)

Next, from [4.3] and [4.39] we conclude that ϕ(d, +∞) (x) > ϕ(x) for large enough initial surpluses provided that R(d, +∞) >

θ . μ(1 + θ)

Let g4 (R) = g2 (R) − g3 (R) on [0, 1/μ). The function g4 (R) is negative on (0, R(d, +∞) ) and positive on (R(d, +∞) , 1/μ). Moreover, g4 (0) = g4 (R(d, +∞) ) = 0 and     θ θd θd − 1. [4.43] g4 = exp − μ(1 + θ) μ(1 + θ) μ

84

Ruin Probabilities

Therefore, ϕ(d, +∞) (x) > ϕ(x) for large enough initial surpluses provided that the expression in the right-hand side of [4.43] is negative. Let



θd g5 (d) = exp μ(1 + θ)

 −

θd − 1. μ

Taking the derivative yields     θd 1 θ  exp g5 (d) = −1 . μ 1+θ μ(1 + θ) Since g5 (0) = 0 and g5 (d) < 0 for d given by [4.37], we prove assertion ii) of the theorem.  4.3. Survival probability in the classical risk model with a liability limit If the insurance company establishes a liability limit only and the claim sizes are exponentially distributed, then equation [4.2] for ϕ(0, L) (x) can be written as  x  (0, L)  (0, L) (0, L) ϕ (x) + = λϕ (x) − λ ϕ(0, L) (x − y) dF (0, L) (y), [4.44] c 0

where

⎧ ⎪ ⎨0 (0, L) (y) = 1 − e−y/μ F ⎪ ⎩ 1

if y < 0, if 0 ≤ y < L, if y ≥ L,

and

 (0, L) c(0, L) = λ(1 + θ) E Yi   L −y/μ   ye dy + Le−L/μ = λμ(1 + θ) 1 − e−L/μ . = λ(1 + θ) μ 0

In section 4.3.1, we derive analytic expressions for ϕ(0, L) (x) in the case of exponentially distributed claim sizes. In section 4.3.2, we investigate how a liability limit changes the survival probability for small enough and large enough initial surpluses.

Classical Risk Model with a Franchise and a Liability Limit

85

4.3.1. Analytic expression for the survival probability To formulate the next theorem, introduce the constants   γ¯1 = 1 − (1 + θ) 1 − e−L/μ ,   γ¯2 = μ(1 + θ) 1 − e−L/μ ,   θ 1 − e−L/μ ¯ C1, 1 = − , γ¯1 C¯1, 2 =

θ , γ¯1 (1 + θ)

   1 γ¯1 θ exp −L + , γ¯1 γ¯2 (1 + θ) μ γ¯2      1 θ −L/μ −L/μ , 1− = − (1 + θ) 1 − e e γ¯1 (1 + θ) γ¯1       1 1 γ¯1 θ L 1+ + = + − 1 exp −L . γ¯1 (1 + θ) γ¯1 γ¯2 μ γ¯2

A¯2, 0 = − C¯2, 1 C¯2, 2

Moreover, let the constants A¯n+1, j , 0 ≤ j ≤ n − 1, be given in a recurrent way by formulas    A¯n, n−2 1 γ¯1 ¯ + An+1, n−1 = − exp −L , n ≥ 2, [4.45] n¯ γ2 μ γ¯2    (j + 2)¯ γ2 A¯n+1, j+1 1 γ¯1 1 exp −L + A¯n+1, j = − − γ¯1 j+1 μ γ¯2  n−3     A¯n, i (i + 2)A¯n, i+1  i−j+1 i + 1 × + (−L) j γ¯2 γ¯1 [4.46] i=j−1   A¯n, n−2 n−1 + (−L)n−j+1 , j γ¯2 1 ≤ j ≤ n − 2,

n ≥ 3,

86

Ruin Probabilities

A¯n+1, 0

   ¯ 2¯ γ2 A¯n+1, 1 Cn, 2 A¯n, 0 1 γ¯1 + =− − exp −L + γ¯1 μ γ¯2 γ¯2 γ¯1 n−3  A¯n, i (i + 2)A¯n, i+1  [4.47] + + (−L)i+1 γ¯2 γ¯1 i=0  A¯n, n−2 n−1 + (−L) , n ≥ 2. γ¯2

Finally, let the constants C¯n+1, 1 and C¯n+1, 2 be given by formulas n−2  γ ¯ 2 C¯n+1, 1 = C¯n, 1 + (i + 1)(A¯n+1, i − A¯n, i )(nL)i γ¯1 i=0 [4.48]  + nA¯n+1, n−1 (nL)n−1 enL¯γ1 /¯γ2 , n ≥ 2, γ¯2 (A¯n, 0 − A¯n+1, 0 ) C¯n+1, 2 = C¯n, 2 + γ¯1 n−2  + A¯n, i−1 − A¯n+1, i−1 i=1

 [4.49] (i + 1)¯ γ2 (A¯n, i − A¯n+1, i ) + (nL)i γ¯1   n¯ γ2 A¯n+1, n−1 ¯ ¯ + An, n−2 − An+1, n−2 − (nL)n−1 γ¯1 − A¯n+1, n−1 (nL)n ,

n ≥ 2.

Note that to compute the constants by the formulas above for any n ≥ 2, we have to know all the constants for n − 1. Moreover, for any fixed n ≥ 2, we start from the computation of A¯n+1, j for j from n − 1 to 0 and after that we can compute C¯n+1, 1 and C¯n+1, 2 . We have introduced all the constants only to formulate the next theorem and will get them in the proof.  (0, L)  (x) t≥0 follow [4.4] under T HEOREM 4.3.– Let the surplus process Xt the above assumptions with d = 0 and 0 < L < +∞, and the claim sizes be exponentially distributed with mean μ. Then, (0, L)

ϕ(0, L) (x) = ϕn+1 (x) for all x ∈ [nL, (n + 1)L),

n ≥ 0,

Classical Risk Model with a Franchise and a Liability Limit

87

where (0, L)

ϕ1

(x) = C¯1, 1 + C¯1, 2 eγ¯1 x/¯γ2 ,

[4.50]

  (x) = C¯2, 1 + C¯2, 2 + A¯2, 0 x eγ¯1 x/¯γ2 , [4.51]   n−1  (0, L) j+1 ¯ ¯ ¯ ϕn+1 (x) = Cn+1, 1 + Cn+1, 2 + An+1, j x eγ¯1 x/¯γ2 , [4.52] j=0 (0, L)

ϕ2

n ≥ 2. P ROOF.– By theorem 2.1 and remark 2.1, ϕ(0, L) (x) is continuous on R+ and continuously differentiable on R+ , except at the point x = L, where there are only one-sided derivatives (see also section 1.1.3 and [ROL 99, pp. 162–163]). (0, L)

Let us introduce the functions ϕn+1 (x), n ≥ 0, in the following way:

(0, L)

ϕn+1 (x) is defined on [nL, (n + 1)L) and coincides with ϕ(0, L) (x) on this interval. If x ∈ [0, L), then substituting c(0, L) and F (0, L) (y) into [4.44] yields  (0, L)   (x) μ(1 + θ) 1 − e−L/μ ϕ1  x 1 (0, L) (0, L) = ϕ1 (x) − ϕ (x − y) e−y/μ dy. μ 0 1

[4.53]

(0, L)

(x) has the continuous second derivative on [0, L). This gives that ϕ1 So, integro-differential equation [4.53] can be reduced to the differential one  (0, L)   μ(1 + θ) 1 − e−L/μ ϕ1 (x)  (0, L)    − 1 − (1 + θ) 1 − e−L/μ ϕ1 (x) = 0

[4.54]

in a standard way (see section 1.1.6 and [ASM 10, BEA 84, BÜH 70, DIC 10, GER 79, GRA 91, ROL 99]).     Let γ¯1 = 1 − (1 + θ) 1 − e−L/μ and γ¯2 = μ(1 + θ) 1 − e−L/μ . Solving [4.54] yields [4.50], where the constants C¯1, 1 and C¯1, 2 can be found

88

Ruin Probabilities

from

⎧ ⎪ ⎪ ⎪ ⎨

(0, L)

ϕ1

(0) =

θ , 1+θ

L) ⎪   ϕ(0, (0) ⎪ L) 1 ⎪ ⎩ ϕ(0, (0) = . 1 γ¯2

[4.55]

The first condition in [4.55] guarantees that the solution is the survival probability and the second one follows from [4.53]. Thus, the constants C¯1, 1 and C¯1, 2 are given before the assertion of the theorem. If x ∈ [L, +∞), then substituting c(0, L) and F (0, L) (y) into [4.44] yields    μ(1 + θ) 1 − e−L/μ ϕ(0, L) (x) = ϕ(0, L) (x)  [4.56] e−x/μ x − ϕ(0, L) (y) ey/μ dy − e−L/μ ϕ(0, L) (x − L). μ x−L Note that we imply the right derivative of ϕ(0, L) (x) at the point x = L. It is easily seen from [4.56] that the second classical derivative of ϕ(0, L) (x) exists on [L, +∞) except at the point x = 2L. So, integro-differential equation [4.56] can be reduced to the differential one     γ¯2 ϕ(0, L) (x) − γ¯1 ϕ(0, L) (x) [4.57]   = −e−L/μ ϕ(0, L) (x − L) , x ∈ [L, +∞), where we imply the right second derivative of ϕ(0, L) (x) at the point x = 2L. If x ∈ [L, 2L), then [4.57] can be rewritten as  (0, L)   (0, L)  γ¯2 ϕ2 (x) − γ¯1 ϕ2 (x)   L γ¯1 (x − L) θ exp − + =− . γ¯2 (1 + θ) μ γ¯2

[4.58]

By theorem 2.1 and remark 2.1, the following conditions must hold to guarantee that the solution is the survival probability: ⎧ (0, L) (0, L) (x) = ϕ2 (L), lim ϕ ⎪ ⎪ ⎨ x↑L 1 [4.59]    (0, L)  ⎪ θe−L/μ L) ⎪ ⎩ lim ϕ(0, , (x) = ϕ2 (L) + 1 x↑L γ¯2 (1 + θ) (see also section 1.1.3 and [ROL 99, pp. 162–163]).

Classical Risk Model with a Franchise and a Liability Limit

89

The solution to [4.58] can be found by writing (0, L)

ϕ2

(0, L)

(0, L)

(x) = ϕ2, gen (x) + ϕ2, part (x),

(0, L)

where ϕ2, gen (x) is a general solution to the linear homogeneous equation (0, L)

corresponding to [4.58] and ϕ2, part (x) is a particular solution to linear heterogeneous equation [4.58]. (0, L)

Since ϕ2, part (x) can be written as (0, L) ϕ2, part (x) = A¯2, 0 x eγ¯1 x/¯γ2 , (0, L)

where A¯2, 0 is a constant, substituting ϕ2, part (x) into [4.58] and applying the method of undetermined coefficients yield A¯2, 0 . Furthermore, we can write (0, L) ϕ2, gen (x) as (0, L) ϕ2, gen (x) = C¯2, 1 + C¯2, 2 eγ¯1 x/¯γ2 , (0, L) (x) satisfies conditions where the constants C¯2, 1 and C¯2, 2 are such that ϕ2 [4.59]. (0, L)

(x) is defined by [4.51] and the constants A¯2, 0 , C¯2, 1 and C¯2, 2 Thus, ϕ2 are given before the assertion of the theorem. (0, L)

(0, L)

In general, if we know ϕn (x), n ≥ 2, we can find ϕn+1 (x) applying considerations similar to those in the proof of theorem 4.1. Since  (0, L)   (0, L)    L) γ¯2 ϕn+1 (x) − γ¯1 ϕn+1 (x) = −e−L/μ ϕ(0, (x − L) , n we get [4.52] by induction on n. Applying the method of undetermined coefficients to  γ¯2

 n−2  γ¯ 2 A¯n+1, j−1 2¯ γ1 A¯n+1, 0 1 ¯ + 2An+1, 1 + γ¯2 γ¯22 j=1

 2(j + 1)¯ γ1 A¯n+1, j ¯ + + (j + 2)(j + 1)An+1, j+1 xj γ¯2

90

Ruin Probabilities



  γ¯12 A¯n+1, n−2 2n¯ γ1 A¯n+1, n−1 γ¯12 A¯n+1, n−1 n n−1 + + + x x γ¯2 γ¯22 γ¯22   n−1  γ¯1 A¯n+1, j−1 ¯ ¯ − γ¯1 An+1, 0 + + (j + 1)An+1, j xj γ¯2 j=1

  ¯ γ¯1 A¯n+1, n−1 n γ¯1 Cn, 2 + x =− + A¯n, 0 γ¯2 γ¯2   n−3  γ¯1 A¯n, i γ¯1 A¯n, n−2 + + (i + 2)A¯n, i+1 (−L)i+1 + (−L)n−1 γ¯2 γ¯2 i=0

+

n−1  n−3  j=1

i=j−1



   γ¯1 A¯n, i i−j+1 i + 1 ¯ + (i + 2)An, i+1 (−L) j γ¯2

      γ¯1 A¯n, n−2 1 γ¯1 n−j+1 n − 1 j + x exp −L + (−L) j γ¯2 μ γ¯2 yields the constants A¯n+1, j , 0 ≤ j ≤ n − 1, in [4.52]. Thus, they are given by [4.45], [4.46] and [4.47]. To find the constants C¯n+1, 1 and C¯n+1, 2 in [4.52], we use the continuity of and its derivative at the point x = nL, which implies ⎧ (0, L) L) ⎪ (x) = ϕn+1 (nL), lim ϕ(0, ⎪ n ⎨ x↑nL

ϕ(0, L) (x)

   (0, L)  ⎪ L) ⎪ (x) = ϕn+1 (nL) , ⎩ lim ϕ(0, n x↑nL

and guarantees that the solution is the survival probability. Thus, C¯n+1, 1 and C¯n+1, 2 are given by [4.48] and [4.49], respectively. The proof is complete.     (0, 30)  (x) t≥0 E XAMPLE 4.2.– Let the surplus processes Xt (x) t≥0 and Xt follow [4.1] and [4.4], respectively, under the above assumptions with d = 0 and L = 30. Furthermore, let ϕ(x) and ϕ(0, 30) (x) be the corresponding survival probabilities, the claim sizes be exponentially distributed with mean μ and θ = 0.1. To compute ϕ(x) and ϕ(0, 30) (x), we apply formula [4.3] and theorem 4.3, respectively. Table 4.2 gives a comparison of ϕ(x) and ϕ(0, 30) (x) for different μ. The computational results show that a liability limit can considerably increase the survival probability when L is too small for

Classical Risk Model with a Franchise and a Liability Limit

91

given μ (L = 2μ = 30 in the example). If L is considerably greater than μ, then a liability limit increases the survival probability very slightly.  x

μ

ϕ(x)

ϕ(0, 30) (x)

1 10 20 30 60 90 120 1 10 20 30 60 90 120 1 10 20 30 60 90 120

3 3 3 3 3 3 3 10 10 10 10 10 10 10 15 15 15 15 15 15 15

0.11804409 0.32856662 0.50409494 0.63373607 0.85243581 0.94054782 0.97604729 0.09913610 0.16990844 0.24204280 0.30790874 0.47311066 0.59887894 0.69462638 0.09640207 0.14436915 0.19468542 0.24204280 0.36805097 0.47311066 0.56070447

0.11804551 0.32859294 0.50416485 0.63385601 0.85257442 0.94064002 0.97609908 0.09958778 0.17602881 0.25754340 0.33560557 0.52122121 0.65518553 0.75054563 0.09729201 0.15573294 0.22281568 0.29223602 0.45896776 0.58698104 0.67711345



ϕ(0, 30) (x) ϕ(x)

 − 1 × 100%

0.0012% 0.0080% 0.0139% 0.0189% 0.0163% 0.0098% 0.0053% 0.4556% 3.6022% 6.4041% 8.9951% 10.1690% 9.4020% 8.0503% 0.9232% 7.8713% 14.4491% 20.7373% 24.7022% 24.0684% 20.7612%

Table 4.2. Comparison of ϕ(x) and ϕ(0, 30) (x) for different μ

4.3.2. Case of small enough and large enough initial surpluses We now investigate how a liability limit changes the survival probability for small enough and large enough initial surpluses.    (0, L)  (x) t≥0 T HEOREM 4.4.– Let the surplus processes Xt (x) t≥0 and Xt follow [4.1] and [4.4], respectively, under the above assumptions with d = 0 and 0 < L < +∞. Moreover, let ϕ(x) and ϕ(0, L) (x) be the corresponding survival probabilities and the claim sizes be exponentially distributed with mean μ. Then, ϕ(0, L) (x) > ϕ(x) for any 0 < L < +∞ and for small enough and large enough initial surpluses.

92

Ruin Probabilities

P ROOF.– By [4.3] and [4.50], we have ϕ(x) ≈

θx θ + 1 + θ μ(1 + θ)2

and (0, L)

ϕ1

(x) ≈

θx θ   + 1 + θ μ(1 + θ)2 1 − e−L/μ

for small enough initial surpluses. This gives the assertion of the theorem for such initial surpluses. By the Cramér–Lundberg approximation (see section 1.1.7 and [ASM 10, BEA 84, BÜH 70, DIC 10, GER 79, GRA 91, ROL 99, SCH 08b]), we have ϕ(0, L) (x) ∼ 1 −

θ R(0, L) μ(0, L) (1

+ θ)

e−R

(0, L) x

[4.60]

for large enough initial surpluses, where R(0, L) is a unique positive solution (if it exists) to  +∞    (0, L) y  1 − F (0, L) (y) dy = μ(1 + θ) 1 − e−L/μ [4.61] eR 0

and μ

(0, L)

1   = μ(1 + θ) 1 − e−L/μ



+∞

y eR

(0, L) y



 1 − F (0, L) (y) dy. [4.62]

0

It is obvious that the improper integral in the right-hand side of [4.62] is finite. If R(0, L) = 1/μ is a solution to [4.61], then   L = μ(1 + θ) 1 − e−L/μ must hold. If R(0, L) = 1/μ, then substituting F (0, L) (y) into [4.61] yields eLR

(0, L)

   = (1 + θ) eL/μ − 1 μR(0, L) − 1 + eL/μ .

[4.63]

Classical Risk Model with a Franchise and a Liability Limit

93

Consider the function

   g¯(R) = eLR − (1 + θ) eL/μ − 1 μR − 1 − eL/μ

¯ ∗ ) and increasing on (R ¯ ∗ , +∞), where on R+ . It is decreasing on [0, R    μ(1 + θ) eL/μ − 1 1 ∗ ¯ . R = ln L L   Moreover, g¯(0) = θ eL/μ − 1 > 0, g¯(1/μ) = 0 and limR→+∞ g¯(R) = +∞. ¯ ∗ = 1/μ if and only if [4.63] holds. Note that R Thus, we conclude that R = 1/μ is a unique zero of g¯(R) if [4.63] is true. This means that [4.61] has the unique solution R(0, L) = 1/μ. Otherwise, if [4.63] is not true, then R = R(0, L) and R = 1/μ are positive zeros of g¯(R), which means that R(0, L) is a unique positive solution to [4.61], such that R(0, L) = 1/μ. Furthermore, in this case, g¯(R) is negative between its zeros R = R(0, L) and R = 1/μ, otherwise it is positive. Since  g¯

θ μ(1 + θ)





θL = exp μ(1 + θ)

 −1>0

and 1 θ < , μ(1 + θ) μ we get θ < R(0, L) . μ(1 + θ) Consequently, by [4.3] and [4.60], we have the assertion of the theorem for large enough initial surpluses.  4.4. Survival probability in the classical risk model with both a franchise and a liability limit If the insurance company establishes both a franchise and a liability limit and the claim sizes are exponentially distributed, then equation [4.2] for ϕ(d, L) (x) can be written as  x  (d, L)  (d, L) (d, L) ϕ (x) + = λϕ (x) − λ ϕ(d, L) (x − y) dF (d, L) (y), [4.64] c 0

94

Ruin Probabilities

where

⎧ ⎪ 0 ⎪ ⎪ ⎪ ⎨1 − e−d/μ F (d, L) (y) = ⎪1 − e−y/μ ⎪ ⎪ ⎪ ⎩1

and

if y < 0, if 0 ≤ y < d, if d ≤ y < L, if y ≥ L,

 (d, L) c(d, L) = λ(1 + θ) E Yi  L −y/μ  ye −L/μ = λ(1 + θ) dy + Le μ d   = λ(1 + θ) (μ + d)e−d/μ − μe−L/μ .

In section 4.4.1, we derive analytic expressions for ϕ(d, L) (x) in the case of exponentially distributed claim sizes. In section 4.4.2, we investigate how a franchise and a liability limit change the survival probability for small initial surpluses. 4.4.1. Analytic expression for the survival probability To formulate the next theorem, introduce the constants   γ˜ = (1 + θ) μ + d − μe(d−L)/μ , C˜1, 1 = θ/(1 + θ), θ e−d/˜γ , (1 + θ)(˜ γ + μ)   θ γ˜ μ + d(˜ γ + μ) −d/˜γ = e , 1+ 1+θ (˜ γ + μ)2 θ˜ γμ ed/μ . =− (1 + θ)(˜ γ + μ)2

A˜2, 0 = − C˜2, 1 C˜2, 2

Moreover, let the constants A˜n+1, j , 0 ≤ j ≤ n − 1, be given in a recurrent way by formulas A˜n, n−2 −d/˜γ e , A˜n+1, n−1 = − n(˜ γ + μ)

n ≥ 2,

[4.65]

Classical Risk Model with a Franchise and a Liability Limit

 1 1 (j + 2)˜ γ μA˜n+1, j+1 + γ˜ + μ j+1     n−2  i−j+1 i + 1 −d/˜ γ ˜ e An, i (−d) , × j

95

A˜n+1, j = −

[4.66]

i=j−1

1 ≤ j ≤ n − 2,

A˜n+1, j

3 ≤ n ≤ m,

and n − m + 1 ≤ j ≤ n − 2, n ≥ m + 1,  1 1 (j + 2)˜ γ μA˜n+1, j+1 + =− γ˜ + μ j+1    n−2  i + 1 i−j+1 e−d/˜γ A˜n, i (−d) × j i=j−1

−

n−m−1  i=j−1

  i−j+1 i + 1 ˜ An−m+1, i (−L) j



d−L L × exp − μ γ˜

A˜n+1, 0

[4.67]

 ,

1 ≤ j ≤ n − m, n ≥ m + 1,  1 2˜ γ μA˜n+1, 1 =− γ˜ + μ    n−2  i+1 −d/˜ γ ˜ ˜ An, i (−d) + Cn, 1 + e ,

[4.68]

i=0

A˜m+1, 0

2 ≤ n ≤ m − 1,  1 2˜ γ μA˜m+1, 1 =− γ˜ + μ   m−2  A˜m, i (−d)i+1 e−d/˜γ + C˜m, 1 + i=0



d−L L θ exp − − 1+θ μ γ˜

 ,

[4.69]

96

Ruin Probabilities

 1 2˜ γ μA˜n+1, 1 γ˜ + μ   n−2  i+1 ˜ ˜ An, i (−d) + Cn, 1 + e−d/˜γ

A˜n+1, 0 = −

i=0

 − C˜n−m+1, 1 +  × exp

n−m−1 

[4.70]

 A˜n−m+1, i (−L)i+1

i=0

d−L L − μ γ˜

 ,

n ≥ m + 1.

˜n+1, j , 0 ≤ j ≤ n − 2, be given in a recurrent way Next, let the constants B by formulas ˜ ˜3, 0 = C2, 2 ed/μ , B γ˜ + μ ˜n, n−3 B ed/μ , n ≥ 3, (n − 1)(˜ γ + μ)  1 ˜n+1, j+1 + 1 (j + 2)˜ γ μB = γ˜ + μ j+1     n−3  i−j+1 i + 1 d/μ ˜ e Bn, i (−d) , × j

˜n+1, n−2 = B ˜n+1, j B

[4.71]

[4.72]

i=j−1

1 ≤ j ≤ n − 3,

˜n+1, j B

4 ≤ n ≤ m,

and n − m ≤ j ≤ n − 3, n ≥ m + 1,  1 ˜n+1, j+1 + 1 (j + 2)˜ γ μB = γ˜ + μ j+1    n−3  i−j+1 i + 1 ˜ Bn, i (−d) × j i=j−1

−

n−m−2  i=j−1



˜n−m+1, i (−L)i−j+1 i + 1 B j

1 ≤ j ≤ n − m − 1,

n ≥ m + 1,



 ed/μ ,

[4.73]

Classical Risk Model with a Franchise and a Liability Limit

˜n+1, 0 = B

˜n+1, 0 B

97

    n−3  1 i+1 d/μ ˜ ˜ ˜ Bn, i (−d) e , 2˜ γ μBn+1, 1 + Cn, 2 + γ˜ + μ [4.74] i=0

3 ≤ n ≤ m,   n−3  1 ˜ ˜n, i (−d)i+1 ˜ 2˜ γ μBn+1, 1 + Cn, 2 + B = γ˜ + μ i=0

− C˜n−m+1, 2 −

n−m−2 

 ˜n−m+1, i (−L) B

i+1

 e

d/μ

,

[4.75]

i=0

n ≥ m + 1. Finally, let the constants C˜n+1, 1 and C˜n+1, 2 be given by formulas ˜3, 0 γ˜ μB ˜3, 0 − 2dB C˜3, 2 = C˜2, 2 + γ˜ + μ

  γ˜ μ(A˜3, 0 − A˜2, 0 + 4dA˜3, 1 ) γ˜ + μ exp 2d + , γ˜ + μ γ˜ μ

n−3 γ˜ μ(A˜n, 0 − A˜n+1, 0 )  ˜ ˜ ˜ + Cn+1, 1 = Cn, 1 + An, i − A˜n+1, i γ˜ + μ i=0

 (i + 2)˜ γ μ(A˜n, i+1 − A˜n+1, i+1 ) (nd)i+1 γ˜ + μ  ˜n+1, n−1  n˜ γ μ A + A˜n, n−2 − A˜n+1, n−2 − γ˜ + μ +

× (nd)n−1 − A˜n+1, n−1 (nd)n n−3  γ˜ μ ˜n, i − B ˜n+1, i )(nd)i + (i + 1)(B γ˜ + μ i=0    γ˜ + μ n−2 ˜ , − (n − 1)Bn+1, n−2 (nd) exp −nd γ˜ μ n ≥ 2,

n = m,

[4.76]

98

Ruin Probabilities

 ˜n, 0 ) n−4 ˜n+1, 0 − B γ˜ μ(B ˜ ˜ ˜n+1, i ˜n, i − B + Cn+1, 2 = Cn, 2 + B γ˜ + μ i=0

˜n, i+1 )  ˜n+1, i+1 − B (i + 2)˜ γ μ(B + (nd)i+1 γ˜ + μ  ˜n+1, n−2  (n − 1)˜ γ μB ˜ ˜ + Bn, n−3 − Bn+1, n−3 + γ˜ + μ ˜n+1, n−2 (nd)n−1 × (nd)n−2 − B n−2  γ˜ μ + (i + 1)(A˜n+1, i − A˜n, i )(nd)i γ˜ + μ i=0    γ ˜ + μ n−1 , + nA˜n+1, n−1 (nd) exp nd γ˜ μ n ≥ 3,

[4.77]

n = m,

m−3 γ˜ μ(A˜m, 0 − A˜m+1, 0 )  ˜ ˜ ˜ + Am, i − A˜m+1, i Cm+1, 1 = Cm, 1 + γ˜ + μ i=0

 (i + 2)˜ γ μ(A˜m, i+1 − A˜m+1, i+1 ) + (md)i+1 γ˜ + μ   m˜ γ μA˜m+1, m−1 ˜ ˜ + Am, m−2 − Am+1, m−2 − γ˜ + μ [4.78] × (md)m−1 − A˜m+1, m−1 (md)m m−3  γ˜ μ ˜m, i − B ˜m+1, i )(md)i + (i + 1)(B γ˜ + μ i=0    γ ˜ + μ m−2 ˜m+1, m−2 (md) − (m − 1)B exp −md γ˜ μ   d−L L θμ exp − , − (1 + θ)(˜ γ + μ) μ γ˜

Classical Risk Model with a Franchise and a Liability Limit

99

 ˜m, 0 ) m−4 ˜m+1, 0 − B γ˜ μ(B ˜ ˜ ˜m+1, i ˜m, i − B + Cm+1, 2 = Cm, 2 + B γ˜ + μ i=0

˜m, i+1 )  ˜m+1, i+1 − B (i + 2)˜ γ μ(B + (md)i+1 γ˜ + μ  ˜m, m−3 − B ˜m+1, m−3 + B +

˜m+1, m−2 (m − 1)˜ γ μB γ˜ + μ

˜m+1, m−2 (md) −B ×

m−2 

m−1

 [4.79]

γ˜ μ + γ˜ + μ

(i + 1)(A˜m+1, i − A˜m, i )(md)i

i=0

+ mA˜m+1, m−1 (md) +

(md)m−2

 m−1



γ˜ + μ exp md γ˜ μ



θμ ed/μ . (1 + θ)(˜ γ + μ)

Note that in order to compute the constants by the formulas above for any n ≥ 2, we have to know all the constants for n − 1. Moreover, for any fixed ˜n+1, j n ≥ 2, we start from the computation of A˜n+1, j for j from n − 1 to 0, B for j from n − 2 to 0 and after that we can compute C˜n+1, 1 and C˜n+1, 2 . We have introduced all the constants only to formulate the next theorem and we will get them in the proof.  (d, L)  (x) t≥0 follow [4.4] under T HEOREM 4.5.– Let the surplus process Xt the above assumptions with 0 < d < +∞ and L = md, where m ≥ 3 is integer, and the claim sizes be exponentially distributed with mean μ. Then, (d, L)

ϕ(d, L) (x) = ϕn+1 (x) for all x ∈ [nd, (n + 1)d),

n ≥ 0,

where (d, L)

ϕ1

(d, L)

ϕ2

(x) = C˜1, 1 ex/˜γ ,

[4.80]

  (x) = C˜2, 1 + A˜2, 0 x ex/˜γ + C˜2, 2 e−x/μ ,

[4.81]

100

Ruin Probabilities

 (d, L) ϕn+1 (x)

=

C˜n+1, 1 +

n−1 

 j+1 ˜ An+1, j x ex/˜γ

j=0

 + C˜n+1, 2 +

n−2 

[4.82]



˜n+1, j xj+1 e−x/μ , B

n ≥ 2.

j=0

P ROOF.– By theorem 2.1 and remark 2.1, ϕ(d, L) (x) is continuous on R+ and continuously differentiable on this interval, except at the point x = L, where there are only one-sided derivatives (see also section 1.1.3 and [ROL 99, pp. 162–163]). (d, L)

Let us introduce the functions ϕn+1 (x), n ≥ 0, in the following way:

(d, L)

ϕn+1 (x) is defined on [nd, (n + 1)d) and coincides with ϕ(d, L) (x) on this interval. If x ∈ [0, d), then substituting c(d, L) and F (d, L) (y) into [4.64] yields  (d, L)   (d, L) (1 + θ) μ + d − μe(d−L)/μ ϕ1 (x) = ϕ1 (x). [4.83]   Let γ˜ = (1 + θ) μ + d − μe(d−L)/μ . Solving [4.83] gives [4.80], where (d, L) the constant C˜1, 1 can be found from ϕ1 (0) = θ/(1 + θ), which guarantees that the solution is the survival probability. Thus, C˜1, 1 = θ/(1 + θ). If x ∈ [d, L), then [4.64] can be rewritten as    (1 + θ) μ + d − μe(d−L)/μ ϕ(d, L) (x)  e(d−x)/μ x−d (d, L) (d, L) =ϕ (x) − ϕ (y) ey/μ dy, μ 0

[4.84]

which shows that ϕ(d, L) (x) has the second derivative on this interval. So, integro-differential equation [4.84] can be reduced to the differential one     γ˜ μ ϕ(d, L) (x) + (˜ γ − μ) ϕ(d, L) (x) − ϕ(d, L) (x) [4.85] = −ϕ(d, L) (x − d). Comparison of [4.84] with [4.19] and [4.80] with [4.14] gives [4.81] for (d, L) and [4.82] for ϕn+1 (x), 2 ≤ n ≤ m − 1, where the constants A˜2, 0 ,

(d, L) ϕ2 (x)

Classical Risk Model with a Franchise and a Liability Limit

101

˜3, 0 , C˜3, 2 are given before the assertion of the theorem and the C˜2, 1 , C˜2, 2 , B constants A˜n+1, n−1 for 2 ≤ n ≤ m − 1, A˜n+1, j for 1 ≤ j ≤ n − 2 and ˜n+1, n−2 for 3 ≤ n ≤ m − 1, 3 ≤ n ≤ m − 1, A˜n+1, 0 for 2 ≤ n ≤ m − 1, B ˜ ˜ Bn+1, j for 1 ≤ j ≤ n − 3, 4 ≤ n ≤ m − 1, Bn+1, 0 for 3 ≤ n ≤ m − 1 and C˜n+1, 1 for 2 ≤ n ≤ m − 1, C˜n+1, 2 for 3 ≤ n ≤ m − 1 are given by [4.65], [4.66], [4.68], [4.71], [4.72], [4.74], [4.76] and [4.77], respectively. If x ∈ [L, +∞), then [4.64] can be rewritten as    (1 + θ) μ + d − μe(d−L)/μ ϕ(d, L) (x)  e(d−x)/μ x−d (d, L) (d, L) =ϕ (x) − ϕ (y) ey/μ dy, μ x−L

[4.86]

which shows that ϕ(d, L) (x) has the second derivative on this interval. Note that we imply the right second derivative of ϕ(d, L) (x) at the point x = L. Therefore, integro-differential equation [4.86] can be reduced to the differential one     γ − μ) ϕ(d, L) (x) − ϕ(d, L) (x) γ˜ μ ϕ(d, L) (x) + (˜ [4.87] = e(d−L)/μ ϕ(d, L) (x − L) − ϕ(d, L) (x − d). (d, L)

From [4.87], it follows that ϕn+1 (x), n ≥ m, are also given by [4.82]. Next, if x ∈ [md, (m + 1)d), then 0 ≤ x − L < d and (m − 1)d ≤ x − d < (d, L) (d, L) md. Hence, ϕ(d, L) (x − L) = ϕ1 (x) and ϕ(d, L) (x − d) = ϕm (x). So, to (d, L) (d, L) (x) and ϕm (x), find the constants in [4.82] for n = m, we substitute ϕ1 which have already been found, into [4.87]. Moreover, we take into account that ⎧ (d, L) L) ⎪ ϕ(d, (x) = ϕm+1 (L) , ⎪ lim m ⎨ x↑L  (d, L)   (d, L)  ⎪ ⎪ ⎩ lim ϕm (x) = ϕm+1 (L) + x↑L

θ e(d−L)/μ , γ˜ (1 + θ)

must hold by theorem 2.1 and remark 2.1 to guarantee that the solution is the survival probability (see also section 1.1.3 and [ROL 99, pp. 162–163]). Arguments similar to those in the proofs of theorems 4.1 and 4.3 show that ˜m+1, m−2 , B ˜m+1, j for A˜m+1, m−1 , A˜m+1, j for 1 ≤ j ≤ m − 2, B ˜ 1 ≤ j ≤ m − 3 and Bm+1, 0 are given by [4.65], [4.66], [4.71], [4.72] and

102

Ruin Probabilities

[4.74] with n = m; A˜m+1, 0 , C˜m+1, 1 and C˜m+1, 2 are given by [4.69], [4.78], [4.79], respectively. To find the constants in [4.82] for n ≥ m + 1, we also apply arguments similar to those in the proofs of theorems 4.1 and 4.3 and take into account that ϕ(d, L) (x) is continuously differentiable on [L, +∞). Thus, A˜n+1, n−1 , A˜n+1, j for n − m + 1 ≤ j ≤ n − 2, A˜n+1, j for 1 ≤ j ≤ n − m, A˜n+1, 0 , ˜n+1, n−2 , B ˜n+1, j for n − m ≤ j ≤ n − 3, B ˜n+1, j for 1 ≤ j ≤ n − m − 1, B ˜n+1, 0 , C˜n+1, 1 and C˜n+1, 2 are given by [4.65], [4.66], [4.67], [4.70], [4.71], B [4.72], [4.73], [4.75], [4.76] and [4.77], respectively. This proves the theorem.  R EMARK 4.2.– In the assertion of theorem 4.5 we assumed that L = md, where m ≥ 3 is integer. Note that our considerations remain true if we remove the restriction that m is integer. But in this case the computations are much more tedious because we have to consider a finer partition of R+ into the intervals where ϕ(d, L) (x) is given by the same analytic expressions.    (1, 3)  (x) t≥0 E XAMPLE 4.3.– Let the surplus processes Xt (x) t≥0 and Xt follow [4.1] and [4.4], respectively, under the above assumptions with d = 1 and L = 3. Furthermore, let ϕ(x) and ϕ(1, 3) (x) be the corresponding survival probabilities, the claim sizes be exponentially distributed with mean μ and θ = 0.1. To compute ϕ(x) and ϕ(1, 3) (x), we apply formula [4.3] and theorem 4.5, respectively. The computational results in Table 4.3 show that a franchise and a liability limit can either decrease or increase the survival probability.  4.4.2. Case of small initial surpluses We now investigate how a franchise and a liability limit change the survival probability for small initial surpluses.    (d, L)  (x) t≥0 T HEOREM 4.6.– Let the surplus processes Xt (x) t≥0 and Xt follow [4.1] and [4.4], respectively, under the above assumptions with 0 < d < +∞ and L = md, where m ≥ 3 is integer. Moreover, let ϕ(x) and ϕ(d, L) (x) be the corresponding survival probabilities and the claim sizes be exponentially distributed with mean μ.

Classical Risk Model with a Franchise and a Liability Limit

i) If d − μe(d−L)/μ > 0, then ϕ(d, L) (x) < ϕ(x) for        θ d − μe(d−L)/μ μ(1 + θ) ln 1 + ,d . x ∈ 0, min θ μ(1 + θ)

103

[4.88]

ii) If d − μe(d−L)/μ < 0, then ϕ(d, L) (x) > ϕ(x) for x ∈ [0, d]. x

μ

ϕ(x)

ϕ(1, 3) (x)

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2

0.13130633 0.16990844 0.20679519 0.24204280 0.27572412 0.30790874 0.33866317 0.36805097 0.11133723 0.13130633 0.15082671 0.16990844 0.18856139 0.20679519 0.22461926 0.24204280

0.11600438 0.14802717 0.18329012 0.21776307 0.25190941 0.28571710 0.31724577 0.35039170 0.11111988 0.13582390 0.16360737 0.19261398 0.22300123 0.25483819 0.28247501 0.31371490



ϕ(1, 3) (x) ϕ(x)

 − 1 × 100%

-11.6536% -12.8783% -11.3664% -10.0312% -8.6371% -7.2072% -6.3241% -4.7981% -0.1952% 3.4405% 8.4737% 13.3634% 18.2645% 23.2322% 25.7573% 29.6113%

Table 4.3. Comparison of ϕ(x) and ϕ(1, 3) (x) for μ = 1 and μ = 2

P ROOF.– Introduce the function g˜(x) =

ϕ(x) ϕ(d, L) (x)

for x ∈ [0, d]. Substituting [4.3] and [4.80] into [4.89] yields    1 θx g˜(x) = 1 + θ − exp − θ μ(1 + θ)   x  .  × exp − (1 + θ) μ + d − μe(d−L)/μ

[4.89]

104

Ruin Probabilities

Taking the derivative gives   1 x  exp −   g˜ (x) =  θ μ + d − μe(d−L)/μ (1 + θ) μ + d − μe(d−L)/μ       μ(1 + θ) + θ d − μe(d−L)/μ θx exp − × −1 . μ(1 + θ) μ(1 + θ) Note that g˜ (x) > 0 for all x that are given by [4.88] if d − μe(d−L)/μ > 0 and g˜ (x) < 0 for x ∈ [0, d] if d − μe(d−L)/μ < 0. Moreover, g˜(0) = 1. Therefore, g˜(x) > 1 for all x that are given by [4.88] if d − μe(d−L)/μ > 0 and g˜(x) < 1 for x ∈ [0, d] if d − μe(d−L)/μ < 0, which proves the theorem. 

5 Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

In this chapter, we consider the classical risk model where an insurance company is able to adjust a franchise amount continuously. The problem of optimal control by the franchise amount is solved from the viewpoint of survival probability maximization. We derive the Hamilton–Jacobi–Bellman equation for the optimal survival probability and prove the existence of the solution to this equation with certain properties. The verification theorem gives the connection between this solution and the optimal survival probability, which differ in a constant multiplier. Then, we concentrate on the case of exponentially distributed claim sizes. Finally, we extend these results to the problem of optimal control by a deductible amount. 5.1. Introduction We deal with the problem of survival probability maximization in the classical risk model, where an insurance company is able to choose franchise and deductible amounts continuously. A franchise, as mentioned before, is a provision in an insurance policy whereby an insurer does not pay unless damage exceeds the franchise amount, whereas, a deductible is a provision whereby an insurer pays any amounts of damage that exceed the deductible amount. As a rule, these provisions are applied when an insured’s losses are

106

Ruin Probabilities

relatively small to prevent a large number of trivial claims. Moreover, a deductible encourages an insured to take more care of the insured property. Normally, a franchise and a deductible also imply reduction of insurance premiums. Thus, changes of claim and premium sizes have an influence on the survival probability of the insurance company. Our problems are maximizing the survival probability adjusting the franchise or deductible amounts. To solve these problems, we apply stochastic control theory. We assume that the net profit condition holds, i.e. c > λμ, in the classical risk model described in section 1.1.1. Moreover, let the insurance company use the expected value principle for premium calculation, which means that c = λμ(1 + θ), where θ > 0 is a safety loading. In addition, we suppose that the insurance company is able to choose either a franchise amount or a deductible amount at every time t ≥ 0. The problems of optimal control by the franchise and deductible amounts are considered in sections 5.2 and 5.3, respectively. Section 5.4 contains some bibliographical notes regarding optimal control problems in insurance. 5.2. Optimal control by the franchise amount In this section, we solve the problem of optimal control by the franchise amount from viewpoint of survival probability maximization. 5.2.1. Problem statement We assume that the insurance company adjusts the franchise amount dt at every time t ≥ 0 on the basis of the information available up to time t, i.e. every admissible strategy (dt )t≥0 of the franchise amount choice is a predictable process w.r.t. the natural filtration generated by (Nt )t≥0 and (Yi )i≥1 . Let 0 ≤ dt ≤ dmax , where dmax is the maximum allowed franchise amount, such that 0 < F (dmax ) < 1. In particular, if dt = 0, then a franchise is not used at time t. In what follows, we will denote any admissible strategy (dt )t≥0 briefly by (dt ). We suppose that the safety loading θ > 0 is constant, but insurance compensation is variable and depends on the franchise amount. Then, in contrast to the classical risk model, in this case the premium intensity

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

107

at time t also depends on the franchise amount at this time and is given by  +∞ c(dt ) = λ(1 + θ) y dF (y) dt

provided that the insurance company applies the expected value principle for premium calculation. (d )

Let Xt t (x) be the surplus of the insurance company at time t provided that its initial surplus is x and the strategy (dt ) is used. Then, the surplus  (d )  process Xt t (x) t≥0 follows (d ) Xt t (x)

 =x+

t 0

c(ds ) ds −

Nt 

Yi

{Yi >dτi } ,

t ≥ 0.

[5.1]

i=1

The ruin time under the admissible strategy (dt ) is defined by   (d ) τ (dt ) (x) = inf t ≥ 0 : Xt t (x) < 0 and the corresponding infinite-horizon survival probability is given by  ϕ(dt ) (x) = P τ (dt ) (x) = ∞ . Our aim is to maximize the survival probability over all admissible strategies (dt ), i.e. to find ϕ∗ (x) = sup ϕ(dt ) (x) (dt )

∗

and show that there is an optimal strategy (d∗t ), such that ϕ∗ (x) = ϕ(dt ) (x) for all x ≥ 0. We will later show that the optimal strategy is a function of an initial surplus only. To solve this problem, we apply approaches of [AND 08, HIP 00a, HIP 03a, HIP 03b, SCH 01, SCH 02b], which were used to solve optimal control problems by investments and/or reinsurance. In section 5.2.2, we derive the Hamilton–Jacobi–Bellman equation for ϕ∗ (x) under the additional assumption that this function is differentiable. In section 5.2.3, we prove the existence of a solution to this equation with certain properties. In section 5.2.4, we establish a connection between this solution and ϕ∗ (x), which allows us to find the optimal survival probability. Section 5.2.5 deals with the case of exponentially distributed claim sizes.

108

Ruin Probabilities

5.2.2. Hamilton–Jacobi–Bellman equation For any δ > 0 and ε > 0, consider the strategy (dt ), such that the franchise amount at time t equals

d if 0 ≤ t ≤ δ ∧ τ1 ,  (d)  dt = ˜ (x) if t > δ ∧ τ1 and δ ∧ τ1 < τ (d) (x), dt−(δ∧τ ) X 1

δ∧τ1

  where (d) is any admissible constant strategy, d˜t (x) ((d˜t ) for short) is an ˜ admissible strategy, such that ϕ(dt ) (x) > ϕ∗ (x) − ε. Note that we do not need to determine the strategy (dt ) for t > δ ∧ τ1 if δ ∧ τ1 = τ (d) (x), i.e. ruin occurs at the time τ1 , and the case δ ∧ τ1 > τ (d) (x) is impossible. For that strategy (dt ), by the law of total probability, we have   ϕ∗ (x) ≥ ϕ(dt ) (x) = e−λδ ϕ(dt ) x + c(d)δ   δ   −λs + λe F (d) ϕ(dt ) x + c(d)s 

0

d∨(x+c(d)s)

+

ϕ

(dt )



x + c(d)s − y dF (y) ds

d

  ≥ e−λδ ϕ∗ x + c(d)δ + 

d∨(x+c(d)s)

+

∗









δ 0

[5.2]



  λe−λs F (d) ϕ∗ x + c(d)s 



ϕ x + c(d)s − y dF (y) ds − ε.

d

Letting ε → 0 and doing elementary calculations in [5.2], we get   ϕ∗ x + c(d)δ − ϕ∗ (x) −λδ 1 − e−λδ ∗ e ϕ (x) − δ δ     1 δ −λs + λe F (d) ϕ∗ x + c(d)s δ 0  d∨(x+c(d)s)   ∗ + ϕ x + c(d)s − y dF (y) ds ≤ 0. d

[5.3]

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

109

Assuming that ϕ∗ (x) is differentiable on R+ and letting δ → 0 in [5.3], we obtain  +∞     (1 + θ) y dF (y) ϕ∗ (x) + F (d) − 1 ϕ∗ (x) d



d∨x

+

[5.4] ∗

ϕ (x − y) dF (y) ≤ 0.

d

Inequality [5.4] is true for all d ∈ [0, dmax ], and equality in [5.4] is attained when d is the value of some optimal strategy at the time t = 0. This yields the Hamilton–Jacobi–Bellman equation for ϕ∗ (x):   +∞   y dF (y) ϕ∗ (x) (1 + θ) sup d∈[0, dmax ]



d



∗



+ F (d) − 1 ϕ (x) +

d∨x

∗

ϕ (x − y) dF (y)

[5.5]

= 0.

d

Note that [5.5] can be rewritten as  d ∗  A ϕ (x) = 0, sup d∈[0, dmax ]

 (d )  where Ad is the infinitesimal generator of Xt t (x) t≥0 as dt = d for all t ≥ 0. Rewrite [5.5] in  a more convenient form. Since 0 < F (dmax ) < 1, expressing ϕ∗ (x) from [5.4] gives 

d∨x   ∗  1 − F (d) ϕ∗ (x) − d ϕ∗ (x − y) dF (y) ϕ (x) = . [5.6] inf

+∞ d∈[0, dmax ] (1 + θ) d y dF (y)  (d )  P ROPOSITION 5.1.– Let the surplus process Xt t (x) t≥0 follow [5.1] under the above assumptions. If ϕ∗ (x) is differentiable on R+ , then it satisfies the Hamilton–Jacobi–Bellman equation [5.5], which is equivalent to [5.6]. R EMARK 5.1.– Note that if there is one solution to [5.5] or [5.6], then there are infinitely many solutions to these equations, which differ with a multiplicative constant.

110

Ruin Probabilities

5.2.3. Existence theorem We now prove the existence of a solution to [5.6] with certain properties. T HEOREM 5.1.– If the r.v.’s Yi , i ≥ 1, have a p.d.f. f (y), then there is a solution G(x) to [5.6] with G(0) = θ/(1 + θ), which is non-decreasing and continuously differentiable on R+ and θ/(1 + θ) ≤ limx→+∞ G(x) ≤ 1.   P ROOF.– Define a sequence of functions Gn (x) n≥0 on R+ in the following

way. Let G0 (x) = ϕ(0) (x), where ϕ(0) (x) is the survival probability provided that dt = 0 for all t ≥ 0, and  

d∨x 1 − F (d) Gn−1 (x) − d Gn−1 (x − y) dF (y) ,

+∞ (1 + θ) d y dF (y)

 Gn (x) =

inf

d∈[0, dmax ]

Gn (0) = θ/(1 + θ),

n ≥ 1. [5.7]

This gives θ + Gn (x) = 1+θ



x 0

Gn (u) du,

n ≥ 1.

[5.8]

Since Yi , i ≥ 1, have a p.d.f. f (y) and E[Yi ] < ∞, the function G0 (x) is continuously differentiable on R+ and satisfies the equation

x G0 (x) − 0 G0 (x − y)f (y) dy θ  , [5.9] , G0 (0) = G0 (x) =

+∞ 1+θ (1 + θ) yf (y) dy 0

by theorem 2.1 and remark 2.1 (see also section 1.1.3 and [ROL 99, pp. 162– 163]). Since θ > 0, there is a unique solution to [5.9], which is a non-decreasing function with limx→+∞ G0 (x) = 1 (see section 1.1.5 and [ASM 10, BEA 84, BÜH 70, DIC 10, GER 79, GRA 91, ROL 99]). Equation [5.9] and equality [5.7] with n = 1 yield G1 (x) ≤ G0 (x) for all x ≥ 0. Therefore, from [5.8] we get G1 (x) ≤ G0 (x). Moreover, by

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

111

equality [5.7] with n = 1 and the properties of G0 (x), we conclude that G1 (x) is non-negative and continuous on R+ . Let Gn (x) ≤ Gn−1 (x) for all x ≥ 0, which implies Gn (x) ≤ Gn−1 (x), and Gn (x) be non-negative and continuous on R+ . Then, for all d ∈ [0, dmax ] and x ≥ 0, we have  d∨x   1 − F (d) Gn−1 (x) − Gn−1 (x − y)f (y) dy 

d

=  ≥

d∨x

Gn−1 (x)f (y) dy + 

+∞ d∨x





+∞

Gn (x)f (y) dy + 



= 1 − F (d) Gn (x) −

d

d∨x  x

d∨x  x

d d∨x

d

x−y

x−y

Gn−1 (u) du

f (y) dy



Gn (u) du

f (y) dy

Gn (x − y)f (y) dy.

This together with [5.7] yields Gn+1 (x) ≤ Gn (x) for all x ≥ 0 (therefore, Gn+1 (x) ≤ Gn (x)) and Gn+1 (x) is non-negative and continuous on R+ .   Thus, by induction, Gn (x) n≥0 is a non-increasing sequence of functions, which are non-negative and continuous on R+ . Hence, there is a pointwise limit of these functions. We denote it by g(x).   From [5.8], we deduce that Gn (x) n≥0 is a non-increasing sequence of non-decreasing functions. Furthermore, it is bounded from below by the value of these functions at x = 0, i.e. θ/(1 + θ). Hence, there is a pointwise  limit of these functions. By the Cauchy criterion, it follows that Gn (˜ x) n≥0 is a Cauchy sequence for any fixed x ˜ ≥ 0.   We now use the Cauchy criterion to show that Gn (x) n≥0 converges uniformly on any [0, x ˜], where 0 < x ˜ < +∞. For all k ≥ 1 and n ≥ 1, we have    1 − F (d)G (x) − G    n−1 (x) k−1     sup Gk (x) − Gn (x) ≤ sup 

+∞  x∈[0, x ˜] (1 + θ) d yf (y) dy x∈[0, x ˜] d∈[0, dmax ]

112

Ruin Probabilities

 

d∨x   G (x − y) − G (x − y) f (y) dy n−1 k−1  − d

+∞   (1 + θ) d yf (y) dy    1 − F (d) ≤ sup sup Gk−1 (x) − Gn−1 (x) [5.10]

+∞ yf (y) dy x∈[0, x˜] d∈[0, dmax ] (1 + θ) d   Gk−1 (˜ x) − Gn−1 (˜ x) . = μ(1 + θ) Here we used the facts that  1 − F (d) sup

+∞ yf (y) dy d∈[0, dmax ] (1 + θ) d is attained at d = 0 and

 x             Gk−1 (u) − Gn−1 (u) du sup Gk−1 (x) − Gn−1 (x) = sup 

x∈[0, x ˜]

x∈[0, x ˜]

0

  x˜        Gk−1 (u) − Gn−1 (u) du = Gk−1 (˜ x) − Gn−1 (˜ x). =  0

  By [5.10], the sequence Gn (x) n≥0 converges uniformly on any interval [0, x ˜]. Thus, g(x) is continuous on any interval [0, x ˜]. Hence, it is continuous on R+ . Let  x θ + g(u) du. [5.11] G(x) = 1+θ 0 From [5.11], we conclude that G(x) is continuously differentiable on R+ and G (x) = g(x). Since 0 ≤ g(x) ≤ G0 (x), from [5.11] we get θ/(1 + θ) ≤ limx→+∞ G(x) ≤ 1.

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

113

Furthermore,  

d∨x   1 − F (d) Gn−1 (x) − d Gn−1 (x − y) dF (y) 

+∞  inf d∈[0, dmax ] (1 + θ) d y dF (y) 

d∨x   1 − F (d) G(x) − d G(x − y) dF (y)  − inf

+∞   d∈[0, dmax ] (1 + θ) d y dF (y)    1 − F (d)G n−1 (x) − G(x)  ≤ sup 

+∞ (1 + θ) d y dF (y) d∈[0, dmax ]   

d∨x   G (x − y) − G(x − y) dF (y) n−1  d −

+∞   (1 + θ) d y dF (y)     1 − F (d)G n−1 (x) − G(x)   ≤ sup 

+∞   (1 + θ) d y dF (y) d∈[0, dmax ]   x    1 Gn−1 (u) − G (u) du ≤

+∞ μ(1 + θ) dmax y dF (y) 0   x ≤ sup Gn−1 (u) − G (u)

+∞ μ(1 + θ) dmax y dF (y) u∈[0, x] and lim

  sup Gn−1 (u) − G (u) = 0.

n→∞ u∈[0, x]

Letting n → ∞ in [5.7] shows that G(x) is a solution to [5.6], which completes the proof.  5.2.4. Verification theorem We now prove that ϕ∗ (x) coincides with

G(x) limx→+∞ G(x) .

 (d )  T HEOREM 5.2.– Let the surplus process Xt t (x) t≥0 follow [5.1] and let G(x) be the solution to [5.6] that satisfies the conditions of theorem 5.1. Then, for any x ≥ 0 and arbitrary admissible strategy (dt ), we have ϕ(dt ) (x) ≤

G(x) , limx→+∞ G(x)

[5.12]

114

Ruin Probabilities

   (d∗ ) and equality in [5.12] is attained under the strategy (d∗t ) = d∗t Xt−t (x) ,   where d∗t (x) minimizes the right-hand side of [5.6], i.e. G(x) . limx→+∞ G(x)

∗

ϕ∗ (x) = ϕ(dt ) (x) =

P ROOF.– We denote by G(x) to [5.6] that satisfies the conditions  ∗the solution  ∗ of theorem 5.1. Let (dt ) = dt (x) be the corresponding strategy and (dt ) be any admissible strategy. Moreover, we suppose that G(x) = 0 for all x < 0 and G (x) at x = 0 means the right derivative of G(x) at this point. Applying Dynkin’s formula (see, 07, pp. 03, p. 127]  and  [ØKS 11–  for  example [ØKS (d∗t ) (dt ) and G X (d∗ ) (x) 12]) to the processes G Xt∧τ (dt ) (x) (x) yields

t

(x)

t≥0

   (d ) E G Xt∧τt (dt ) (x) (x)  t∧τ(dt ) (x) = G(x) + E



t∧τ(dt ) (x) 

y dF (y) G



ds



Xs(dt ) (x)





ds

 F (ds ) − 1 G(Xs(dt ) (x))



+ 0



(dt )

ds ∨Xs

(x)



Xs(dt ) (x)

G

+ ds

  (d∗ ) E G X t (d∗ ) t∧τ

t

 − y dF (y) ds 

 (x)

(x)

 t∧τ(d∗t ) (x) = G(x) + E

 (1 + θ)

0 ∗ t∧τ (dt ) (x)



+ 

d∗s

y dF (y) G

 (d∗ ) F (d∗s ) − 1 G(Xs t (x))

(d∗ t)

d∗s ∨Xs

d∗s

+∞

 

0

+

+∞

(1 + θ) 0

and

t∧τ

t≥0

(x)

G



(d∗ ) Xs t (x)

 − y dF (y) ds . 

 



 (d∗ ) Xs t (x)

ds

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

115

Since G(x) is a solution to [5.6], it is also a solution to [5.5]. Therefore, it follows from the above that for all x ≥ 0 and t ≥ 0,       (d∗ ) (d ) [5.13] E G Xt∧τt (dt ) (x) (x) ≤ G(x) = E G X t (d∗ ) (x) . t∧τ

t

(x)

We now show that if ruin does not occur, then for any admissible strategy  (d )  (dt ), the process Xt t (x) t≥0 is unbounded from above a.s., i.e. for all x ˜ > 0,  (d ) ˜ for all t ≥ 0, τ (dt ) (x) = ∞ = 0. [5.14] P Xt t (x) ≤ x Let c be the premium intensity when the franchise amount equals 0. The probability of more than (˜ x + c)/dmax claims of size larger than dmax on any unit time interval is positive. In addition, the claim process has stationary independent increments. Hence, by the Borel–Cantelli lemma, there is a random integer value t˜ > 0, such that there are more than (˜ x + c)/dmax (d ) claims of this kind on the time interval [t˜, t˜ + 1]. If in addition Xt t (x) ≤ x ˜ ˜ for all t ∈ [0, t], then   (d ) x + c)/dmax = 0, ˜ + c − dmax (˜ Xt˜ t (x) < x i.e. τ (dt ) (x) < ∞. This yields [5.14]. We now fix any admissible strategy (dt ) and ε > 0, which is small enough. Recall that ϕ(0) (x), which was introduced at the beginning of the proof of theorem 5.1, is the survival probability provided that dt = 0 for all t ≥ 0. Let xε > x be a large enough number, such that 1 − ϕ(0) (xε ) < ε. Set   (d ) τˆxε = inf t > 0 : Xt t (x) = xε .   By [5.14], τˆxε < ∞ a.s. if τ (dt ) (x) = ∞ a.s. Define the strategy dεt in the following way:

dt if t ≤ τˆxε , ε dt = 0 if t > τˆxε . Then,  ε P τ (dt ) (x) = ∞, τ (dt ) (x) < ∞ ≤ 1 − ϕ(0) (xε ) < ε.

[5.15]

116

Ruin Probabilities

 (0) Furthermore, since P[ˆ τxε < ∞] = 1 and P limt→+∞ Xt (x) = ε  (d ) = 1, we get P limt→+∞ Xt t (x) = +∞, τ (0) (x) = ∞ ε +∞, τ (dt ) (x) = ∞ = 1.  τˆ  Applying [5.13] to the strategy dt xε gives      (dεt ) (d∗ ) E G X (dεt ) (x) ≤ G(x) = E G X t (d∗ ) (x)

t∧τ

t∧τ

t

 (x)

(x)

.

[5.16]

Letting t → ∞ in [5.16] yields   ε P τ (dt ) (x) = ∞ − ε ≤ P τ (dt ) (x) = ∞, τ (dt ) (x) = ∞ ≤

 ∗ G(x) ≤ P τ (dt ) (x) = ∞ , limx→+∞ G(x)

[5.17]

where  θ/(1  +ε θ) ≤ lim  x→+∞ G(x)  ≤ 1.∗ Here we used [5.15] and the facts (dt ) (dt ) that G X (dεt ) (x) = 0 and G X (d∗ ) (x) = 0. τ

(x)

τ

t

(x)

Since ε > 0 is arbitrary in [5.17], the proof is complete.



R EMARK 5.2.– In the proof of theorem 5.2, we used any solution to [5.6] that satisfies the conditions of theorem 5.1. However, theorem 5.2 also implies uniqueness of such a solution. The corresponding strategy (d∗t ) may not be unique in general. 5.2.5. Exponentially distributed claim sizes We now deal with the case of exponentially distributed claim sizes.  (d )  T HEOREM 5.3.– Let the surplus process Xt t (x) t≥0 follow [5.1], the claim sizes be exponentially distributed with mean μ and dmax = μ. Then, the strategy (dt ) with dt = 0 for all t ≥ 0 is not optimal. P ROOF.– If the strategy (dt ) with dt = 0 for all t ≥ 0 is optimal, then ϕ(0) (x) is the solution to [5.6] that satisfies the conditions of theorem 5.1, i.e. G1 (x) = G0 (x) for all x ≥ 0. We now show that this is not true. Indeed, note that (0)

G0 (x) = ϕ

 θx 1 exp − (x) = 1 − , 1+θ (1 + θ)μ

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

  exp − μd 1−

  θx exp − (1+θ)μ   G1 (x) = inf ⎝ d∈[0, μ] (1 + θ)(d + μ) exp − μd     ⎞

d∨x  θ(x−y) y 1 1 1 − dy exp − exp − 1+θ μ μ d (1+θ)μ ⎠.   − d (1 + θ)(d + μ) exp − μ ⎛

1 1+θ

Let x ≥ 0 be any fixed number. Consider two cases. 1) If d ≥ x (in addition, here x ∈ [0, μ]), then   ⎞  ⎛ θx θx 1 + θ − exp − (1+θ)μ 1 + θ − exp − (1+θ)μ ⎠= inf ⎝ (1 + θ)2 (d + μ) 2μ(1 + θ)2 d∈[x,μ] as d = μ. 2) If d < x, then     ⎛ θx θd exp − (1+θ)μ exp (1+θ)μ − inf ⎝ (1 + θ)(d + μ) d∈[0, μ∧x]

1 1+θ

⎞ ⎠

 θ θx = exp − μ(1 + θ)2 (1 + θ)μ as d = 0 because for all d > 0,       ⎛ ⎞ θd 1 θd θd exp (1+θ)μ − 1+θ − 1 exp (1+θ)μ + 1 ⎝ ⎠ = μ   > 0. d+μ (1 + θ)(d + μ)2 d

Notice that

  θx 1 + θ − exp − (1+θ)μ 2μ(1 + θ)2

for x ≤ x ˆ, where x ˆ= (1 + θ)μ x ˆ= θ

(1+θ)μ θ



 θ θx ≤ exp − μ(1 + θ)2 (1 + θ)μ   θ . Moreover, ln 1 + 1+θ ∞

 (−1)i θ + 1+θ i i=1



θ 1+θ

i  < μ.

117

118

Ruin Probabilities

Here we used the fact that its sum is negative.

∞

(−1)i i=1 i



θ 1+θ

i

is a Leibniz series; hence,

Therefore, we get   ⎧ θx ⎪ 1 + θ − exp − (1+θ)μ ⎪ ⎪ ⎨ if x ∈ [0, x ˆ], 2  2μ(1 + θ) G1 (x) =  ⎪ θ θx ⎪ ⎪ ⎩ if x ∈ (ˆ x, +∞). exp − 2 μ(1 + θ) (1 + θ)μ This shows that the assertion G1 (x) = G0 (x) for all x ∈ [0, x ˆ] is not true, which completes the proof.  R EMARK 5.3.– Theorem 5.3 implies that we can always increase the survival probability adjusting the franchise amount if the claim sizes are exponentially distributed. E XAMPLE 5.1.– Let the claim sizes be exponentially distributed with mean μ = 10, dmax = μ and θ = 0.1. By theorem 1.5, we have ϕ(0) (x) ≈ 1 − 0.90909090 e−x/110 , Applying theorems 5.1 and 5.2 yields

0.11104877 ex/22 ∗ ϕ (x) ≈ 1 − 0.90382792 e−x/110

d∗t (x)

=

x ≥ 0.

if if

x ≤ 8.93258, x > 8.93258,

10 if x ≤ 8.93258, 0 if x > 8.93258.

  Note that the optimal strategy d∗t (x) is simple enough in this example. Some computational results are given in Table 5.1. They show that we can considerably increase the survival probability for small initial surpluses applying the optimal strategy of the franchise amount choice. On the contrary, the difference between ϕ(0) (x) and ϕ∗ (x) is very slight for large initial surpluses. 

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

x

ϕ(0) (x)

ϕ∗ (x)

0 1 2 3 4 5 10 20 30 40 50 60 70 80 90 100 200 300 400 500

0.09090909 0.09913610 0.10728866 0.11536744 0.12337311 0.13130633 0.16990844 0.24204280 0.30790874 0.36805097 0.42296689 0.47311066 0.51889696 0.56070447 0.59887894 0.63373607 0.85243581 0.94054782 0.97604729 0.99034969

0.11104877 0.11621292 0.12161722 0.12727283 0.13319146 0.13938532 0.17471407 0.24643083 0.31191545 0.37170951 0.42630750 0.47616097 0.52168220 0.56324768 0.60120114 0.63585648 0.85329010 0.94089200 0.97618596 0.99040555



ϕ∗ (x) ϕ(0) (x)

119

 − 1 × 100%

22.1536% 17.2256% 13.3551% 10.3195% 7.9583% 6.1528% 2.8284% 1.8129% 1.3013% 0.9940% 0.7898% 0.6447% 0.5368% 0.4536% 0.3878% 0.3346% 0.1002% 0.0366% 0.0142% 0.0056%

Table 5.1. Comparison of ϕ(0) (x) and ϕ∗ (x) for different initial surpluses

5.3. Optimal control by the deductible amount In this section, we solve the problem of optimal control by the deductible amount from the viewpoint of survival probability maximization. 5.3.1. Problem statement We now assume that the insurance company adjusts the deductible amount d¯t at every time t ≥ 0 on the basis of the information available up to time t, i.e. every admissible strategy (d¯t )t≥0 of the deductible amount choice is a predictable process w.r.t. the natural filtration generated by (Nt )t≥0 and (Yi )i≥1 . Let 0 ≤ d¯t ≤ d¯max , where d¯max is the maximum allowed deductible amount, such that 0 < F (d¯max ) < 1. In particular, if d¯t = 0, then a deductible is not used at time t. In what follows, we will denote any admissible strategy (d¯t )t≥0 briefly by (d¯t ). We suppose that the safety loading θ > 0 is constant, but insurance compensation is variable and depends on the

120

Ruin Probabilities

deductible amount. Then, the premium intensity at time t also depends on the deductible amount at this time and is given by  +∞ ¯ c(dt ) = λ(1 + θ) (y − d¯t ) dF (y). d¯t

(d¯ )

Let Xt t (x) be the surplus of the insurance company at time t if its initial surplus is x and the strategy (d¯t ) is used. Then, the surplus process  (d¯t )  Xt (x) t≥0 follows (d¯t )

Xt

 (x) = x +

t 0

c(d¯s ) ds −

Nt 

(Yi − d¯τi )+ .

[5.18]

i=1

The ruin time under the admissible strategy (d¯t ) is defined by   ¯ (d¯ ) τ (dt ) (x) = inf t ≥ 0 : Xt t (x) < 0 and the corresponding infinite-horizon survival probability is given by  ¯ ¯ ϕ(dt ) (x) = P τ (dt ) (x) = ∞ . Our aim is to maximize the survival probability over all admissible strategies (d¯t ), i.e. to find ¯

ϕ∗ (x) = sup ϕ(dt ) (x) (d¯t )

¯∗ and show that there is an optimal strategy (d¯∗t ), such that ϕ∗ (x) = ϕ(dt ) (x) for all x ≥ 0.

To solve this problem, we apply techniques similar to those in section 5.2. In section 5.3.2, we derive the Hamilton–Jacobi–Bellman equation for ϕ∗ (x) under the additional assumption that this function is differentiable. In section 5.3.3, we formulate the existence and verification theorems. Section 5.3.4 is devoted to the case of exponentially distributed claim sizes.

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

121

5.3.2. Hamilton–Jacobi–Bellman equation For any δ > 0 consider the strategy (d¯t ), such that the deductible amount at time t equals

d¯ if 0 ≤ t ≤ δ ∧ τ1 , d¯t = ˜  (d)  ¯ ¯ ¯ dt−(δ∧τ1 ) Xδ∧τ1 (x) if t > δ ∧ τ1 and δ ∧ τ1 < τ (d) (x), ¯ is an arbitrary admissible constant strategy, where (d) ˜ ¯

 ˜ d¯t (x) ((d˜¯t ) for

brevity) is an admissible strategy, such that ϕ(dt ) (x) > ϕ∗ (x) − ε. Here ε > 0 is any fixed number. Applying the law of total probability gives  ¯ ¯  ¯ ϕ∗ (x) ≥ ϕ(dt ) (x) = e−λδ ϕ(dt ) x + c(d)δ   δ   −λs ¯ ¯ ϕ(d¯t ) x + c(d)s + λe F (d)  + ≥e

0

¯ x+c(d)s+ d¯



∗





 ¯ ¯ x + c(d)s + d − y dF (y) ds

 ¯ + ϕ x + c(d)δ

−λδ

+

ϕ

d¯

(d¯t )

¯ x+c(d)s+ d¯ d¯



δ

λe 0

−λs



  ¯ ¯ ϕ∗ x + c(d)s F (d)

 ¯ ¯ ϕ x + c(d)s + d − y dF (y) ds − ε. ∗



Letting ε → 0 and doing elementary calculations in [5.19] yield   ¯ − ϕ∗ (x) ϕ∗ x + c(d)δ 1 − e−λδ ∗ e−λδ − ϕ (x) δ δ     1 δ −λs ¯ ¯ ϕ∗ x + c(d)s + λe F (d) δ 0  x+c(d)s+ ¯ d¯   ∗ ¯ ¯ + ϕ x + c(d)s + d − y dF (y) ds ≤ 0. d¯

[5.19]

[5.20]

122

Ruin Probabilities

Assuming that ϕ∗ (x) is differentiable on R+ and letting δ → 0 in [5.20], we get  +∞     ¯ dF (y) ϕ∗ (x)  + F (d) ¯ − 1 ϕ∗ (x) (y − d) (1 + θ)  +

d¯

x+d¯

d¯

[5.21] ϕ∗ (x + d¯ − y) dF (y) ≤ 0.

Inequality [5.21] is true for all d¯ ∈ [0, d¯max ] and equality in [5.21] is attained when d¯ is the value of some optimal strategy at the time t = 0. Thus, we have the Hamilton–Jacobi–Bellman equation   +∞   ¯ dF (y) ϕ∗ (x)  sup (y − d) (1 + θ) d¯

¯ d∈[0, d¯max ]





¯ − 1 ϕ∗ (x) + + F (d)



x+d¯ d¯

ϕ∗ (x + d¯ − y) dF (y)

[5.22]

= 0.

To rewrite [5.22] in a more convenient form, we express from [5.21] and get  ∗  ϕ (x) =



ϕ∗ (x)



 inf

¯ d∈[0, d¯max ]



¯ ϕ∗ (x) − ¯x+d¯ ϕ∗ (x + d¯ − y) dF (y) 1 − F (d) d .

+∞ ¯ dF (y) (1 + θ) d¯ (y − d) [5.23]

 (d¯ )  P ROPOSITION 5.2.– Let the surplus process Xt t (x) t≥0 follow [5.18] under the above assumptions. If ϕ∗ (x) is differentiable on R+ , then it satisfies the Hamilton–Jacobi–Bellman equation [5.22], which is equivalent to [5.23]. 5.3.3. Existence and verification theorems We now formulate the existence and verification theorems. T HEOREM 5.4.– If the r.v.’s Yi , i ≥ 1, have a p.d.f. f (y), then there is a solution G(x) to [5.23] with G(0) = θ/(1 + θ), which is non-decreasing and continuously differentiable on R+ and θ/(1 + θ) ≤ limx→+∞ G(x) ≤ 1.

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

123

 (d¯ )  T HEOREM 5.5.– Let the surplus process Xt t (x) t≥0 follow [5.18] and let G(x) be the solution to [5.23] that satisfies the conditions of theorem 5.4. Then, for any x ≥ 0 and arbitrary admissible strategy (d¯t ), we have ¯

ϕ(dt ) (x) ≤

G(x) , limx→+∞ G(x)

[5.24]

  ¯∗  (d ) and equality in [5.24] is attained under the strategy (d¯∗t ) = d¯∗t Xt−t (x) ,   where d¯∗ (x) minimizes the right-hand side of [5.23], i.e. t

¯∗

ϕ∗ (x) = ϕ(dt ) (x) =

G(x) . limx→+∞ G(x)

The proofs of these theorems are similar to the corresponding ones in section 5.2. We only note that the solution G(x) to [5.23] that satisfies the conditions   of theorem 5.4 can be found as the limit of the sequence Gn (x) n≥0 on R+ , where G0 (x) = ϕ(0) (x) is the survival probability provided that d¯t = 0 for all t ≥ 0 and   

¯ Gn−1 (x) − x+d¯ Gn−1 (x + d¯ − y) dF (y) 1 − F ( d) d Gn (x) = , inf

+∞ ¯ dF (y) ¯ d∈[0, d¯max ] (1 + θ) ¯ (y − d) d

Gn (0) = θ/(1 + θ),

n ≥ 1.

5.3.4. Exponentially distributed claim sizes  (d¯ )  T HEOREM 5.6.– Let the surplus process Xt t (x) t≥0 follow [5.18] and the ¯

claim sizes be exponentially distributed. Then ϕ∗ (x) = ϕ(dt ) (x) for every admissible strategy (d¯t ), i.e. every admissible strategy is optimal. P ROOF.– If the claim sizes are exponentially distributed with mean μ, then G0 (x) is given in the proof of theorem 5.3 and     ⎛ ¯ 1 θx exp − μd 1 − 1+θ exp − (1+θ)μ ⎝   G1 (x) = inf 

+∞  ¯ d∈[0, d¯max ] y − d¯ μ1 exp − μy dy (1 + θ) d¯     ⎞

x+d¯  ¯ θ(x+d−y) y 1 1 1 − dy exp − exp − ¯ 1+θ μ μ d (1+θ)μ ⎠   − 1

+∞  y y − d¯ μ exp − μ dy (1 + θ) d¯

124

Ruin Probabilities

    ¯ 1 θx exp − μd 1 − 1+θ exp − (1+θ)μ ⎝      = inf ¯ ¯ ¯ d∈[0, d¯max ] (1 + θ) (d¯ + μ) exp − μd − d¯ exp − μd     ⎞ ¯ 1 θx 1 − 1+θ exp − (1+θ)μ exp − μd   ⎠    − ¯ ¯ (1 + θ) (d¯ + μ) exp − μd − d¯ exp − μd  ⎞   ⎛ θ θx d¯ exp − exp − μ (1+θ)μ ⎠ ⎝ 1+θ   = inf ¯ ¯ d∈[0, d¯max ] (1 + θ)μ exp − μd  θ θx = exp − = G0 (x), (1 + θ)2 μ (1 + θ)μ ⎛

where the infimum is attained at any d¯ ∈ [0, d¯max ]. This yields G1 (x) = G0 (x). Hence, G(x) = ϕ(0) (x), which is our assertion.  R EMARK 5.4.– Theorem 5.6 implies that we cannot increase the survival probability adjusting the deductible amount for exponentially distributed claim sizes. 5.4. Bibliographical notes The problems of minimization of the ruin probability, which is equivalent to maximization of the survival probability, were mainly considered when investment and/or reinsurance are available. In the problems of optimal control by investments the price of a risky asset usually follows a geometric Brownian motion. The problem of optimal control by investments was studied by Browne [BRO 95] in the diffusion setting. He focused on obtaining investment strategies which are optimal in the sense of minimizing the ruin probability. The problem of optimal control by investments in the classical risk model was solved by Hipp and Plum [HIP 00a] without borrowing constraints and by Azcue and Muler [AZC 09] under borrowing constraints. Their aim was also to find the investment strategies which minimize the ruin probability. Liu and Yang [LIU 04] generalized results of [HIP 00a] by including a risk-free asset and also dealt with the cases where borrowing constraints or reinsurance are present.

Optimal Control by the Franchise and Deductible Amounts in the Classical Risk Model

125

In the classical risk model, Hipp [HIP 04a] obtained the asymptotics of the ruin probability under the optimal investment strategy in the case of light-tailed distribution of the claim sizes. For sub-exponentially distributed claim sizes, the asymptotics of the ruin probability under the optimal investment strategy was obtained by Schmidli [SCH 05]. Some results concerning the asymptotics of the ruin probabilities in optimal investment problems were obtained by Gaier and Grandits [GAI 02], Gaier, Grandits and Schachermayer [GAI 03] and Grandits [GRA 04a]; see also [GAI 04, GRA 05] for the case where a risk-free asset is available. Optimal investment problems in the risk models where the premium intensity depends on the current surplus were solved in [AND 08, HIP 03a]. Optimal investment strategies for an insurer with state-dependent constraints were studied by Edalati and Hipp [EDA 13]. The problem of optimal proportional reinsurance was solved by Schmidli [SCH 01] in a diffusion setup and the classical risk model in the sense of minimizing the ruin probability. The problem of optimal control by excess of loss reinsurance in the classical risk model was investigated by Hipp and Vogt [HIP 03b]. Asymptotics of the ruin probabilities under optimal reinsurance policies were found by Schmidli [SCH 02a, SCH 04b, SCH 08a]. Hald and Schmidli [HAL 04] and Waters [WAT 79, WAT 83] considered the problem of minimizing the ruin probability in an asymptotic sense and maximized the adjustment coefficient. The problem of minimizing the ruin probability when both investment and reinsurance are available was considered in [HIP 10, LUO 08, PRO 05, SCH 02b, SCH 04a, TAK 03]. Luo and Taksar [LUO 11] minimized the probability of absolute ruin, i.e. the probability that the inferior limit of the surplus process is −∞, by investment and reinsurance in a diffusion setting. Bai and Guo [BAI 08] solved the problem of minimizing the ruin probability when the basic claim process follows a Brownian motion with drift and the insurer is allowed to invest in a risk-free asset and several risky assets and to purchase proportional reinsurance. Hipp and Taksar [HIP 00b] considered the possibility of investing in new business to minimize the probability of technical ruin. A systematic description of optimal control methods applied to insurance problems was provided by Schmidli [SCH 08b]; see also [HIP 04b] for a

126

Ruin Probabilities

short survey in this direction. Azcue and Muler [AZC 14] applied the viscosity approach to control problems in insurance. Note that optimality does not necessarily imply minimizing the ruin probability. One of the most widely distributed approaches is the maximization of dividends (see for example [AZC 14, SCH 08b] and references therein). For general methods and approaches in continuous-time stochastic control theory, we refer the reader, for instance, to [DAV 93, FLE 06, ØKS 07]. For more information on various benefits that can be provided by insurance policies, see, for example [PIT 14] and references therein.

6 Risk Models with Investments in Risk-Free and Risky Assets

In this chapter, we deal with generalizations of the classical risk model and the risk model with stochastic premiums where an insurance company invests all surplus in risk-free and risky assets proportionally. The price of the risky asset follows a jump process. We get upper and lower bounds for the infinite-horizon survival probability and investigate the continuity and differentiability of the infinite- and finite-horizon survival probabilities in the generalization of the classical risk model. Moreover, we extend these results to the generalization of the risk model with stochastic premiums. Finally, we obtain relations connecting accuracy and reliability of uniform approximations of the survival probabilities by their statistical estimates. 6.1. Description of the models We consider generalizations of the classical risk model and the risk model with stochastic premiums described in sections 1.1.1 and 1.2.1, respectively: an insurance company invests all surplus in risk-free and risky assets proportionally. The price of the risk-free asset equals Bt = B0 ert at time t, where B0 is the price of the asset at the time t = 0 and r > 0 is a risk-free N˜t ˜  Yi interest rate. The price of the risky asset equals St = S0 exp r˜t + i=1 at time t, where S0 is the price of the asset at the time t = 0, r˜ > 0 is a constant, (Y˜i )i≥1 is a sequence of i.i.d. r.v.’s with c.d.f. F˜ (y) = P[Y˜i ≤ y], ˜t )t≥0 is a Poisson process with such that P[Y˜i > 0] > 0 and P[Y˜i < 0] > 0, (N ˜ ˜t )t≥0 constant intensity λ > 0. We denote by τ˜i the time of the ith jump of (N

128

Ruin Probabilities

 ˜t = 0. All the r.v.’s and processes in the models and set 0i=1 Y˜i = 0 if N considered are mutually independent. We suppose that all surplus is invested in the assets proportionally at any time t ≥ 0: a part α is invested in the risky asset and a part 1 − α is invested in the risk-free asset, where 0 < α ≤ 1. The case α = 0 is considered in Chapters 2 and 3. Set rˆ = α˜ r + (1 − α)r.   Then, the surplus process Xt (x) t≥0 follows either the equation  Xt (x) = x +

−

Nt 

t



rˆXs (x) + c ds + α

0

˜t N 

 ˜  Xτ˜i− (x) eYi − 1

i=1

Yi ,

[6.1]

t ≥ 0,

i=1

if we deal with the classical risk model, or the equation  Xt (x) = x + rˆ

+

¯t N  i=1

t 0

Xs (x) ds + α

Y¯i −

˜t N 

 ˜  Xτ˜i− (x) eYi − 1

i=1 Nt 

Yi ,

[6.2]

t ≥ 0,

i=1

if we deal with the risk model with stochastic premiums. The rest of the chapter is organized as follows. In section 6.2, we obtain upper and lower power bounds for the infinite-horizon survival probability and investigate the continuity and differentiability of the infinite- and finite-horizon survival probabilities in the generalization of the classical risk model described above. In section 6.3, we extend these results to the generalization of the risk model with stochastic premiums. Section 6.4 is devoted to derivation of relations connecting accuracy and reliability of uniform approximations of the survival probabilities by their statistical estimates. In section 6.5, we give some bibliographical notes regarding different risk models with investments in a risky asset.

Risk Models with Investments in Risk-Free and Risky Assets

129

6.2. Classical risk model with investments in risk-free and risky assets In section 6.2.1, we obtain upper and lower bounds for the infinite-horizon survival probability. In sections 6.2.2 and 6.2.3, we investigate the continuity and differentiability of the infinite- and finite-horizon survival probabilities, respectively. 6.2.1. Upper and lower bounds for the infinite-horizon survival probability We now obtain upper and lower power bounds for the infinite-horizon survival probability using some techniques of [FRO 02, KAL 02]. To this end, we introduce the discrete-time process Xnx = Xτn (x), n ≥ 0, where X0x = x. Since ruin can occur only when a claim arrives, we conclude that

ϕ(x) = P Xnx ≥ 0, n ≥ 1 . For n ≥ 1, we define ⎧ ⎨min{˜ τi : τ˜i > τn−1 } τ˜1, n = i≥1 ⎩ τn

τ˜2, n =

if

min{˜ τi : τ˜i > τn−1 } < τn ,

if

min{˜ τi : τ˜i > τn−1 } ≥ τn ,

i≥1

i≥1

⎧ ⎨max{˜ τi : τ˜i ≤ τn }

if

max{˜ τi : τ˜i ≤ τn } > τn−1 ,

⎩ τn

if

max{˜ τi : τ˜i ≤ τn } ≤ τn−1 .

i≥1

i≥1

i≥1

From [6.1], it follows that x + P2, n − Yn , Xnx = P1, n Xn−1

n ≥ 1,

[6.3]

where P1, n = erˆ(τn −τn−1 )

 i : τn−1 0 and P[Y˜i > 0] > 0, i ≥ 1, from [6.4] we deduce that 0 < P[P1, n ≤ β3 ] < 1,

n ≥ 1,

for any β3 ∈ (0, 1). Consequently, there are constants β3 ∈ (0, 1), β4 > 0 and p˜1 ∈ (0, 1), such that P[P1, n ≤ β3 , P2, n − Yn ≤ β4 ] = p˜1 ,

n ≥ 1.

[6.8]

Risk Models with Investments in Risk-Free and Risky Assets

131

Furthermore, there are constants β5 > 0 and p˜2 ∈ (0, 1), such that P[P1, n ≤ β5 , P2, n − Yn ≤ −2β4 β5 /(1 − β3 )] = p˜2 ,

n ≥ 1.

[6.9]

Here we used the facts that P1, n and P2, n , n ≥ 1, can be small enough positive and we can choose small enough positive β5 in [6.9]. For x ≥ β4 (1 − β3 ), we define     ln β4 (1 − β3 ) − ln x . κ(x) = 1 + ln β3 Next, we introduce the following events: κ(x)

E1 =



P1, n ≤ β3 , P2, n − Yn ≤ β4



[6.10]

n=1

and   E2 = P1, κ(x)+1 ≤ β5 , P2, κ(x)+1 − Yκ(x)+1 ≤ −2β4 β5 /(1 − β3 ) . [6.11]   Note that the events P1, n ≤ β3 , P2, n − Yn ≤ β4 , 1 ≤ n ≤ κ(x), and E2 are mutually independent. Next, if E1 occurs, then by [6.6] and [6.10], we have x Xκ(x)

≤

κ(x) β3 x

κ(x)

+ β4



n=1

κ(x)−n

β3

κ(x)

< β3

x + β4 /(1 − β3 )

a.s.

[6.12]

A simple computation shows that κ(x) β3 x

ln(β4 (1−β3 ))−ln x ln β3

≤ x β3

    ln β4 (1 − β3 ) − ln x = exp ln β3 ln β3 

= β4 (1 − β3 ) < β4 /(1 − β3 ). [6.13]

132

Ruin Probabilities

Moreover, if E1 ∩ E2 occurs, then by [6.3] and [6.11]–[6.13], we get x x ≤ β5 Xκ(x) − 2β4 β5 /(1 − β3 ) Xκ(x)+1  κ(x)  < β5 β3 x + β4 /(1 − β3 ) − 2β4 /(1 − β3 )  κ(x)  = β5 β3 x − β4 /(1 − β3 ) < 0 a.s.,

[6.14]

which implies ruin. Hence, from [6.8]–[6.11] and [6.14], we obtain κ(x)

ϕ(x) ≤ 1 − P[E1 ∩ E2 ] = 1 − p˜2 p˜1 ≤1− 1+

where β1 = p˜2 p˜1

1+ p˜2 p˜1

ln(β4 (1−β3 ))−ln x ln β3

ln(β4 (1−β3 )) ln β3

= 1 − β1 x−β2 ,

and β2 = ln p˜2 / ln β3 . The proof is complete. 

R EMARK 6.1.– Bound [6.7] implies that if the current surplus is invested in the risk-free and risky assets proportionally, then the survival probability grows no faster than a power function when initial surpluses are large enough. Hence, such investments may decrease the survival probability in comparison to the case without any investments. In particular, exponential asymptotics for the corresponding ruin probability, which holds in the classical risk model without any investments, is not true in this case. To formulate and prove the next theorem, we define the sets Z and Z ∗ as follows:   ˜ −z

< ∞, Z = z : E 1 − α + αeY1  

   ˜ E 1 − α + αeY˜1 −z − 1 > 0 λ + rˆz − λ and

   ˜ −z

< ∞, 0 < z ≤ 1 . Z ∗ = z : E 1 − α + αeY1

Note that Z is always non-empty since it contains 0.

Risk Models with Investments in Risk-Free and Risky Assets

133

  T HEOREM 6.2.– Let the surplus process Xt (x) t≥0 follow [6.1] under the above assumptions. If there is z0 ∈ Z ∗ , such that 

   ˜ E 1 − α + αeY˜1 −z0 − 1 − rˆz0 < 0, [6.15] λ then there is a constant β0 > 0, such that ϕ(x) ≥ 1 − β0 x−z0

[6.16]

for all x > 0. P ROOF.– Let z ∈ Z ∗ . A simple computation shows that       −z −z

Y˜i rˆτ1 1 − α + αe E P1, 1 = E e i : τ˜i ≤τ1



+∞

=E

λe

0



+∞

=

 

−(λ+ˆ r z)s

1 − α + αe

 Y˜i −z

 ds

i : τ˜i ≤s

λe

−(λ+ˆ r z)s

0

 ∞  ˜ i −λs ˜    i  (λs) e Y˜1 −z E 1 − α + αe ds i! i=0



+∞

=λ

    ˜ −z

˜ + λE ˜ s ds exp −λ − rˆz − λ 1 − α + αeY1

0

=

λ 

.   ˜ E 1 − α + αeY˜1 −z − 1 λ + rˆz − λ



Obviously, E P1,−z1 is well-defined on Z. Since z0 ∈ Z ∗ and [6.15] holds,

from the above we conclude that z0 ∈ Z and 0 < E P1,−z10 < 1. Furthermore,

since E Y1 < ∞ and 0 < z0 ≤ 1, we see that E Y1z0 < ∞. By [6.6] and the fact that P2, i , i ≥ 1, are non-negative, we have Xnx ≥ Pn∗ x − Pn∗

∞  Yi Pi∗

a.s.,

n ≥ 1.

i=1

Therefore, we get   ∞ Yi ϕ(x) ≥ 1 − P >x . Pi∗ i=1

[6.17]

134

Ruin Probabilities

Since P1, i , i ≥ 1, are i.i.d. r.v.’s, Yi , i ≥ 1, are also i.i.d. r.v.’s, which are independent of P1, i , i ≥ 1, and 0 < z0 ≤ 1, applying Chebyshev’s inequality yields  z0     ∞ ∞  Yi Yi z0 E  E  ∗ ∞ P Pi∗ Yi i=1 i i=1 >x ≤ ≤ P P∗ x z0 xz 0 [6.18] i=1 i −z0 z0

z0 ∞  E P1, 1 E Y1 E Y1  −z0 i

 = β0 x−z0 , E P1, 1 = = z  z 0 0 1 − E P −z0 x x 1, 1 i=1 where



E P1,−z10 E Y1z0

. β0 = 1 − E P1,−z10

Finally, [6.17] and [6.18] give [6.16], which is the desired conclusion. R EMARK 6.2.– Theorem 6.2 gives a sufficient condition limx→+∞ ϕ (x) = 1. We will use this condition in theorem 6.3 below.

 for

We now consider two examples, which show that the condition of theorem 6.2 can be fulfilled, i.e. there is z0 ∈ Z ∗ , such that [6.15] holds. E XAMPLE 6.1.– Let α = 1 and Y˜i , i ≥ 1, be normally distributed with parameters 0 and σ ˜ 2 . For all z ∈ R, we have  2 2 ˜ −z

= eσ˜ z /2 < ∞, E 1 − α + αeY1   which gives Z ∗ = z : 0 < z ≤ 1 . Next, for all z ∈ R, we get 

     ˜ eσ˜ 2 z 2 /2 − 1 − r˜z. ˜ E 1 − α + αeY˜1 −z − 1 − rˆz = λ λ Moreover, for small enough z > 0, we have ˜σ2z2   ˜ eσ˜ 2 z 2 /2 − 1 − r˜z ∼ λ˜ − r˜z, λ 2 which is negative. Thus, there is z0 ∈ Z ∗ , such that [6.15] holds.



Risk Models with Investments in Risk-Free and Risky Assets

135

E XAMPLE 6.2.– Let α = 1 and Y˜i , i ≥ 1, have the Laplace distribution with −|y|/μ ˜ p.d.f. f˜(y) = e 2˜μ , where y ∈ R and μ ˜ > 0. For all |z| < 1/˜ μ, we have  ˜ −z

= E 1 − α + αeY1

1 < ∞, 1−μ ˜2 z 2

  which gives Z ∗ = z : − 1/˜ μ < z < 1/˜ μ, 0 < z ≤ 1 . For all |z| < 1/˜ μ, we obtain ˜ μ2 z 2 

   ˜ E 1 − α + αeY˜1 −z − 1 − rˆz = λ˜ λ − r˜z. 1−μ ˜2 z 2 It is easy to check that   ˜ μ2 z 2 − r˜z 1 − μ λ˜ ˜2 z 2 < 0 if either z
0; rˆx + c

[6.19]

iii) the function ϕ(x) satisfies the integro-differential equation  x ˜ (ˆ rx + c)ϕ+ (x) = (λ + λ)ϕ(x) −λ ϕ(x − y) dF (y) 0

 ˜ −λ

+∞ −∞

   ϕ (1 − α) + αey x dF˜ (y)

[6.20]

on R+ with the boundary condition limx→+∞ ϕ(x) = 1; iv) for all x∗ > 0, we have ˜    ϕ+ (x) ≤ λ + λ . rˆx∗ + c x∈[x∗ ,+∞) sup

[6.21]

P ROOF.– It follows immediately from [6.1] that Xt (x) = (x + c/ˆ r) erˆt − c/ˆ r up to the time τ1 ∧ τ˜1 . Let τ1 ∧ τ˜1 = s. Moreover, let Y1 = y if τ1 ∧ τ˜1 = τ1 and Y˜1 = y if τ1 ∧ τ˜1 = τ˜1 . It is clear that ruin does not occur up to the time τ1 ∧ τ˜1 and it will not take place after that time if and only if one of the following two conditions holds: – τ1 ∧ τ˜1 = τ1 , y ≤ (x + c/ˆ r) erˆt − c/ˆ r and ruin does not occur on the time interval (s, +∞) with the initial surplus (x + c/ˆ r) erˆt − c/ˆ r − y; – τ1 ∧ τ˜1 = τ˜1and ruin does not time interval (s, +∞) with  occur on the  y r ˆ t r) e − c/ˆ the initial surplus (1 − α) + αe (x + c/ˆ r .

Risk Models with Investments in Risk-Free and Risky Assets

137



˜ P τ1 ∧ τ˜1 = τ˜1 = λ/(λ ˜ ˜ and + λ) Since P τ1 ∧ τ˜1 = τ1 = λ/(λ + λ), ˜ by the law of total τ1 ∧ τ˜1 is exponentially distributed with mean 1/(λ + λ), probability, for all x ≥ 0, we have  +∞ ˜ e−(λ+λ)s ϕ(x) = 0

  × λ  ˜ +λ

(x+c/ˆ r) erˆs −c/ˆ r 0

+∞ −∞



ϕ

  ϕ (x + c/ˆ r) erˆs − c/ˆ r − y dF (y)

(1 − α) + αe

y

[6.22]



   rˆs ˜ × (x + c/ˆ r) e − c/ˆ r dF (y) ds. r = u in the outer integral in the Changing the variable (x + c/ˆ r) erˆs − c/ˆ right-hand side of [6.22], we get ˜

ϕ(x) = (ˆ rx + c)(λ+λ)/ˆr +∞ 

 ×

x



˜ +λ

+∞ −∞

1 rˆu + c

(λ+λ)/ˆ ˜ r+1   λ

u 0

ϕ(u − y) dF (y)

[6.23]

   ˜ ϕ (1 − α) + αe u dF (y) du. 

y

Next, the inner integrals in the right-hand side of [6.23] are non-decreasing and bounded functions of u. Hence, the integrand in the outer integral is integrable on any [x, +∞), where x ≥ 0. Thus, ϕ(x) is continuous on R+ and the first assertion of the theorem is proved. By Lebesgue’s decomposition theorem, we have F (y) = F c (y) + F d (y), where F c (y) and F d (y) are continuous and discrete components of F (y), respectively. Denote by (yk )k≥1 a sequence of all points of discontinuity of F (y), if any. Set pk = F (yk ) − limy↑yk F (y). Then, we can rewrite [6.23] as ϕ(x) = ϕ1 (x) + ϕ2 (x),

[6.24]

138

Ruin Probabilities

where ˜

rx + c)(λ+λ)/ˆr ϕ1 (x) = (ˆ +∞ 

 ×

x

1 rˆu + c

(λ+λ)/ˆ ˜ r+1



 ˜ I(u) ˜ λI(u) + λ du,

[6.25]

˜

ϕ2 (x) = λ(ˆ rx + c)(λ+λ)/ˆr +∞ 

 × 

u

I(u) =  ˜ I(u) =

x

1 rˆu + c

(λ+λ)/ˆ ˜ r+1  

 pk ϕ(u − yk ) du,

[6.26]

k : yk ≤u

ϕ(u − y) dF c (y),

0 +∞ −∞

   ϕ (1 − α) + αey u dF˜ (y).

We showed the continuity of I(u) in the proof of theorem 2.1. Next, for all u ≥ 0, u0 ≥ 0 and M1 < M2 , we have   ˜ − I(u ˜ 0 ) I(u)  +∞         y y ≤ u − ϕ (1 − α) + αe u0  dF˜ (y) ϕ (1 − α) + αe  −∞

  ≤ F˜ (M1 ) + 1 − F˜ (M2 )         + sup ϕ (1 − α) + αey u − ϕ (1 − α) + αey u0 . y∈[M1 , M2 ]

[6.27]

Choosing M1 < 0 and M2 > 0 with large enough absolute values, we can first make F˜ (M1 ) and 1 − F˜ (M2 ) small enough, and then we can make the third summand in [6.27] small enough when u is close to u0 . Hence, the ˜ follows on R+ . continuity of I(u) Thus, the integrand in the right-hand side of [6.25] is continuous on R+ w.r.t. u. By [6.25], theorem A.1 and remark A.1, ϕ1 (x) is differentiable on R+ and ˜ ˜ I(x) ˜ λ+λ λI(x) + λ ϕ1 (x) − . [6.28] ϕ1 (x) = rˆx + c rˆx + c

Risk Models with Investments in Risk-Free and Risky Assets

139

Note that the integrand in the right-hand side of [6.26] is right-continuous on R+ w.r.t.u. Consequently, there is a right derivative of ϕ2 (x) on R+ (we  denote it by ϕ2 (x) + ) and 

ϕ2 (x)

 +

=

˜ λ+λ λ ϕ2 (x) − rˆx + c rˆx + c



pk ϕ(x − yk ).

[6.29]

k : yk ≤x

Since the Riemann integral does not depend on values of the integrand at a countable number of points, we can rewrite [6.26] as ˜

ϕ2 (x) = λ(ˆ rx + c)(λ+λ)/ˆr +∞ 

 ×

x

1 rˆu + c

(λ+λ)/ˆ ˜ r +1  

 pk ϕ(u − yk ) du.

[6.30]

k : yk 0. Thus, from [6.32], we conclude that ϕ2 (x) is differentiable on R+ , except at positive points yk , k ≥ 1. Therefore, by [6.24], ϕ(x) is also differentiable

140

Ruin Probabilities

on R+ , except at positive points of discontinuity of F (y), and [6.19] holds. Moreover, from [6.24], [6.28] and [6.29], we obtain ϕ+ (x) =

˜   λ+λ ϕ1 (x) + ϕ2 (x) rˆx + c    ˜ λ λ ˜ I(x) − p ϕ(x − y ) − I(x) + k k rˆx + c rˆx + c k : yk ≤x

 x ˜ λ λ+λ ϕ(x) − = ϕ(x − y) dF (y) rˆx + c rˆx + c 0  +∞ ˜    λ ϕ (1 − α) + αey x dF˜ (y), − rˆx + c −∞ which yields [6.20]. Since the right-hand side of [6.20] is continuous on R+ , except at positive points of discontinuity of F (y), we conclude that ϕ+ (x) is also continuous on R+ , except at these points. Bound [6.21] follows immediately from [6.20], which completes the proof.  6.2.3. Continuity of the finite-horizon survival probability and existence of its partial derivatives   T HEOREM 6.4.– Let the surplus process Xt (x) t≥0 follow [6.1] under the above assumptions. 1) If Yi , i ≥ 1, have a p.d.f. f (y), which is continuous on R+ , and Y˜i , i ≥ 1, have a p.d.f. f˜(y), which is continuous on R, then ϕ(x, t) is continuous on (0, +∞) × [0, +∞) as a function of two variables. 2) If the p.d.f. f (y) of Yi , i ≥ 1, has the derivative f  (y) on R+ , such that is integrable and bounded on R+ , and the p.d.f. f˜(y) of Y˜i , i ≥ 1, have the derivative f˜ (y) on R, such that |f˜ (y)|, f˜(y) e−y and |f˜ (y)| e−y are integrable and bounded on R, then: |f  (y)|

i) ϕ(x, t) has partial derivatives w.r.t. x and t on (0, +∞) × [0, +∞), which are continuous as functions of two variables;

Risk Models with Investments in Risk-Free and Risky Assets

141

ii) ϕ(x, t) satisfies the partial integro-differential equation ∂ϕ(x, t) ∂ϕ(x, t) − (ˆ rx + c) ∂t ∂x  x ˜ + (λ + λ) ϕ(x, t) − λ ϕ(x − y, t) dF (y) 0

 ˜ −λ

+∞ −∞

[6.33]

   ϕ (1 − α) + αey x, t dF˜ (y) = 0

on (0, +∞) × [0, +∞) with the boundary conditions ϕ(x, 0) = 1 and limx→+∞ ϕ(x, t) = 1; iii) for all x∗ > 0 and T > 0, we have      ∂ϕ(x, t)   ≤ C 1 C2 + C3 , sup  ∂x  x∗ x∈[x∗ ,+∞),

[6.34]

t∈[0,T ]

where

⎧ ˜ ⎪ ⎨ 1 − e(ˆr−λ−λ)T ˜ − rˆ C1 = λ+λ ⎪ ⎩T 



C2 = λ f (0) + C3 =

+∞

if

˜ = rˆ, λ+λ

if

˜ = rˆ, λ+λ





|f (y)| dy ,

0

  +∞  +∞ ˜   λ f˜(y) e−y dy + |f˜ (y)| (1−α) e−y +α dy . (1−α) α −∞ −∞

P ROOF.– Arguments similar to those in the proofs of theorems 2.2, 3.2 and 6.3 show that   (x+c/ˆr) erˆs −c/ˆr  t  ˜ −(λ+λ)s e ϕ (x + c/ˆ r) erˆs λ ϕ(x, t) = 0

0

 ˜ − c/ˆ r − y, t − s dF (y) + λ



+∞ −∞

  (1 − α) + αey

ϕ

    ˜ rˆs ˜ r , t − s dF (y) ds + e−(λ+λ)t × (x + c/ˆ r) e − c/ˆ for all x ≥ 0 and t ≥ 0.

[6.35]

142

Ruin Probabilities

Changing the variable t − s = v in the outer integral on the right-hand side of [6.35], we get ϕ(x, t) = e

˜ −(λ+λ)t



t

e

˜ (λ+λ)v

  λ

0

×e

rˆ(t−v)

(x+c/ˆ r) erˆ(t−v) −c/ˆ r 0



˜ − c/ˆ r − y, v dF (y) + λ



 ϕ (x + c/ˆ r)

  ϕ (1 − α) + αey

+∞ −∞

[6.36]

    ˜ rˆ(t−v) × (x + c/ˆ r) e − c/ˆ r , v dF˜ (y) dv + e−(λ+λ)t . We now use that Yi and Y˜i , i ≥ 1, have p.d.f.’s f (y) and f˜(y), respectively. the variables (x +c/ˆ r) erˆ(t−v) − c/ˆ r − y = u and  For x > 0, changing y r ˆ (t−v) r) e (1 − α) + αe (x + c/ˆ − c/ˆ r = u in the first and second inner integrals, respectively, on the right-hand side of [6.36], we obtain   (x+c/ˆr) erˆ(t−v) −c/ˆr  t ˜ ˜ −(λ+λ)t (λ+λ)v ϕ(x, t) = e e ϕ(u, v) λ 0



0

× f (x + c/ˆ r) e  +∞ ˜ +λ 

rˆ(t−v)

 − c/ˆ r − u du  ϕ(u, v)

r ) erˆ(t−v) −c/ˆ r (1−α) (x+c/ˆ

1  u + (1 − α) (x + c/ˆ r) erˆ(t−v) − c/ˆ r   u ˜  × f ln  α (x + c/ˆ r) erˆ(t−v) − c/ˆ r   α−1 ˜ + du dv + e−(λ+λ)t . α 

×

[6.37]

Note that we excluded the case x = 0 to avoid division by zero in [6.37]. Let



t

I(x, t) =   × λ

˜

e(λ+λ)v

0 (x+c/ˆ r) erˆ(t−v) −c/ˆ r 0

  ϕ(u, v)f (x + c/ˆ r) erˆ(t−v) − c/ˆ r − u du

Risk Models with Investments in Risk-Free and Risky Assets



143

ϕ(u, v)   r) erˆ(t−v) − c/ˆ r r) erˆ(t−v) −c/ˆ r u + (1 − α) (x + c/ˆ (1−α) (x+c/ˆ     u α−1 ˜   du dv. × f ln + α α (x + c/ˆ r) erˆ(t−v) − c/ˆ r

˜ +λ

+∞





If f (y) is continuous on R+ and f˜(y) is continuous on R, then applying arguments similar to those in the proof of theorem 2.2 establishes the continuity of ϕ(x, t) as a function of two variables. Indeed, for any fixed (x0 , t0 ) ∈ (0, +∞) × (0, +∞) and all 0 < x ≤ x0 and 0 ≤ t ≤ t0 , we have   I(x, t) − I(x0 , t0 )   (x+c/ˆr) erˆ(t−v) −c/ˆr  t    ˜ (λ+λ)v f (x0 + c/ˆ ≤ r) erˆ(t0 −v) − c/ˆ r−u e λ 0

0

  r − u  du − f (x + c/ˆ r) erˆ(t−v) − c/ˆ  (x0 +c/ˆr) erˆ(t0 −v) −c/ˆr   f (x0 + c/ˆ r) erˆ(t0 −v) − c/ˆ r − u du +λ r (x+c/ˆ r) erˆ(t−v) −c/ˆ

  1      r ˆ (t −v) r) erˆ(t0 −v) − c/ˆ r r) e 0 −c/ˆ r u + (1 − α) (x0 + c/ˆ (1−α) (x0 +c/ˆ    u α−1 ˜ + × f ln  α α (x0 + c/ˆ r) erˆ(t0 −v) − c/ˆ r 

˜ +λ

+∞

1   u + (1 − α) (x + c/ˆ r) erˆ(t−v) − c/ˆ r     u α − 1  du + × f˜ ln   r ˆ (t−v) α α (x + c/ˆ r) e − c/ˆ r −



(1−α)((x0 +c/ˆ r) erˆ(t0 −v) −c/ˆ r)

1   u + (1 − α) (x + c/ˆ r) erˆ(t−v) − c/ˆ r (1−α)((x+c/ˆ r) erˆ(t−v) −c/ˆ r)     u α−1 ˜ + × f ln  du dv α α (x + c/ˆ r) erˆ(t−v) − c/ˆ r  t0 ˜ ˜ e(λ+λ)v dv. + (λ + λ) ˜ +λ

t

144

Ruin Probabilities

From this we get   I(x, t) − I(x0 , t0 )    t ˜ (λ+λ)v ≤ e λ 0

(x+c/ˆ r) erˆ(t−v) −c/ˆ r 0

 f (x0 + c/ˆ r) erˆ(t0 −v)

   − c/ˆ r − u − f (x + c/ˆ r) erˆ(t−v) − c/ˆ r − u  du   + λF (x0 + c/ˆ r) erˆ(t0 −v) − (x + c/ˆ r) erˆ(t−v)  M2   1 ˜   +λ  u + (1 − α)(x + c/ˆ r) erˆ(t0 −v) − c/ˆ r 0 M1 (v)    u α−1 + × f˜ ln  α α (x0 + c/ˆ r) erˆ(t0 −v) − c/ˆ r 1 [6.38]  u + (1 − α) (x + c/ˆ r) erˆ(t−v) − c/ˆ r     u α−1 ˜  du + × f ln   α α (x + c/ˆ r) erˆ(t−v) − c/ˆ r      (1 − α) (x0 + c/ˆ r) erˆ(t0 −v) − c/ˆ r α−1 ˜ ˜   + λF ln + α α (x + c/ˆ r) erˆ(t−v) − c/ˆ r     M2 α−1 ˜ ˜ + + 2λ 1 − F ln  α α (x0 + c/ˆ r) erˆ(t0 −v) − c/ˆ r    M1 (v) α−1 + + F˜ ln  dv r ˆ (t −v) 0 α α (x0 + c/ˆ r) e − c/ˆ r 

−

˜

˜

+ e(λ+λ)t0 − e(λ+λ)t , where the function M1 (v) and the number M2 are such that   r) erˆ(t0 −v) − c/ˆ r < M1 (v) < M2 < ∞ (1 − α) (x0 + c/ˆ for all v ∈ [0, t0 ]. Therefore, choosing M1 (v), such that     r) erˆ(t0 −v) − c/ˆ r − M1 (v) sup (1 − α) (x0 + c/ˆ v∈[0,t0 ]

Risk Models with Investments in Risk-Free and Risky Assets

145

is small enough, and large enough M2 we can first make    M2 α−1 ˜ + 1 − F ln  α α (x0 + c/ˆ r) erˆ(t0 −v) − c/ˆ r    M1 (v) α−1 ˜ + + F ln  α r) erˆ(t0 −v) − c/ˆ α (x0 + c/ˆ r small enough and, then, by the continuity of f (y) and f˜(y), we can make the right-hand side of [6.38] small enough when x is close to x0 and t is close to t0 . Hence, ϕ(x, t) is continuous at (x0 , t0 ) for x ≤ x0 and t ≤ t0 . Similarly, we can consider other cases. Thus, we proved the continuity of ϕ(x, t) on (0, +∞) × [0, +∞) as a function of two variables. If f (y) and f˜(y) have the derivatives f  (y) and f˜ (y), which satisfy conditions of the theorem, then we can show the existence of partial derivatives of ϕ(x, t) as in the proof of theorem 2.2. Furthermore, from [6.37] we get   t  ∂ϕ(x, t) ˜ ˜ (r−λ−λ)t (λ+λ−r)v =e e λf (0) ϕ (x + c/ˆ r) erˆ(t−v) ∂x 0  (x+c/ˆr) erˆ(t−v) −c/ˆr  − c/ˆ r, v + λ ϕ(u, v) 

0

 × f (x + c/ˆ r) erˆ(t−v) − c/ˆ r − u du  +∞ ˜ −λ   ϕ(u, v) 

r ) erˆ(t−v) −c/ˆ r (1−α) (x+c/ˆ

×



1

2 u + (1 − α) (x + c/ˆ r) erˆ(t−v) − c/ˆ r    u  × (α − 1) f˜ ln  α (x + c/ˆ r) erˆ(t−v) − c/ˆ r  α−1 u + + α (x + c/ˆ r) erˆ(t−v) − c/ˆ r   u  × f˜ ln  α (x + c/ˆ r) erˆ(t−v) − c/ˆ r   α−1 + du dv α

[6.39]

146

Ruin Probabilities

and

 x ∂ϕ(x, t) ˜ ϕ(x, t) + λ = − (λ + λ) ϕ(u, t) f (x − u) du ∂t 0     +∞ u ϕ(u, t) α−1 ˜ ˜ f ln + du +λ αx α (1−α)x u + (α − 1)x  t ˜ ˜ (r−λ−λ)t + (ˆ rx + c) e e(λ+λ−r)v 0



  × λf (0) ϕ (x + c/ˆ r) erˆ(t−v) − c/ˆ r, v 

(x+c/ˆ r) erˆ(t−v) −c/ˆ r

+λ

ϕ(u, v)

0

  × f  (x + c/ˆ r) erˆ(t−v) − c/ˆ r − u du  +∞ ˜ −λ   ϕ(u, v)

[6.40]

r ) erˆ(t−v) −c/ˆ r (1−α) (x+c/ˆ

1  2 u + (1 − α) (x + c/ˆ r) erˆ(t−v) − c/ˆ r    u ˜  × (α − 1) f ln  α (x + c/ˆ r) erˆ(t−v) − c/ˆ r  u α−1 + + α (x + c/ˆ r) erˆ(t−v) − c/ˆ r   u  × f˜ ln  α (x + c/ˆ r) erˆ(t−v) − c/ˆ r   α−1 du dv. + α ×

Combining [6.39] and [6.40], we have  x ∂ϕ(x, t) ˜ = −(λ + λ) ϕ(x, t) + λ ϕ(u, t) f (x − u) du ∂t 0     +∞ u ϕ(u, t) α−1 ˜ ˜ f ln + +λ du αx α (1−α)x u + (α − 1)x

Risk Models with Investments in Risk-Free and Risky Assets

147

∂ϕ(x, t) ∂x  x ∂ϕ(x, t) ˜ − (λ + λ) ϕ(x, t) + λ ϕ(x − y, t) dF (y) = (ˆ rx + c) ∂x 0  +∞    ˜ +λ ϕ (1 − α) + αey x, t dF˜ (y), + (ˆ rx + c)

−∞

which gives [6.33]. From [6.39], it follows that     t  +∞  ∂ϕ(x, t)  ˜ ˜ (λ+λ−r)v   ≤ e(r−λ−λ)t e |f  (y)| dy λf (0) + λ  ∂x  0 0  +∞ ˜ − α) λ(1  f˜(y) e−y dy +  α (x + c/ˆ r) erˆ(t−v) − c/ˆ r −∞ ˜ λ  +  α (x + c/ˆ r) erˆ(t−v) − c/ˆ r   +∞    −y ˜ |f (y)| (1 − α) e + α dy dv × −∞

for all x > 0 and t ≥ 0, which yields [6.34]. As in theorem 2.2, the boundary condition ϕ(x, 0) = 1 is obvious and the condition limx→+∞ ϕ(x, t) = 1 follows immediately from the fact that there can be only a finite number of claims on any finite time interval a.s. This completes the proof.  R EMARK 6.3.– As in theorem 2.2, the requirements that we impose on the distributions of Y¯i , Yi and Y˜i , i ≥ 1, in theorem 6.4 are not necessary for the continuity of ϕ(x, t) or the existence of its partial derivatives (see also remark 2.4). On the other hand, ϕ(x, t) may not have partial derivatives in some cases. 6.3. Risk model with stochastic premiums and investments in risk-free and risky assets We will now deal with the generalization of the risk model with stochastic premiums considered in section 6.1. As in section 6.2.1, we introduce the discrete-time process Xnx = Xτn (x), n ≥ 0, where X0x = x. Applying

148

Ruin Probabilities

arguments similar to those in section 6.2.1 shows that [6.3] is also true for this model, where P1, n , n ≥ 1, are given by [6.4] and       Y˜i rˆ(τn −¯ τi ) ¯ 1 − α + αe , n ≥ 1. P2, n = Yi e i : τn−1 0; rˆx

iv) the function ϕ(x) satisfies the integro-differential equation  +∞ ¯ + λ + λ)ϕ(x) ˜ ¯ rˆxϕ+ (x) = (λ −λ ϕ(x + y) dF¯ (y) 

x

−λ

 ˜ ϕ(x − y) dF (y) − λ

0

0

+∞ −∞

   ϕ (1 − α) + αey x dF˜ (y)

on (0, +∞) with the boundary condition limx→+∞ ϕ(x) = 1 and its value at the point x = 0 is given by ¯  +∞ λ ϕ(0) = ¯ ϕ(y)dF¯ (y); λ+λ 0 v) for all x∗ > 0, we have ˜  ¯   ϕ+ (x) ≤ λ + λ + λ . rˆx∗ x∈[x∗ ,+∞) sup

[6.41]

Risk Models with Investments in Risk-Free and Risky Assets

149

  T HEOREM 6.6.– Let the surplus process Xt (x) t≥0 follow [6.2] under the above assumptions. 1) If Y¯i and Yi , i ≥ 1, have p.d.f.’s f¯(y) and f (y), respectively, which are continuous on R+ , and Y˜i , i ≥ 1, have a p.d.f. f˜(y), which is continuous on R, then ϕ(x, t) is continuous on (0, +∞) × [0, +∞) as a function of two variables. 2) If the p.d.f.’s f¯(y) and f (y) of Y¯i and Yi , i ≥ 1, respectively, have the derivatives f¯ (y) and f  (y) on R+ , such that |f¯ (y)| and |f  (y)| are integrable and bounded on R+ , and the p.d.f. f˜(y) of Y˜i , i ≥ 1, have the derivative f˜ (y) on R, such that |f˜ (y)|, f˜(y) e−y and |f˜ (y)| e−y are integrable and bounded on R, then: i) ϕ(x, t) has partial derivatives w.r.t. x and t on (0, +∞) × [0, +∞), which are continuous as functions of two variables; ii) ϕ(x, t) satisfies the partial integro-differential equation ∂ϕ(x, t) ∂ϕ(x, t) ¯ + λ + λ) ˜ ϕ(x, t) − rˆx + (λ ∂t ∂x  +∞  x ¯ ¯ −λ ϕ(x + y, t) dF (y) − λ ϕ(x − y, t) dF (y)  ˜ −λ

0

0 +∞

−∞

   ϕ (1 − α) + αey x, t dF˜ (y) = 0

on (0, +∞) × [0, +∞) with the boundary conditions ϕ(x, 0) = 1 and limx→+∞ ϕ(x, t) = 1; iii) for all x∗ > 0 and T > 0, we have      ∂ϕ(x, t)  C3   ≤ C 1 C2 + , sup  ∂x  x∗ x∈[x∗ ,+∞),

[6.42]

t∈[0,T ]

where

⎧ ¯ ˜ ⎪ ⎨ 1 − e(ˆr−λ−λ−λ)T ¯+λ+λ ˜ − rˆ C1 = λ ⎪ ⎩T   ¯ λf¯(0) − λf (0) + λ C2 = ¯

if

¯+λ+λ ˜ = rˆ, λ

if

¯+λ+λ ˜ = rˆ, λ



+∞ 0

|f¯ (y)| dy + λ

 0

+∞

|f  (y)| dy,

150

Ruin Probabilities

  +∞  +∞ ˜   λ −y  −y ˜ ˜ f (y) e dy + C3 = |f (y)| (1−α) e +α dy . (1−α) α −∞ −∞ We leave out proofs of these theorems. The proof of theorem 6.5 is analogous to the proofs of theorems 2.1, 3.1 and 6.3. The proof of theorem 6.6 is analogous to the proofs of theorems 2.2, 3.2 and 6.4. 6.4. Accuracy and reliability of uniform approximations of the survival probabilities by their statistical estimates In this section, we obtain relations connecting accuracy and reliability of uniform approximations of the survival probabilities by their statistical estimates. We deal with the finite- and infinite-horizon survival probabilities in sections 6.4.1 and 6.4.2, respectively. 6.4.1. Finite-horizon survival probability   Let now the surplus process Xt (x) t≥0 follow either [6.1] or [6.2] under the above assumptions. For any fixed T > 0, the function ϕ(x, T ) is defined on R+ . Let x∗ and x∗ be fixed and such that x∗ > x∗ > 0. Our aim is to derive a relation connecting accuracy and reliability of a uniform approximation of the finite-horizon survival probability by its statistical estimate for all x ∈ [x∗ , x∗ ]. To this end, we will use bounds [6.34] and [6.42]. Note that we can apply similar considerations to the risk models investigated in Chapters 2 and 3 and use the corresponding bounds for the partial derivative of ϕ(x, t) w.r.t. x obtained in those chapters. Fix h > 0, such that h x∗ − x∗ . Next, let h1 = h, hi = h

x1 = x ∗ + h 1 ;

C2 x∗ xi−1 + C3 xi−1 , C2 x∗ xi−1 + C3 x∗

xi = xi−1 + hi ,

2 ≤ i ≤ k − 1,

where C2 and C3 are defined in theorems 6.4 and 6.6 and k = k(x∗ , x∗ , h, C2 , C3 ) is such that xk−1 < x∗ ≤ xk . For simplicity of notation, we set x0 = x∗ , xk = x∗ and hk = x∗ − xk−1 .

Risk Models with Investments in Risk-Free and Risky Assets

151

Let the piecewise-linear function ϕ(x, ¯ T ) be defined on [x∗ , x∗ ] by   x − xi−1 , ϕ(x, ¯ T ) = ϕ(xi−1 , T ) + ϕ(xi , T ) − ϕ(xi−1 , T ) xi − xi−1 x ∈ [xi−1 , xi ],

1 ≤ i ≤ k.

  We simulate paths of the surplus processes Xt (xi ) t≥0 , 0 ≤ i ≤ k, N times on [0, T ] using Monte Carlo methods. In the end, ruin does not occur νxi (N, T ) times. Let the piecewise-linear function ϕ(x, ˆ T ) be defined on [x∗ , x∗ ] as follows. First, we define ϕ(x, ˆ T ) at points xi as a statistical estimate for ϕ(x, T ) by ϕ(x ˆ i, T ) =

νxi (N, T ) , N

0 ≤ i ≤ k.

Moreover, set   x − xi−1 ϕ(x, ˆ T ) = ϕ(x ˆ i−1 , T ) + ϕ(x ˆ i , T ) − ϕ(x ˆ i−1 , T ) , xi − xi−1 x ∈ (xi−1 , xi ),

1 ≤ i ≤ k.   T HEOREM 6.7.– Let the surplus process Xt (x) t≥0 follow either [6.1] or [6.2] under the above assumptions. Moreover, let the conditions of either theorem 6.4 or theorem 6.6, respectively, hold that are sufficient for the existence of partial derivatives of ϕ(x, t). Then, for any x∗ > x∗ > 0, h ∈ (0, x∗ − x∗ ) and ε > 0, we have      C 3 ˆ T )  ≤ ε + C 1 h C2 + P sup ϕ(x, T ) − ϕ(x, x∗ x∈[x∗ , x∗ ] [6.43] 2

≥ 1 − 2(k + 1) e−2N ε . P ROOF.– In what follows, ϕ1 (·, ·) denotes the partial derivative of ϕ(x, t) w.r.t. the first argument. By the definition of ϕ(x, T ), Lagrange’s mean value theorem, bounds [6.34] or [6.42] and the definition of hi , we get     ϕ(x, T ) − ϕ(x, xi+1 , T ) hi+1 ¯ T ) ≤ ϕ(xi+1 , T ) − ϕ(xi , T ) = ϕ1 (ˆ     C3 C3 ≤ C1 hi+1 C2 + = C1 h C 2 + , xi x∗ x ∈ [xi , xi+1 ],

0 ≤ i ≤ k − 1,

where x ˆi+1 ∈ [xi , xi+1 ], 0 ≤ i ≤ k − 1.

152

Ruin Probabilities

Hence, we have     C 3 sup ϕ(x, T ) − ϕ(x, . ¯ T )  ≤ C1 h C 2 + x∗ x∈[x∗ , x∗ ] Next, we introduce the following events    ˆ i , T ) ≤ ε , Ei, ε = ϕ(x ¯ i , T ) − ϕ(x

[6.44]

0 ≤ i ≤ k.

By Hoeffding’s inequality (see section A.2), for any ε > 0, we have

¯i, ε ≤ 2e−2N ε2 , 0 ≤ i ≤ k. P E Consequently, we get     k     P sup ϕ(x, Ei, ε ¯ T ) − ϕ(x, ˆ T) ≤ ε =P x∈[x∗ , x∗ ]

 =1−P

i=0



k

¯i, ε E i=0

k 

¯i, ε ≥ 1 − 2(k + 1) e−2N ε2 . P E ≥1−

[6.45]

i=0

Relation [6.43] follows immediately from [6.44] and [6.45], which completes the proof.  R EMARK 6.4.– Relation [6.43] connects accuracy and reliability of the uniform approximation of the finite-horizon survival probability by its statistical estimate for all x ∈ [x∗ , x∗ ]. It is easily seen that any given accuracy of the approximation can be attained by small enough ε > 0 and h > 0 and for the given accuracy, any required reliability is provided by large enough N . Nevertheless, even pointwise accuracy of Monte Carlo methods is −l in fact limited. Indeed, if the " reliability is 1 − 10 , where l is a ! l lnrequired 10+ln 2 , which grows rapidly as ε → 0. For positive integer, then N = 2ε2 example, N = 2 649 159 if l = 2 and ε = 0.001. Hence, it is difficult to estimate the survival probability appropriately when an initial surplus is large enough. In this case, we have to simulate too large number of paths.

Risk Models with Investments in Risk-Free and Risky Assets

153

R EMARK 6.5.– We now give an upper bound for k = k(x∗ , x∗ , h, C2 , C3 ). Note that   C2 x∗ xi−1 + C3 xi−1 C3 (xi−1 − x∗ ) hi = h =h 1+ C2 x∗ xi−1 + C3 x∗ x∗ (C2 xi−1 + C3 )   (i − 1)hC3 >h 1+ , 1 ≤ i ≤ k − 1. x∗ (C2 x∗ + C3 ) Let hi



 (i − 1)hC3 =h 1+ , x∗ (C2 x∗ + C3 )

1 ≤ i ≤ k ,

  where k  is the least integer number, such that ki=1 hi ≥ x∗ − x∗ . Since    k k    hC3 k  (k  − 1) (i − 1)hC3   =h k + , hi = h 1+ x∗ (C2 x∗ + C3 ) 2x∗ (C2 x∗ + C3 ) i=1

we get

i=1

#

k =

+

hC3 − 2x∗ (C2 x∗ + C3 ) 2hC3  2 $ 2x∗ (C2 x∗ + C3 ) − hC3 + 8C3 x∗ (C2 x∗ + C3 )(x∗ − x∗ ) 2hC3

.

It is easily seen that k ≤ k  . We will now give two examples. E XAMPLE 6.3.– Let the claim sizes be exponentially distributed with mean μ −|y|/μ ˜ and Y˜i , i ≥ 1, have the Laplace distribution with p.d.f. f˜(y) = e 2˜μ , where y ∈ R and μ ˜ ∈ (0, 1). In spite of the fact that f˜(y) is not differentiable at the

point x = 0, it is easy to show that theorems 6.4 and 6.6 are applicable in this case. Moreover,  let the premium sizes be exponentially distributed with mean μ ¯ when Xt (x) t≥0 follows [6.2].   A simple computation shows that C2 = 2λ/μ if Xt (x) t≥0 follows [6.1]     ¯ μ + λ/μ if Xt (x) and C2 = ¯ follows [6.2]. Since λ/¯ μ − λ/μ + λ/¯ t≥0  +∞ 1 f˜(y) e−y dy = , 1 − μ ˜2 −∞

154

Ruin Probabilities



+∞



−∞ +∞ −∞

|f˜ (y)| dy =

1 , μ ˜

|f˜ (y)| e−y dy =

1 μ(1 − μ ˜2 )

for μ ˜ ∈ (0, 1), we get C3 =

˜ − α) ˜ λ(1 λ + . μ ˜ αμ(1 − μ ˜)

Set T = 1, x∗ = 1, x∗ = 8, h = 0.00001, λ = 1, μ = 5, ˜ = 2 and μ r = 0.001, r˜ = 0.004, λ ˜ = 0.08. Moreover, set c = 6 if  ¯ = 6 and μ Xt (x) t≥0 follows [6.1] and set λ ¯ = 1 if Xt (x) t≥0 follows [6.2]. Let the required accuracy and reliability be 0.001 and 0.99, respectively,i.e.     P sup ϕ(x, 1) − ϕ(x, ˆ 1) ≤ 0.001 ≥ 0.99. x∈[1, 8]

Tables 6.1 and 6.3 give comparisons ˆ for integer initial surpluses  of ϕ(x)  from [1, 8] and different α if Xt (x) t≥0 follows [6.1] and [6.2], respectively. The corresponding values of rˆ, ε, k and N , which depend on α, are given in Tables 6.2 and 6.4. x 1 2 3 4 5 6 7 8

α = 0.25 0.561270 0.610120 0.677954 0.721247 0.760956 0.786624 0.816830 0.843835

α = 0.50 0.561093 0.609695 0.678559 0.720687 0.758686 0.788435 0.817659 0.844012

α = 0.75 0.561981 0.610136 0.677589 0.718702 0.760075 0.787550 0.817152 0.845231

α = 0.99 0.559461 0.609609 0.676448 0.717148 0.759429 0.787254 0.815577 0.845337   Table 6.1. Comparison of ϕ(x) ˆ for different α if Xt (x) t≥0 follows [6.1]. Exponentially distributed claim sizes

E XAMPLE 6.4.– Let the claim sizes have the Pareto distribution with p.d.f.  γ+1 μ(γ − 1) γ , f (y) = μ(γ − 1) μ(γ − 1) + y

Risk Models with Investments in Risk-Free and Risky Assets

155

where y ∈ R+ , μ > 0, γ > 1, and Y˜i , i ≥ 1, have the Laplace distribution −|y|/μ ˜ ˜ ∈ (0, 1). Furthermore, let the with p.d.f. f˜(y) = e 2˜μ , where y ∈ R and μ premium sizes have the Pareto distribution with p.d.f.  γ¯+1 μ ¯(¯ γ − 1) γ¯ ¯ , f (y) = μ ¯(¯ γ − 1) μ ¯(¯ γ − 1) + y   where y ∈ R+ , μ ¯ > 0, γ¯ > 1, when Xt (x) t≥0 follows [6.2]. Note that we write the p.d.f.’s of the Pareto distributions in such a form in order to get E[Yi ] = μ and E[Y¯i ] = μ ¯, i ≥ 1. α 0.25 0.50 0.75 0.99

N 20 075 749 12 643 204 11 072 168 10 416 840   Table 6.2. Values of rˆ, ε, k and N for different α if Xt (x) t≥0 follows [6.1]. Exponentially distributed claim sizes x 1 2 3 4 5 6 7 8

rˆ 0.00175 0.00250 0.00325 0.00397

α = 0.25 0.660976 0.698463 0.738687 0.773619 0.809001 0.829187 0.842306 0.875158

ε 0.000661 0.000833 0.000891 0.000919

k 209 786 211 689 213 658 215 612

α = 0.75 α = 0.99 0.660778 0.661011 0.701007 0.703572 0.738998 0.739811 0.778589 0.777929 0.809151 0.810166 0.828774 0.830821 0.848979 0.849062 0.871285 0.874557   Table 6.3. Comparison of ϕ(x) ˆ for different α if Xt (x) t≥0 follows [6.2]. Exponentially distributed premium and claim sizes

It is easy to check that C2 =

α = 0.50 0.660751 0.699885 0.740966 0.773273 0.809035 0.830122 0.844995 0.873945

2λγ μ(γ−1)

  if Xt (x) t≥0 follows [6.1] and

  ¯γ ¯γ  λ¯ λ¯ λγ  λγ  − + + C2 =   μ ¯(¯ γ − 1) μ(γ − 1) μ ¯(¯ γ − 1) μ(γ − 1)   if Xt (x) t≥0 follows [6.2]. The constant C3 is given in example 6.3.

Set T = 1, x∗ = 1, x∗ = 8, h = 0.00001, λ = 1, μ = 5, γ =  2, ˜ r = 0.001, r˜ = 0.004, λ = 2 and μ ˜ = 0.08. Moreover, set c = 6 if Xt (x) t≥0

156

Ruin Probabilities

  ¯ = 6, μ follows [6.1] and set λ ¯ = 1 and γ¯ = 2 if Xt (x) t≥0 follows [6.2]. Let the required accuracy and reliability be 0.001 and 0.99, respectively. α 0.25 0.50 0.75 0.99

rˆ 0.00175 0.00250 0.00325 0.00397

ε 0.000868 0.000929 0.000949 0.000959

k 257 765 299 956 336 146 366 356

N 11 776 810 10 382 981 10 010 400 9 854 452   Table 6.4. Values of rˆ, ε, k and N for different α if Xt (x) t≥0 follows [6.2]. Exponentially distributed premium and claim sizes

Tables 6.5 and 6.7 give comparisons ˆ for integer initial surpluses   of ϕ(x) from [1, 8] and different α if Xt (x) t≥0 follows [6.1] and [6.2], respectively. The corresponding values of rˆ, ε, k and N are given in Tables 6.6 and 6.8. x 1 2 3 4 5 6 7 8

α = 0.75 α = 0.99 0.680689 0.679641 0.727982 0.728079 0.766246 0.766806 0.790657 0.791278 0.814265 0.814473 0.833948 0.833822 0.854873 0.854690 0.870616 0.870087   Table 6.5. Comparison of ϕ(x) ˆ for different α if Xt (x) t≥0 follows [6.1]. Pareto distributed claim sizes α 0.25 0.50 0.75 0.99

α = 0.25 0.682065 0.729443 0.766450 0.789548 0.812283 0.834393 0.855646 0.872234

rˆ 0.00175 0.00250 0.00325 0.00397

α = 0.50 0.681105 0.729544 0.765893 0.789777 0.813272 0.834281 0.855007 0.871462

ε 0.000660 0.000832 0.000889 0.000917

k 211 614 215 377 219 238 223 043

N 20 162 910 12 694 234 11 120 053 10 465 800   Table 6.6. Values of rˆ, ε, k and N for different α if Xt (x) t≥0 follows [6.1]. Pareto distributed claim sizes

In the above examples, the survival probabilities do not depend considerably on α. If we take other values of parameters, e.g. larger T , then dependence can be more significant. Moreover, from these examples, we see

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that the accuracy of Monte Carlo methods is limited. We had to simulate a large number of paths to provide accuracy 0.001. x 1 2 3 4 5 6 7 8

α = 0.75 α = 0.99 0.680748 0.680560 0.730631 0.730929 0.764757 0.762156 0.792025 0.788622 0.810828 0.809768 0.836113 0.835144 0.846782 0.847044 0.868036 0.868128   Table 6.7. Comparison of ϕ(x) ˆ for different α if Xt (x) t≥0 follows [6.2]. Pareto distributed premium and claim sizes α 0.25 0.50 0.75 0.99

α = 0.25 0.681357 0.728993 0.766374 0.795067 0.812547 0.834758 0.846817 0.870882

α = 0.50 0.681070 0.729162 0.765045 0.792973 0.811227 0.835565 0.846612 0.870119

rˆ 0.00175 0.00250 0.00325 0.00397

ε 0.000855 0.000915 0.000935 0.000945

k 298 423 362 976 411 351 447 609

N 12 247 208 10 801 518 10 413 216 10 246 613   Table 6.8. Values of rˆ, ε, k and N for different α if Xt (x) t≥0 follows [6.2]. Pareto distributed premium and claim sizes

6.4.2. Infinite-horizon survival probability   Let the surplus process Xt (x) t≥0 follow either [6.1] or [6.2] under the above assumptions. Our next aim is to get a relation connecting accuracy and reliability of a uniform approximation of the infinite-horizon survival probability by its statistical estimate for all x ∈ [x∗ , x∗ ]. We assume that x∗ ∗ ∗ and  x are also such that x > x∗ > 0. Nevertheless, we can take x∗ = 0 if Xt (x) t≥0 follows [6.1]. To get that relation, we use bounds [6.21] and [6.41]. Consequently, it is true for a much wider class of distributions than [6.43]. We also fix h > 0, such that h x∗ − x∗ . Moreover, let h1 = h, hi =

x1 = x ∗ + h 1 ;

h(ˆ rxi−1 + c) rˆx∗ + c

or

hi =

hxi−1 x∗

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Ruin Probabilities

  if Xt (x) t≥0 follows [6.1] or [6.2], respectively, and xi = xi−1 + hi ,

2 ≤ i ≤ k − 1,

where k = k(x∗ , x∗ , h) is such that xk−1 < x∗ ≤ xk . To simplify notation, we set x0 = x∗ , xk = x∗ and hk = x∗ − xk−1 . Let C0 =

˜ λ+λ rˆx∗ + c

or

C0 =

¯+λ+λ ˜ λ rˆx∗

  if Xt (x) t≥0 follows [6.1] or [6.2], respectively. Next, we define the piecewise-linear function ϕ(x) ¯ on [x∗ , x∗ ] by   x − xi−1 , ϕ(x) ¯ = ϕ(xi−1 ) + ϕ(xi ) − ϕ(xi−1 ) xi − xi−1 x ∈ [xi−1 , xi ],

1 ≤ i ≤ k.

  When we simulate paths of the surplus processes Xt (xi ) t≥0 , 0 ≤ i ≤ k, N times on the infinite time interval using Monte Carlo methods, ruin does not occur νxi (N ) times. Note that in this case we actually simulate paths on a sufficiently long finite time interval. Let the piecewise-linear function ϕ(x) ˆ be defined on [x∗ , x∗ ] as follows: ϕ(x ˆ i) =

νxi (N ) , N

0 ≤ i ≤ k;

  x − xi−1 ϕ(x) ˆ = ϕ(x ˆ i−1 ) + ϕ(x ˆ i ) − ϕ(x ˆ i−1 ) , xi − xi−1 x ∈ (xi−1 , xi ),

1 ≤ i ≤ k.

  T HEOREM 6.8.– Let the surplus process Xt (x) t≥0 follow either [6.1] or [6.2] under the above assumptions. Moreover, let there be z0 ∈ Z ∗ , such that [6.15] holds. Then, for any x∗ > x∗ > 0, h ∈ (0, x∗ − x∗ ) and ε > 0, we have     2   ≤ ε + C0 h ≥ 1 − 2(k + 1) e−2N ε . ˆ P sup ϕ(x) − ϕ(x) [6.46] x∈[x∗ , x∗ ]

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The proof of this theorem is analogous to the proof of theorem 6.7. We only note that we use bounds [6.21] and [6.41] to get the required analogue of [6.44]. Furthermore, to apply Lagrange’s mean value theorem on [xi , xi+1 ], 0 ≤ i ≤ k − 1, we can divide these intervals by points where ϕ(x) is not differentiable, if any. 6.5. Bibliographical notes Risk models where investment income is random are of great interest from the mathematical point of view. Since it is usually assumed that such income can be negative, the investment risk is added to the insurance risk. In 1993, Paulsen [PAU 93] introduced a risk model that allows a stochastic rate of return on investments as well as a stochastic level of inflation. Integro-differential equations for ψ(x) were derived, but the differentiability of this function was not studied. These equations were used to obtain analytic expressions for ψ(x) in some cases and inequalities for it in other cases. Examples showed that stochastic economic factors may have a serious impact on the infinite-horizon ruin probability. For some early results in this direction, see also [PAU 98, PAU 97]. Paulsen and Gjessing [PAU 97] considered a risk process with stochastic interest rate and showed that the infinite-horizon ruin probability can be found by solving a certain boundary value problem involving an integro-differential equation. In [PAU 98], the classical risk process compounded by another independent process was considered and both of these components were assumed to be Lévy processes. Sharp conditions were given on the parameters of these two components when ruin is certain. The fact that risky investments can be dangerous was first justified mathematically by Kalashnikov and Norberg [KAL 02]. They modeled the basic surplus process due to insurance activity and the logarithm of the price of the risky asset by Lévy processes, and obtained upper and lower power bounds for the infinite-horizon ruin probability when the initial surplus is large enough. These bounds turned out to be power functions. Some similar results were obtained independently by Paulsen [PAU 02]. Yuen, Wang and Wu [YUE 06] generalized results of [KAL 02] for the renewal risk process with stochastic interest. Frolova, Kabanov and Pergamenshchikov [FRO 02] used the bounds obtained in [KAL 02] to show that ruin occurs with probability 1 in the

160

Ruin Probabilities

classical risk model if all surplus is invested in the risky asset, the price of which is modeled by a geometric Brownian motion, and some additional conditions for parameters of the geometric Brownian motion hold. They also showed that if these conditions are not fulfilled, a power asymptotics is true for ψ(x) when the claim sizes are exponentially distributed. The power asymptotics for ψ(x) was obtained by Cai and Xu [CAI 06] in the case where the classical risk process is perturbed by Brownian motion. Moreover, Pergamenshchikov and Zeitouny [PER 06] considered the risk model where the premium intensity is a bounded non-negative random function, and generalized results of [FRO 02]. On the other hand, numerous results indicate that risky investments can be used to improve the solvency of the insurance company. Gaier, Grandits and Schachermayer [GAI 03] considered the classical risk model under the additional assumptions that the company is allowed to borrow and invest in the risky asset, the price of which follows a geometric Brownian motion. They obtained an upper exponential bound for ψ(x) when the claim sizes have exponential moments and a fixed quantity, which is independent of the current surplus, is invested in the risky asset. It appeared that this bound is better than the classical one. Mishura [MIS 06] obtained an exponential bound for the infinite-horizon ruin probability when the price of the risky asset is modeled by a semimartingale with absolutely continuous characteristics w.r.t. the Lebesgue measure. Ma and Sun [MA 03] introduced a new type of exponential martingale parameterized by a general rate function to get various exponential bounds for ψ(x) in an extension of the classical risk model, which contains stochastic interest rate, reserve-dependent expense loading, diffusion perturbed models and many others as special cases. The continuity and differentiability of the infinite-horizon survival probability in models with risky investments were first investigated by Wang and Wu [WAN 01]. They dealt with the classical risk model under the additional assumption that investment income follows a Brownian motion with drift. The authors established the continuity of ϕ(x), obtained sufficient conditions when ϕ(x) is twice continuously differentiable and derived an integro-differential equation for this function, which involves its second derivative. Those sufficient conditions turned out to be too limiting. In particular, it was assumed that the p.d.f. of claim sizes has the second derivative.

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Yuen, Wang and Ng [YUE 04] generalized some results of [WAN 01] for the case when investment income follows a compound Poisson process plus a Brownian motion with positive drift. Grandits [GRA 04b] and Gaier and Grandits [GAI 04] extended results obtained in [WAN 01] to the classical risk model when the insurance company invests all surplus in risk-free and risky assets proportionally and the price of the risky asset follows a geometric Brownian motion. Moreover, they investigated the asymptotics of ψ(x) for claim sizes with tails of regular variation. In all the works mentioned above, the sufficient conditions assumed that the p.d.f. of claim sizes has the second derivative. Cai [CAI 04] considered the case when the classical risk process is invested in a stochastic interest process, which is a Lévy process, derived recursive and integral equations for the ruin probabilities and got lower and upper bounds for ψ(x) using these equations. Cai and Yang [CAI 05] considered the perturbed classical risk process under stochastic or constant interest force and showed that ψ(x) is twice continuously differentiable provided that claim sizes have continuous p.d.f. using stochastic control techniques. Paulsen, Kasozi and Steigen [PAU 05] dealt with the classical risk process compounded by a linear Brownian motion. They gave sufficient conditions for ϕ(x) to be four times continuously differentiable, which in particular implies that ϕ(x) is a solution to a second order integro-differential equation. The authors also analyzed properties of the numerical solution to this equation, transforming it into an ordinary Volterra integral equation of the second kind. Gerber and Yang [GER 07] considered the classical risk model perturbed by diffusion, where the insurance company invests all surplus in risk-free and risky assets proportionally and the price of the risky asset follows a geometric Brownian motion. An integro-differential equation for the absolute ruin probability was derived and solved in some special cases, but the differentiability of this function was not investigated further. For further references in this direction, see also the survey by Paulsen [PAU 08], which deals with the problem of ruin in risk models where assets earn investment income. Finally, numerous investigations are devoted to solving optimal investment problems (see references given in section 5.4). Though a geometric Brownian motion is often used to model prices of risky assets, financial models based on jump processes have acquired

162

Ruin Probabilities

increasing popularity recently. A self-contained overview of the theoretical, numerical and empirical aspects involved in using jump processes in financial modeling was provided in [CON 04]; see also, for example [GEM 02, MER 76] for motivation. For general aspects of simulation and Monte Carlo methods, we refer the reader to, for example [ASM 07, FIS 96, KOR 10, RIP 87, RUB 08]. Since the ruin probabilities are too small for large initial surpluses, rare events simulation is of great interest when we evaluate the ruin probabilities. For a survey on rare events simulation, see, for example [ASM 07]. For evaluation of the ruin probabilities in the classical risk model by the Monte Carlo method, see references given in section 1.1.9.

PART 2

SUPERMARTINGALE ApPROACH TO THE ESTIMATION OF RUIN PROBABILITIES

7 Risk Model with Variable Premium Intensity and Investments in One Risky Asset

In this chapter, we consider a generalization of the classical risk model where premium intensity depends on a current surplus of an insurance company. All surplus is invested in one risky asset, the price of which follows a geometric Brownian motion. Our main aim is to show that if the premium intensity grows rapidly with increasing surplus, then an upper exponential bound for the ruin probability holds under certain conditions in spite of the fact that all surplus is invested in the risky asset. To this end, we apply the supermartingale approach and allow the surplus process to explode. To be more precise, we let the premium intensity be a quadratic function. In addition, we investigate the question concerning the probability of explosion of the surplus process between claim arrivals in detail. 7.1. Description of the model We assume that an insurance company has a non-negative initial surplus x and denote by Xt (x) its surplus at time t ≥ 0. For simplicity of notation, we also write Xt instead of Xt (x) when no confusion can arise. Let c : R → R+ \{0} be a measurable function, such that c(u) = c(0) for all u < 0, and let c(Xt ) be a premium intensity, which depends on the surplus at time t. As for claim arrivals, we make the same assumptions and notation as for the classical risk model in section 1.1.1. Moreover, we suppose that there exists

166

Ruin Probabilities

the shifted moment generating function h : R+ → R+ of the claim sizes Yi , i ≥ 1, introduced in section 1.1.7 and satisfying the classical assumption that is also formulated in this section. In addition, we assume that all surplus is invested in one risky asset, the price of which equals St at time t. We model the process (St )t≥0 by a geometric Brownian motion. Thus, dSt = St (a dt + b dWt ),

[7.1]

where a ∈ R, b > 0 and (Wt )t≥0 is a standard Brownian motion. We suppose that the r.v.’s (Yi )i≥1 and the processes (Nt )t≥0 and (Wt )t≥0 are mutually independent. Let (Ft )t≥0 be the filtration generated by (Yi )i≥1 , (Nt )t≥0 and (Wt )t≥0 , i.e.   Ft = σ (Ns )0≤s≤t , (Ws )0≤s≤t , Y1 , Y2 , . . . , YNt . Under the above assumptions, the surplus process (Xt )t≥0 follows the equation  Xt = x +



t 0

c(Xs ) ds +

t 0

N

t  Xs dSs − Yi , Ss

t ≥ 0.

[7.2]

i=1

Substituting [7.1] into [7.2] yields  Xt = x +

t 0



c(Xs ) + aXs ds + b



t 0

Xs dWs −

Nt 

Yi ,

t ≥ 0.

[7.3]

i=1

To simplify notation, we also let τ stand for τ (x) in this chapter. The rest of the chapter is organized as follows. Section 7.2 contains some auxiliary results and deals with the detailed investigation of the question concerning the probability of explosion of the surplus process between two successive claim arrivals. In section 7.3, we formulate and prove the existence and uniqueness theorem for the stochastic differential equation that describes the surplus process. In section 7.4, we establish the supermartingale property

Risk Model with Variable Premium Intensity and Investments in One Risky Asset

167

for an auxiliary exponential process. This property allows us to obtain an exponential bound for the ruin probability under certain conditions. Finally, in section 7.5, we consider the case where the premium intensity is a quadratic function and obtain an upper exponential bound for the ruin probability. Section 7.6 contains some bibliographical notes concerning the problem considered. 7.2. Auxiliary results Consider now the following stochastic differential equation:  t  t Xt = x + q(Xs ) ds + b Xs dWs , t ≥ 0, 0

0

[7.4]

where x > 0, b > 0, (Wt )t≥0 is a standard Brownian motion, q : R → R+ is a locally Lipschitz continuous function, such that q(u) is strictly increasing on R+ and q(u) = q(0) for all u < 0. Equation [7.4] describes the surplus process between two successive jumps of (Nt )t≥0 up to the first exit time of (Xt )t≥0 from [0, +∞) in the model considered above provided that we impose the corresponding restrictions on c(u), set q(u) = c(u) + au for u ≥ 0 and take the surplus at the time of the last jump of (Nt )t≥0 instead of x. 7.2.1. Sufficient conditions for the solution to explode We now give some results which show that (Xt )t≥0 goes to +∞ either with probability 1 or with positive probability, which is less than 1, under certain conditions. Let t∗ be a possible explosion time of (Xt )t≥0 , i.e. / (−∞, +∞)}. t∗ = inf{t ≥ 0 : Xt ∈ Moreover, let t∗(0,+∞) be the first exit time from (0, +∞) for (Xt )t≥0 , i.e. / (0, +∞)}. t∗(0,+∞) = inf{t ≥ 0 : Xt ∈

168

Ruin Probabilities

By theorem A.8, equation [7.4] has a unique strong solution up to the explosion time t∗ . Note that here, and subsequently, we imply the pathwise uniqueness of solutions. For x > 0, we define     +∞ 2 v q(u) exp − 2 du dv I1 = b x u2 x and  I2 = −



x

exp

0

2 b2



x v

[7.5]

 q(u) du dv. u2

[7.6]

The following lemmas provide sufficient conditions for I1 being finite and I2 being infinite. L EMMA 7.1.– If    2 v q(u) du < +∞ lim sup (1 + ε) ln v − 2 b x u2 v→+∞

for some

ε > 0, [7.7]

then I1 < +∞. P ROOF.– Since show that

 +∞ x

1 v 1+ε

dv < +∞ for all ε > 0 and x > 0, it suffices to



v exp − b22 x q(u) du 2 u   < +∞ lim sup v→+∞ exp −(1 + ε) ln v

for some

ε>0

[7.8]

in order to obtain I1 < +∞. We can rewrite [7.8] as    2 v q(u) du < +∞ for some ε > 0, lim sup exp (1 + ε) ln v − 2 b x u2 v→+∞ which gives [7.7]. L EMMA 7.2.– If q(0) > 0, then I2 = −∞.



Risk Model with Variable Premium Intensity and Investments in One Risky Asset

169

P ROOF.– It is easily seen that     x 2q(0) x 1 exp du dv −I2 ≥ 2 b2 0 v u  x    2q(0) 2q(0) dv = exp − 2 exp b x b2 v 0      +∞ 2q(0)u 2q(0) 1 = exp − 2 exp du = +∞, 2 b x b2 1/x u 

which proves the lemma. P ROPOSITION 7.1.– If q(0) > 0 and condition [7.7] holds, then  P limt↑t∗(0,+∞) Xt = +∞ = 1.

P ROOF.– Note that in this case I1 < +∞ and I2 = −∞ by lemmas 7.1 and 7.2. Thus, the assertion of the proposition follows immediately from theorem A.10.  R EMARK 7.1.– If q(0) = 0, then I2 may be finite (see examples 7.1–7.3 below). By theorem A.10, if I1 < +∞ and I2 > −∞, then limt↑t∗(0,+∞) Xt   exists a.s., 0 < P limt↑t∗(0,+∞) Xt = +∞ < 1 and P limt↑t∗(0,+∞) Xt = 0  = 1 − P limt↑t∗(0,+∞) Xt = +∞ . R EMARK 7.2.– Proposition 7.1 does not give us whether the exit time t∗(0,+∞) is finite. It is well known that Feller’s test for explosions (see, for example, theorem 5.29 in [KAR 91, p. 348]) gives precise conditions for whether or not a one-dimensional diffusion process explodes in finite time. This test is very useful when we want to show that a diffusion process does not explode in finite time (see, for example [MIJ 10]), but it does not solve our problem. We will now give a few examples. E XAMPLE 7.1.– Let

q1 u + q0 q(u) = q0

if if

u ≥ 0, u < 0.

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Ruin Probabilities

The function q(u) has the asserted properties provided that q0 ≥ 0 and q1 > 0. Since

  2 v q 1 u + q0 exp − 2 du dv I1 = b x u2 x     +∞

2q0 1 1 x 2q1 /b2 − dv, = exp v b2 v x x 

+∞



we have I1 = +∞ for 2q1 ≤ b2 and I1 < +∞ for 2q1 > b2 . A.10 and lemma 7.2, we We first consider  ∗ the case q0 > 0. From theorem 2 conclude that P t(0,+∞) = ∞ = 1 if 2q1 ≤ b and P limt↑t∗(0,+∞) Xt = +∞ = 1 if 2q1 > b2 . Consider now the case q0 = 0. Since   x   x  x

2 q1 u x 2q1 /b2 exp 2 du dv = − dv, I2 = − b v u2 v 0 0 2 2 we obtain  I2 > −∞ for 2q1 < b and I2 = −∞ 2for 2q  ∗1 ≥ b . Theorem A.10 yields P limt↑t∗(0,+∞) Xt = 0 = 1 if 2q1 < b , P t(0,+∞) = ∞ = 1 if   2q1 = b2 and P limt↑t∗(0,+∞) Xt = +∞ = 1 if 2q1 > b2 .

E XAMPLE 7.2.– Let

q1 (u + q2 )β q(u) = q1 q2β

if if

u ≥ 0, u < 0.

We impose the following restrictions on the parameters of q(u): β > 1, q1 > 0 and q2 ≥ 0. Since

  2 v q1 (u + q2 )β du lim sup (1 + ε) ln v − 2 b x u2 v→+∞    2q1 v β−2 ≤ lim sup (1 + ε) ln v − 2 u du b v→+∞ x    2q1 v β−1 − xβ−1 (1 + ε) ln v − = −∞ = lim v→+∞ b2 (β − 1) 

for all ε > 0, lemma 7.1 gives I1 < +∞.

Risk Model with Variable Premium Intensity and Investments in One Risky Asset

171

 If q2 > 0, then P limt↑t∗(0,+∞) Xt = +∞ = 1 by proposition 7.1. For q2 = 0, we have     x 2q1 v β−2 exp u du dv I2 = − b2 x 0     x 2q1 v β−1 − xβ−1 exp =− dv > −∞. b2 (β − 1) 0  Hence, in this case limt↑t∗(0,+∞) Xt exists a.s., 0 < P limt↑t∗(0,+∞) Xt =   +∞ < 1 and P limt↑t∗(0,+∞) Xt = 0 = 1− P limt↑t∗(0,+∞) Xt = +∞ by theorem A.10.  E XAMPLE 7.3.– Let

q2 u2 + q1 u + q0 q(u) = q0

if if

u ≥ 0, u < 0.

[7.9]

If q0 ≥ 0, q1 ≥ 0 and q2 > 0, then q(u) has all the properties required. For all ε > 0, we have    2 v q 2 u2 + q 1 u + q 0 du lim sup (1 + ε) ln v − 2 b x u2 v→+∞    2 v ≤ lim sup (1 + ε) ln v − 2 q2 du b x v→+∞   2q2 (v − x) = lim (1 + ε) ln v − = −∞. v→+∞ b2 Hence, I1 < +∞ by lemma 7.1.  If q0 > 0, then P limt↑t∗(0,+∞) Xt = +∞ = 1 by proposition 7.1. For q0 = 0, we obtain    x  x 2 q 2 u2 + q 1 u I2 = − du dv exp 2 b v u2 0    x

2q2 (x − v) x 2q1 /b2 =− exp dv. v b2 0

172

Ruin Probabilities

This gives I2 > −∞ for 2q1 < b2 and I2 = −∞ for 2q1 ≥ b2 . Consequently, if 2q1 < b2 , then limt↑t∗(0,+∞) Xt exists a.s., 0 < P limt↑t∗(0,+∞)   Xt = +∞ < 1 and P limt↑t∗(0,+∞) Xt = 0 = 1 − P limt↑t∗(0,+∞)  Xt = +∞ ; if 2q1 ≥ b2 , then P limt↑t∗(0,+∞) Xt = +∞ = 1.  7.2.2. Finiteness of the first exit time One question which is still unanswered is whether t∗(0,+∞) is finite. We will now study it under the conditions of example 7.3. T HEOREM 7.1.– Let (Xt )t≥0 be a strong solution to [7.4] and q(u) be defined by [7.9] with q0 ≥ 0, q1 ≥ 0 and q2 > 0. 1 i) If q0 = 0 and 2q < 1, then b2 ∗ P t(0,+∞) < ∞, limt↑t∗(0,+∞) Xt = +∞

 x −2q /b2 2q2 v 1 dv v exp − 0 b2

. = +∞ −2q1 /b2 v exp − 2qb22 v dv 0

[7.10]

1 ≥ 1 or q0 > 0, then ii) If either q0 = 0 and 2q b2 ∗ P t(0,+∞) < ∞, limt↑t∗(0,+∞) Xt = +∞ = 1.

[7.11]

P ROOF.– Let n0 = min{n ∈ N : 1/n < x}. For all integer n, such that n ≥ n0 , we denote by t∗(1/n,+∞) the first exit time from (1/n, +∞) for (Xt )t≥0 , i.e. t∗(1/n,+∞) = inf{t ≥ 0 : Xt ∈ / (1/n, +∞)}. Note that the sequence of events  

ω ∈ Ω : t∗(1/n,+∞) (ω) < ∞, limt↑t∗(1/n,+∞) Xt (ω) = +∞

n≥n0

is non-decreasing. Hence,   lim ω ∈ Ω : t∗(1/n,+∞) (ω) < ∞, limt↑t∗(1/n,+∞) Xt (ω) = +∞ n→∞

=

∞   n=n0

 ω ∈ Ω : t∗(1/n,+∞) (ω) < ∞, limt↑t∗(1/n,+∞) Xt (ω) = +∞ .

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173

Furthermore, ∞   n=n0



ω ∈ Ω : t∗(1/n,+∞) (ω) < ∞, limt↑t∗(1/n,+∞) Xt (ω) = +∞

  = ω ∈ Ω : t∗(0,+∞) (ω) < ∞, limt↑t∗(0,+∞) Xt (ω) = +∞ .

Therefore, by the continuity of probability measures, we conclude that  P t∗(0,+∞) < ∞, limt↑t∗(0,+∞) Xt = +∞    = P lim t∗(1/n,+∞) < ∞, limt↑t∗(1/n,+∞) Xt = +∞ [7.12] n→∞  = lim P t∗(1/n,+∞) < ∞, limt↑t∗(1/n,+∞) Xt = +∞ . n→∞

From the considerations given in [KAR 91, pp. 343–344] it follows that E[t∗(1/n,+∞) ] = Mn (x) for all n ≥ n0 , where Mn (x) is a solution to the boundary value problem 1 2 2  b x Mn (x) + (q2 x2 + q1 x + q0 )Mn (x) = −1, 2   1 = 0, Mn (+∞) = 0, Mn n

[7.13]

which can be solved by the usual technique (see, for example [AGA 08]). Here and subsequently, the value of a function at +∞ stands for its limit as the value of the argument tends to +∞. Boundary value problem [7.13] has the unique solution  +∞ mn (+∞) − mn (z) 2mn (x) dz Mn (x) = 2 b mn (+∞) 1/n z 2 mn (z)  2 x mn (x) − mn (z) dz, − 2 b 1/n z 2 mn (z) where 

x



2 exp − 2 mn (x) = b 1/n



v 1/n

 q 2 u2 + q 1 u + q 0 du dv. u2

174

Ruin Probabilities

Note that mn (+∞) < +∞. Furthermore, mn (+∞) − mn (z) mn (z)  +∞  v q2 u2 +q1 u+q0

2 exp − du dv z b2 b2 1/n u2

= lim < +∞ = z 2 z→+∞ 2q2 1 u+q0 exp − b22 1/n q2 u +q du 2 u

lim

z→+∞

 +∞ (here we applied L’Hopital’s rule) and 1/n obtain  +∞ mn (+∞) − mn (z) dz < +∞. z 2 mn (z) 1/n

1 z2

dz < +∞. Therefore, we

Thus, E[t∗(1/n,+∞) ] < ∞ for all n ≥ n0 . This gives P[t∗(1/n,+∞) < ∞] = 1 for all n ≥ n0 . Moreover, by the considerations given in [KAR 91, pp. 343– 344], we have  P limt↑t∗(1/n,+∞) Xt = +∞  v q2 u2 +q1 u+q0

x 2 exp − du dv 2 1/n 1/n b u2

=  +∞ v q2 u2 +q1 u+q0 2 exp − du dv [7.14] 1/n b2 1/n u2

x −2q1 /b2 exp 2q0 − 2q2 v dv 1/n v b2 v b2

. = +∞ −2q1 /b2 2q0 2q2 v dv v exp − 2 2 1/n b v b Consequently, [7.12] and [7.14] yield  P t∗(0,+∞) < ∞, limt↑t∗(0,+∞) Xt = +∞

x −2q1 /b2 exp 2q0 − 2q2 v dv v 2 2 1/n b v b

. = lim  2q0 2q2 v n→∞ +∞ −2q1 /b2 exp b2 v − b2 dv 1/n v

[7.15]

Risk Model with Variable Premium Intensity and Investments in One Risky Asset

175

Consider the following two cases: 1 1) If q0 = 0 and 2q < 1, then both of the integrals in the right-hand b2 side of [7.15] are finite as n → ∞. This yields [7.10]. Note that in this case ∗ ∗ 0 < P t(0,+∞) < ∞, limt↑t(0,+∞) Xt = +∞ < 1. 1 2) If either q0 = 0 and 2q ≥ 1 or q0 > 0, then both of the integrals in the b2 right-hand side of [7.15] are infinite as n → ∞. Applying L’Hopital’s rule, we obtain [7.11].



The theorem is thus proved.

R EMARK 7.3.– Since c(u) is positive by our assumption, the surplus of the insurance company becomes infinitely large in finite time a.s. if the premium intensity is a quadratic function and the claims do not arrive. Note that the time interval between two successive claims can be large enough with positive probability. Hence, the process Xt (x) t≥0 that follows [7.3] goes to +∞ with positive probability. It is clear that ruin  does  not occur in this case. Consequently, from now on we can consider Xt (x) t≥0 up to the minimum from the ruin time and its possible explosion. 7.3. Existence and uniqueness theorem now equation [7.3]. Let t∗ (x) be a possible explosion time of  Consider  Xt (x) t≥0 , i.e. / (−∞, +∞)}. t∗ (x) = inf{t ≥ 0 : Xt (x) ∈ To shorten notation, we let t∗ stand for t∗ (x). T HEOREM 7.2.– If c(u) is a locally Lipschitz continuous function on R, then [7.3] has a unique strong solution up to the time τ ∧ t∗ . P ROOF.– Since the process (Nt )t≥0 is homogeneous, it has only a finite number of jumps on any finite time interval a.s. To prove the theorem, we study [7.3] between two successive jumps of (Nt )t≥0 . Let us first consider [7.3] on the time interval [τ0 , τ1 ), where τ0 = 0. It can be rewritten as  t  t   X t = Xτ0 + c(Xs ) + aXs ds + b Xs dWs , τ0 ≤ t < τ1 . [7.16] τ0

τ0

176

Ruin Probabilities

By theorem A.8, the locally Lipschitz continuity of c(u) + au and bu on R implies the existence of a unique strong solution to [7.16] on [τ0 , τ1 ∧ t∗ ). Moreover, the comparison theorem (see section A.3.2) shows that this solution is not less than the solution to the equation  t  t Xt = Xτ0 + a Xs ds + b Xs dWs , τ0 ≤ t ≤ τ1 ∧ t∗ , [7.17] τ0

τ0

a.s. Since the solution to [7.17] is positive, so is the solution to [7.16] on [τ0 , τ1 ∧ t∗ ). Hence, limt↑t∗ Xt = +∞ if t∗ ≤ τ1 . Thus, ruin does not occur up to the time τ1 ∧ t∗ . If t∗ ≤ τ1 , then the theorem follows. Otherwise Xτ1− < +∞ and we set Xτ1 = Xτ1− − Y1 . Next, if Xτ1 < 0, then τ = τ1 , which completes the proof. Otherwise we consider [7.3] on the time interval [τ1 , τ2 ). We rewrite it as  t  t   c(Xs ) + aXs ds + b Xt = Xτ1 + Xs dWs , τ1 ≤ t < τ2 . [7.18] τ1

τ1

Repeating the same arguments, we conclude that [7.18] has a unique strong solution on [τ1 , τ2 ∧ t∗ ) and ruin does not occur up to the time τ2 ∧ t∗ . Thus, we have proved that [7.3] has a unique strong solution on [0, τ2 ∧ t∗ ), which is our assertion if t∗ ≤ τ2 . For the case t∗ > τ2 , we set Xτ2 = Xτ2− −Y2 . Next, if Xτ2 < 0, then τ = τ2 , which proves the theorem. Otherwise we continue in this fashion and prove the theorem by induction.  R EMARK 7.4.– Note that if t∗ < ∞, then the proof of theorem 7.2 implies limt↑t∗ Xt = +∞ and [7.3] also holds for t = t∗ provided that we let both of its sides be formally equal to +∞. In this case, we formally set Xt∗ = +∞. In addition, if τ < ∞, then we set Xτ = Xτi− − Yi , where i is the number of the claim that caused the ruin, and [7.3] also holds for t = τ . 7.4. Supermartingale property for the exponential process   ˜ t (x) ˜ t (x) = Xt∧τ ∧t∗ (x). Let the stopped process X be defined by X t≥0   ˜ t (x) is a solution to [7.3] provided that so is Note that X t≥0   ˜ t (x) = +∞ for Xt (x) 0≤t