Monte Carlo Methods Utilizing Mathematica®: Applications in Inverse Transform and Acceptance-Rejection Sampling 3031232933, 9783031232930
This book provides practical demonstrations of how to carry out definite integrals with Monte Carlo methods using Mathem
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Table of contents :
1
Preface
Contents
1 (1)
Part I Monte Carlo Methods Using Inverse Transform Sampling Utilizing Mathematica
978-3-031-23294-7_1
1 Introduction for Part I
1.1 Random Variable
1.2 Continuous Random Variable
1.3 Uniform Random Variable
1.4 Normal or Gaussian Random Variable
1.5 Generating Random Variates by Inverse Transform Method
1.6 Variance Reduction and Importance Sampling
978-3-031-23294-7_2
2 Evaluation of Definite Integrals Using Inverse Transform Sampling Utilizing Mathematica
2.1 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example I
2.2 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example II
2.3 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example III
2.4 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example IV
2.5 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example V
1 (2)
Part II Monte Carlo Methods Using Acceptance-Rejection Sampling Utilizing Mathematica
978-3-031-23294-7_3
3 Introduction for Part II
3.1 Random Variable
3.2 Continuous Random Variable
3.3 Uniform Random Variable
3.4 Normal or Gaussian Random Variable
3.5 Generating Random Variates by Acceptance-Rejection Method
3.6 Variance Reduction and Importance Sampling
978-3-031-23294-7_4
4 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling Utilizing Mathematica
4.1 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example I
4.2 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example II
4.3 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example III
4.4 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example IV
4.5 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example V
978-3-031-23294-7_5
5 Variational Monte Carlo Method Applied to Ground State of Simple Harmonic Oscillator Using Acceptance-Rejection Sampling Utilizing Mathematica
5.1 The Variational Method of Quantum Mechanics Applied to Ground State of Any Quantum Mechanical System
5.2 The Variational Method of Quantum Mechanics Applied to Ground State of Simple Harmonic Oscillator
5.3 Ground State Energy of Simple Harmonic Oscillator Using Variational Quantum Monte Carlo Method and Acceptance-Rejection Sampling
1 (3)
Concluding Remarks
Appendix: Handouts for Computational Lab and Automated Collection of Accepted Random Variates
A.1 Experiment I: Monte Carlo Integration Using Inverse Transform Sampling Using (a) Mathematica and (b) MS Excel
A.2 Experiment II: Monte Carlo Integration Using Acceptance-Rejection Sampling, Gaussian Pdf and Mathematica
A.3 Automated Collection of Accepted Random Variates
References
Citation preview
Synthesis Lectures on Mathematics & Statistics Series Editor Steven G. Krantz, Department of Mathematics, Washington University, Saint Louis, MO, USA
This series includes titles in applied mathematics and statistics for cross-disciplinary STEM professionals, educators, researchers, and students. The series focuses on new and traditional techniques to develop mathematical knowledge and skills, an understanding of core mathematical reasoning, and the ability to utilize data in specific applications.
Sujaul Chowdhury
Monte Carlo Methods Utilizing Mathematica® Applications in Inverse Transform and Acceptance-Rejection Sampling
Sujaul Chowdhury Department of Physics Shahjalal University of Science and Technology Sylhet, Bangladesh
ISSN 1938-1743 ISSN 1938-1751 (electronic) Synthesis Lectures on Mathematics & Statistics ISBN 978-3-031-23293-0 ISBN 978-3-031-23294-7 (eBook) https://doi.org/10.1007/978-3-031-23294-7 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland
Preface
This book on Monte Carlo Methods Using Inverse Transform and Acceptance-Rejection Sampling Utilizing Mathematica is intended for undergraduate students of Mathematics, Statistics, and Physics who wish to know how Monte Carlo methods work with calculations done using a user-friendly program Mathematica. All treatments have been done as less manually as practicable. All calculations except for gathering accepted values of random variate in acceptance-rejection sampling have been done using Mathematica. Definite integrals of a number of functions F(x) have been evaluated using uniform, linear, Gaussian and exponential probability density functions p(x). It is shown that results agree with known exact values better if p(x) is proportional to F(x). Deviation from the proportionality has been shown to result in worse agreement. All necessary background materials are covered. A separate chapter has been dedicated to variational Monte Carlo method applied to ground state of simple harmonic oscillator. Necessary background on variational method has been covered so that readers need not consult Quantum Mechanics textbooks. The book is on Monte Carlo methods. These are numerical methods for Computational Physics. These are parts of the syllabus for undergraduate students of Mathematics and Physics for the course titled “Computational Physics”. Besides the four referenced books, this is the only book to teach how basic Monte Carlo methods work. This is possibly the 1st Mathematica-based book on the topic. Pedagogical features: the book contains practical demonstrations of how to carry out definite integrals using Monte Carlo methods using Mathematica and using random variates sampled by inverse transform method and acceptance-rejection method. Moreover, the book contains a chapter on Variational Quantum Monte Carlo method applied to simple harmonic oscillator.
v
vi
Preface
Key words for the book are: Monte Carlo methods, Mathematica, basic Monte Carlo integration, Variational Quantum Monte Carlo method, simple harmonic oscillator, inverse transform sampling, and acceptance-rejection sampling.
Sylhet, Bangladesh 2022
Sujaul Chowdhury
Contents
Part I
Monte Carlo Methods Using Inverse Transform Sampling Utilizing Mathematica
1 Introduction for Part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Continuous Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Uniform Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Normal or Gaussian Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Generating Random Variates by Inverse Transform Method . . . . . . . . . . . . 1.6 Variance Reduction and Importance Sampling . . . . . . . . . . . . . . . . . . . . . . . 2 Evaluation of Definite Integrals Using Inverse Transform Sampling Utilizing Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Evaluation of Definite Integrals Using Inverse Transform Sampling: Example V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Part II
3 3 3 4 4 6 9 13 13 17 23 27 31
Monte Carlo Methods Using Acceptance-Rejection Sampling Utilizing Mathematica
3 Introduction for Part II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Continuous Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Uniform Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Normal or Gaussian Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45 45 45 46 46 vii
viii
Contents
3.5 Generating Random Variates by Acceptance-Rejection Method . . . . . . . . . 3.6 Variance Reduction and Importance Sampling . . . . . . . . . . . . . . . . . . . . . . . 4 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling Utilizing Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Variational Monte Carlo Method Applied to Ground State of Simple Harmonic Oscillator Using Acceptance-Rejection Sampling Utilizing Mathematica . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 The Variational Method of Quantum Mechanics Applied to Ground State of Any Quantum Mechanical System . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The Variational Method of Quantum Mechanics Applied to Ground State of Simple Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Ground State Energy of Simple Harmonic Oscillator Using Variational Quantum Monte Carlo Method and Acceptance-Rejection Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49 51 53 53 67 81 96 102
111 111 113
116
Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
133
Appendix: Handouts for Computational Lab and Automated Collection of Accepted Random Variates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
135
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
149
Part I Monte Carlo Methods Using Inverse Transform Sampling Utilizing Mathematica
1
Introduction for Part I
This chapter provides background for Chap. 2. It introduces readers to continuous, uniform and Gaussian random variables. Inverse transform method of generating random variate for given probability distribution function, and concept of importance sampling are also introduced.
1.1
Random Variable
In the context of Monte Carlo methods, to define a random variable, we must indicate the values that the variable can assume and probabilities of occurrence of these values. We do not know value of the variable in any given case; but we do know the values that the variable can assume and the probabilities of occurrence of these values.
1.2
Continuous Random Variable
A random variable x is called continuous if it can assume any fractional value in a certain interval, say a to b. Besides specifying the interval containing all its possible values, we need to state a function p(x) called probability density function or probability distribution function. We have (1) (2) (3) (4)
p(x) ≥ 0 for a ≤ x ≤ b, the product p(x)dx is probability that x ≤ x ≤ x + dx, b p(x)d x is probability that a ≤ x ≤ b where a < a and b < b, ab a p(x)d x = 1 indicating the surety that x lies in the interval a to b.
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Chowdhury, Monte Carlo Methods Utilizing Mathematica®, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-23294-7_1
3
4
1 Introduction for Part I
As such, average or so called expectation value of x is given by A x = b and average value of any function of x, say f (x), is A f = a f (x) p(x)d x.
1.3
b a
x p(x)d x,
Uniform Random Variable
A random variable u in the interval 0 ≤ x ≤ 1 having constant, say C, probability density function p(x) = C is said to be a uniform random variable. The requirement that 1 1 0 p(x)d x = 1 gives 0 Cd x = 1 which gives C = 1. Thus probability density function of uniform random variable in the interval 0 to 1 is p(x) = 1. Program number 1.1 n=654321 SeedRandom[n] Table[{i=i+1,u[i] = RandomReal[]},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","u[i]"}}] ListPlot[Table[{u[i],1},{i,0,99,1}],Frame->True, FrameLabel->{"u [i]","1"}] Program number 1.1 in Mathematica 6.0 can provide us 100 uniform random numbers ui ’s in the interval 0–1 and the result is shown in Table 1.1 by rounding the numbers ui ’s to 4 decimal place in Microsoft Excel. In the program, value of n has been chosen by trial and error so that the random numbers ui ’s are almost uniformly distributed in the interval 0–1 as vindicated by Fig. 1.1 which displays the numbers ui ’s against a stagnant integer 1. This is a quality check of random numbers ui ’s. A benefit of using the command SeedRandom[n] in Program number 1.1 is that it enables us to get the values of ui ’s reproducibly.
1.4
Normal or Gaussian Random Variable
Probability density function p(x) for normal random variable x is given by 1 1 x −a 2 p(x) = √ exp − 2 σ σ 2π
(1.1)
defined for −∝ < x < +∝ . Average value of x is a and variance of x is σ 2 . According a+3σ to so called rule of 3 sigma, a−3σ p(x)d x ≈ 0.997.
1.4
Normal or Gaussian Random Variable
5
Table 1.1 Showing 100 uniform random numbers ui ’s (in the interval 0–1) obtained using Program number 1.1 in Mathematica i
ui
i
ui
i
ui
i
ui
1
0.8618
26
0.9276
51
0.7833
76
0.8258
2
0.4287
27
0.0504
52
0.7877
77
0.4101
3
0.8596
28
0.3510
53
0.7012
78
0.6742
4
0.0342
29
0.1510
54
0.5937
79
0.2088
5
0.7597
30
0.1761
55
0.6616
80
0.8631
6
0.4609
31
0.3796
56
0.5202
81
0.2042
7
0.4189
32
0.5239
57
0.9024
82
0.3027
8
0.4746
33
0.3586
58
0.0050
83
0.0504
9
0.3799
34
0.4202
59
0.2183
84
0.8463
10
0.1687
35
0.5641
60
0.2361
85
0.4134
11
0.1899
36
0.1372
61
0.4477
86
0.2451
12
0.0364
37
0.0774
62
0.7308
87
0.5565
13
0.3335
38
0.8924
63
0.3429
88
0.0084
14
0.4257
39
0.9848
64
0.2260
89
0.3532
15
0.3860
40
0.4854
65
0.1079
90
0.7547
16
0.2445
41
0.3080
66
0.4879
91
0.1188
17
0.5881
42
0.4945
67
0.1655
92
0.5403
18
0.2322
43
0.9390
68
0.1218
93
0.5235
19
0.6743
44
0.5923
69
0.3544
94
0.8228
20
0.9089
45
0.1921
70
0.3523
95
0.5526
21
0.8759
46
0.3830
71
0.3895
96
0.8279
22
0.3658
47
0.6754
72
0.0911
97
0.6770
23
0.2945
48
0.2944
73
0.9631
98
0.7266
24
0.6719
49
0.9737
74
0.5184
99
0.4398
25
0.9104
50
0.4501
75
0.6328
100
0.2243
Program number 1.2 n=654321 SeedRandom[n] Table[{i=i+1,g[i] = RandomReal[NormalDistribution[0,1]]}, {i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","g[i]"}}]
6
1 Introduction for Part I
2.0
1
1.5
1.0
0.5
0.0 0.0
0.2
0.4
0.6
0.8
1.0
Fig. 1.1 Uniform random numbers ui ’s of Table 1.1 plotted against a stagnant integer 1 to vindicate that ui ’s are indeed almost uniformly distributed in the interval 0–1
ListPlot[Table[{g[i],1},{i,0,99,1}],Frame->True, FrameLabel->{"g [i]","1"}] For a = 0, σ = 1, Program number 1.2 in Mathematica 6.0 can provide us 100 Gaussian random numbers gi ’s in the interval −3 to +3 and the result is shown in Table 1.2 by rounding the numbers gi ’s to 4 decimal place in Microsoft Excel. In the program, value of n has been chosen by trial and error so that the random numbers gi ’s are so distributed in the interval −3 to +3 that most of the numbers are close to the average value x = 0 as vindicated by Fig. 1.2 which displays the numbers gi ’s against a stagnant integer 1. This is a quality check of random numbers gi ’s. A benefit of using the command SeedRandom[n] in Program number 1.2 is that it enables us to get the values of gi ’s reproducibly.
1.5
Generating Random Variates by Inverse Transform Method
Values of random variable η corresponding to a given probability density function p(x) defined for a ≤ x ≤ b can be obtained by solving Eq. (1.2): η p(x)d x = u (1.2) a
1.5
Generating Random Variates by Inverse Transform Method
7
Table 1.2 Showing 100 Gaussian random numbers gi ’s (in the interval − 3 to + 3) obtained using program number 1.2 in Mathematica. For a = 0, σ = 1 i
gi
i
gi
i
gi
i
1
−0.4915
26
0.1642
51
0.0250
76
0.2147
2
0.5374
27
−0.7006
52
0.0436
77
−0.1304
3
−0.7192
28
−0.9017
53
0.6939
78
1.2606
4
−1.3024
29
0.4529
54
−0.5178
79
1.8331
5
0.6802
30
0.1524
55
0.9525
80
1.2424
6
1.7751
31
−0.1529
56
−1.5620
81
−0.6542
7
−1.3234
32
0.2197
57
0.7447
82
−0.6587
8
0.0478
33
−2.1041
58
0.5326
83
−0.7827
9
0.1149
34
1.3676
59
0.3137
84
1.1034
10
0.7463
35
−0.8635
60
0.5035
85
−0.1103
11
−0.3424
36
1.1543
61
−1.0072
86
−0.1061
12
−0.7370
37
−0.2724
62
1.6373
87
−0.2752
13
0.3892
38
0.4385
63
−0.4160
88
0.6360
14
−1.4486
39
−0.6121
64
−0.2576
89
0.6611
15
0.8703
40
1.1545
65
2.3177
90
0.6861
16
−1.3761
41
−0.5795
66
0.0741
91
1.1652
17
−1.2559
42
1.3903
67
−0.5673
92
2.6803
18
0.6966
43
0.0406
68
−0.0757
93
−0.8103
19
1.7650
44
1.0811
69
0.1267
94
−1.3781
20
−0.1744
45
0.0427
70
−1.0962
95
−0.2635
21
−1.5339
46
−1.9984
71
−0.2686
96
−0.3182
22
−0.2967
47
0.5023
72
0.7321
97
0.7618
23
−1.3473
48
0.5118
73
2.0250
98
0.9291
24
−0.2439
49
−0.1296
74
−0.5563
99
−0.9932
25
−0.2195
50
0.2063
75
0.7393
100
gi
0.4787
where u is uniform random variable distributed in the interval 0 ≤ x ≤ 1, see Table 1.1. We now verify this. x Let
y(x) =
p(x)d x a
Since p(x) > 0, and
b a
p(x)d x = 1, we have
(1.3)
8
1 Introduction for Part I
2.0
1
1.5
1.0
0.5
0.0 2
1
0
1
2
g Fig. 1.2 Gaussian random numbers gi ’s of Table 1.2 plotted against a stagnant integer 1 to vindicate that gi ’s are so distributed in the interval −3 to +3 that most of the numbers are close to the average value x = 0. For a = 0, σ = 1
a y(a) =
p(x)d x = 0 a
b p(x)d x = 1 emphasizing that p(x) is normalized,
y(b) = a
and y (x) = p(x) > 0. This means that y monotonically increases from 0 to 1 for the interval a < x < b. See Fig. 1.3. Straight line y = u where 0 < u < 1 intersects the curve y(x) at Q giving us the η value of η which satisfies Eq. (1.2): a p(x)d x = u. p(a < η < b ) = p(y(a ) < u < y(b )) = y(b ) − y(a ) b
a p(x)d x −
= a
b p(x)d x =
a
p(x)d x. a
1.6 Variance Reduction and Importance Sampling
9
Fig. 1.3 Figure to help explain inverse transform method of obtaining random variate
η Thus random variable η which obeys Eq. (1.2): a p(x)d x = u has the probability density function p(x). We need to evaluate this integral and obtain values of η for values of uniform random variable u.
1.6
Variance Reduction and Importance Sampling
Let us consider a definite integral b I =
F(x)d x
(1.4)
a
b Let us re-write Eq. (1.4) as I = a
F(x) p(x)d x = p(x)
b f (x) p(x)d x
(1.5)
a
where f (x) = F(x)/p(x). As discussed in Sect. 1.2, average value of any function of x, say f (x), is given by b A f = a f (x) p(x)d x where p(x) is a normalized probability density function. Thus according to Eq. (1.5), value I of the integral is average value of F/p. We can choose any functional form for p(x), but as proved below, choosing p(x) as proportional to F(x) ensures that the variance of f (x) = F/p is small. This will ensure a b better value of the average and hence a better value of the integral I = a F(x)d x.
10
1 Introduction for Part I
Variance of f (x) is square of standard deviation. Variance of f (x) is given by V [ f (x)] = A[ ( f − A f )2 ]
(1.6)
where A stands for average value. Thus V f (x) = A[ f 2 − 2 f A f + A2f ] = A f 2 − 2 A f A f + A2f = A f 2 − A2f b or, V f (x) =
⎛ f 2 (x) p(x)d x − ⎝
a
b
⎞2 f (x) p(x)d x ⎠ =
b
a
a
⎞2 ⎛ b b 2 F = dx − ⎝ F dx⎠ p a
⎛ b ⎞2 F2 F pd x − ⎝ pd x ⎠ p2 p a
a
b
F2 dx − I2 p
V f (x) =
Thus the variance is
(1.7)
a
If p(x) is taken as
p(x) =
|F(x)| b
(1.8)
|F(x)|d x
a
the variance given by Eq. (1.7) becomes V f (x) = ⎛ or, V f (x) = ⎝
b a
⎞⎛ |F(x)|d x ⎠ ⎝
b a
⎞
b a
F2
|F(x)| b a |F(x)|d x
⎛
|F(x)|d x ⎠ − I 2 = ⎝
b
dx − I2 ⎞2 |F(x)|d x ⎠ − I 2
(1.9)
a
b Equation (1.9) in conjunction with Eq. (1.4): I = a F(x)d x reveals that the variance vanishes if the integrand F(x) does not change sign. If it does, the variance will be small if condition laid by Eq. (1.8) is met. Thus the probability density function p(x) should be proportional to the integrand |F(x)|. This is the so called importance sampling. Looking back at Eq. (1.5), we find that value of the integral is approximately deter N F(xi ) mined by the value of the average I = N1 i=1 p(xi ) . As recommended by importance sampling, we need to take p(x) = C F(x) where C is a constant; we need to normalb b . Thus p(x) = ize p(x) first; a p(x)d x = 1 gives a C F(x)d x = 1 or, C = b 1 a F(x)d x b N F(xi ) N F(xi ) F(x) 1 b . The sum N1 i=1 = a F(x)d x. Thus the F(xi ) i=1 p(xi ) becomes N a
F(x)d x
b a F(x)d x
1.6 Variance Reduction and Importance Sampling
11
N F(xi ) sum N1 i=1 p(xi ) equals the integral if probability density function p(x) is taken proportional to the integrand |F(x)|. Slight variation of p(x) from proportionality with |F(x)| N F(xi ) will result in slight difference of N1 i=1 p(xi ) from actual value of the integral I = b a F(x)d x.
2
Evaluation of Definite Integrals Using Inverse Transform Sampling Utilizing Mathematica
After getting required background in Chap. 1, this chapter provides practical demonstrations of how to carry out definite integrals using Monte Carlo method using random variates obtained by inverse transform method. Random variates have been obtained corresponding to probability distribution functions uniform, linear and exponential in the variable. Programs written in Mathematica have been used in this regard.
2.1
Evaluation of Definite Integrals Using Inverse Transform Sampling: Example I
As Example I, we now take up the integral b I =
π F(x)d x =
a
Sin(x)d x
(2.1)
0
where F(x) = Sin(x). We re-write Eq. (2.1) as b I =
F(x) p(x)d x p(x)
(2.2)
a
which, as discussed in Sect. 1.6, implies that average value of F/p is the value of the integral, i.e. I =
N 1 F(xi ) N p(xi )
(2.3)
i=1
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Chowdhury, Monte Carlo Methods Utilizing Mathematica®, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-23294-7_2
13
14
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
where x i ’s are random values of x in the interval a < x < b obeying probability density function p(x). Here p(x) is a suitable probability density function which, as demonstrated in Sect. 1.6, at least should follow F(x), if not be proportional to F(x). See Fig. 2.1 in which we have plotted F(x) = Sin(x) along with normalized probability density function of a uniform random variable in the interval 0 to π given by p(x) = 1/π. Let p(x) = C, a constant. π Normalization requires that 0 Cd x = 1 which gives C = 1/π. Uniform p(x) = 1/π is not too different from a slowly varying function of x like Sin(x). Hence p(x) = 1/π is a good choice. We now generate random variate η obeying p(x) = 1/π in the interval 0 < x < π using η inverse transform sampling. Values of η are obtainable from Eq. (1.2): a p(x)d x = u. As such η = π u where u is uniform random variable of Table 1.1 in the interval 0 < x < 1. Using Program number 2.1 and resulting Table 2.1, we now evaluate the integral with the result 2.127 rather than 2. The difference is attributable to the fact that F(x) and p(x)
1.0
0.8
0.6
0.4
0.2
0.0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
x Fig. 2.1 Showing F(x) = Sin(x) as undashed curve and p(x) = 1/π as dashed curve obtained using the command Plot[{Sin[x],1/Pi},{x,0,Pi},Frame->True, FrameLabel->{"x","F(x) = Sin(x) or p(x) = 1/ \[Pi]" }, PlotStyle->{{Black},{Dashed,Black}}] in Mathematica 6.0.
2.1
Evaluation of Definite Integrals Using Inverse Transform …
15
used are not proportional to each other. That random variate η is uniformly distributed in the interval 0 to π is evidenced by Fig. 2.2 which displays values of η against stagnant integer 1. Program number 2.1 n=654321 SeedRandom[n] Table[{i=i+1,u[i]=RandomReal[],eta[i]=Pi*u[i], y[i]=Sin[eta[i]]},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","u[i]","eta[i]","y[i]"}}] Integral=Sum[Pi*y[i]/100,{i,1,100}] ListPlot[Table[{eta[i],1},{i,0,99,1}],Frame->True, FrameLabel->{"\[Eta] [i]","1"}] We now take up a linear variation for p(x) given by p(x) = C x where C is a constant. π Normalization requires that 0 C x d x = 1 which gives C = 2/π2 . Thus normalized p(x) = 2x/π2 . See Fig. 2.3 in which we have plotted F(x) = Sin(x) along with the normalized linear probability density function p(x) = 2x/π2 . The linear probability density function p(x) = 2x/π2 is not too different from a slowly varying function of x like Sin(x). Hence it is a good choice. We now generate random variate η obeying p(x) = 2x/π2 in the interval 0 < x < π using η inverse transform sampling. Values of η are obtainable from Eq. (1.2): a p(x)d x = u. As such η = π u1/2 where u is uniform random variable of Table 1.1 in the interval 0 < x < 1. Using Program number 2.2 and resulting Table 2.2, we now evaluate the integral with the result 2.104 rather than 2. The difference is attributable to the fact that F(x) and p(x) used are not proportional to each other. That random variate η is distributed as p(x) = 2x/π2 in the interval 0 to π is evidenced by Fig. 2.4 which displays values of η against stagnant integer 1. We find that indeed there are more variates for larger values of η. Program number 2.2 n=654321 SeedRandom[n] Table[{i=i+1,u[i]=RandomReal[],eta[i]=Pi*Sqrt[u[i]], y[i]=(Sin[eta[i]])/eta[i]},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","u[i]","eta[i]","y[i]"}}] Integral=Sum[Pi^2*y[i]/(2*100),{i,1,100}] ListPlot[Table[{eta[i],1},{i,0,99,1}],Frame->True, FrameLabel->{"\[Eta] [i]","1"}]
16
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
N F(xi ) Table 2.1 Showing ui , ηi = π ui , yi = Sin(ηi ) from which we get I = N1 i=1 p(xi ) = 1 N F(ηi ) = π N Sin(η ) as an approximate value of the integral I = b F(x)d x = i i=1 a N Nπ i=1 1/π Sin(x)d x = 2.127 whereas exact value is 2. Using p(x) = 1/π. Using prog. 2.1 0 i
ui
η i = π ui
yi = Sin(ηi )
i
ui
η i = π ui
yi = Sin(ηi )
1
0.8618
2.7075
0.4206
26
0.9276
2.9141
0.2255
2
0.4287
1.3469
0.9750
27
0.0504
0.1583
0.1577
3
0.8596
2.7005
0.4269
28
0.3510
1.1026
0.8924
4
0.0342
0.1076
0.1074
29
0.1510
0.4745
0.4569
5
0.7597
2.3865
0.6853
30
0.1761
0.5534
0.5256
6
0.4609
1.4478
0.9924
31
0.3796
1.1926
0.9293
7
0.4189
1.3160
0.9677
32
0.5239
1.6460
0.9972
8
0.4746
1.4909
0.9968
33
0.3586
1.1266
0.9030
9
0.3799
1.1935
0.9297
34
0.4202
1.3201
0.9687
10
0.1687
0.5300
0.5055
35
0.5641
1.7723
0.9798
11
0.1899
0.5967
0.5619
36
0.1372
0.4309
0.4177
12
0.0364
0.1145
0.1142
37
0.0774
0.2431
0.2407
13
0.3335
1.0476
0.8662
38
0.8924
2.8037
0.3315
14
0.4257
1.3373
0.9729
39
0.9848
3.0938
0.0478
15
0.3860
1.2128
0.9366
40
0.4854
1.5248
0.9989
16
0.2445
0.7681
0.6947
41
0.3080
0.9675
0.8235
17
0.5881
1.8477
0.9619
42
0.4945
1.5535
0.9999
18
0.2322
0.7295
0.6665
43
0.9390
2.9501
0.1904
19
0.6743
2.1184
0.8538
44
0.5923
1.8607
0.9583
20
0.9089
2.8555
0.2822
45
0.1921
0.6035
0.5675
21
0.8759
2.7517
0.3801
46
0.3830
1.2032
0.9332
22
0.3658
1.1492
0.9125
47
0.6754
2.1218
0.8520
23
0.2945
0.9253
0.7988
48
0.2944
0.9248
0.7985
24
0.6719
2.1108
0.8577
49
0.9737
3.0590
0.0825
25
0.9104
2.8601
0.2778
50
0.4501
1.4139
0.9877
51
0.7833
2.4609
0.6293
76
0.8258
2.5943
0.5204
52
0.7877
2.4748
0.6185
77
0.4101
1.2882
0.9603
53
0.7012
2.2028
0.8068
78
0.6742
2.1181
0.8539
54
0.5937
1.8653
0.9569
79
0.2088
0.6560
0.6099
55
0.6616
2.0785
0.8739
80
0.8631
2.7115
0.4170 (continued)
2.2
Evaluation of Definite Integrals Using Inverse Transform …
17
Table 2.1 (continued) i
ui
η i = π ui
yi = Sin(ηi )
i
ui
η i = π ui
yi = Sin(ηi )
56
0.5202
1.6341
0.9980
81
0.2042
0.6416
0.5984
57
0.9024
2.8351
0.3017
82
0.3027
0.9510
0.8140
58
0.0050
0.0157
0.0157
83
0.0504
0.1582
0.1575
59
0.2183
0.6857
0.6332
84
0.8463
2.6586
0.4644
60
0.2361
0.7417
0.6755
85
0.4134
1.2986
0.9632
61
0.4477
1.4065
0.9865
86
0.2451
0.7701
0.6962
62
0.7308
2.2957
0.7485
87
0.5565
1.7483
0.9843
63
0.3429
1.0771
0.8806
88
0.0084
0.0265
0.0265
64
0.2260
0.7100
0.6519
89
0.3532
1.1096
0.8955
65
0.1079
0.3390
0.3326
90
0.7547
2.3710
0.6966
66
0.4879
1.5328
0.9993
91
0.1188
0.3732
0.3646
67
0.1655
0.5199
0.4968
92
0.5403
1.6973
0.9920
68
0.1218
0.3827
0.3735
93
0.5235
1.6447
0.9973
69
0.3544
1.1133
0.8972
94
0.8228
2.5849
0.5284
70
0.3523
1.1068
0.8943
95
0.5526
1.7361
0.9864
71
0.3895
1.2236
0.9403
96
0.8279
2.6008
0.5148
72
0.0911
0.2863
0.2824
97
0.6770
2.1269
0.8493
73
0.9631
3.0256
0.1157
98
0.7266
2.2826
0.7572
74
0.5184
1.6286
0.9983
99
0.4398
1.3816
0.9822
75
0.6328
1.9880
0.9142
100
0.2243
0.7046
0.6477
2.2
Evaluation of Definite Integrals Using Inverse Transform Sampling: Example II
As Example II, we now take up the integral +π/2
b I =
F(x)d x =
Cos(x)d x
(2.4)
−π/2
a
where F(x) = Cos(x). We re-write Eq. (2.4) as b I = a
F(x) p(x)d x p(x)
(2.5)
18
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
2.0
1
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
Fig. 2.2 Showing that random variate η is uniformly distributed in the interval 0 to π
which, as discussed in Sect. 1.6, implies that average value of F/p is the value of the integral, i.e. N 1 F(xi ) I = N p(xi )
(2.6)
i=1
where x i ’s are random values of x in the interval a < x < b obeying probability density function p(x). Here p(x) is a suitable probability density function which, as demonstrated in Sect. 1.6, at least should follow F(x), if not be proportional to F(x). See Fig. 2.5 in which we have plotted F(x) = Cos(x) along with normalized probability density function of a uniform random variable in the interval −π/2 to +π/2 given by p(x) = 1/π. Let p(x) = C, a +π/2 constant. Normalization requires that −π/2 Cd x = 1 which gives C = 1/π. Uniform p(x) = 1/π is not too different from a slowly varying function of x like Cos(x). Hence p(x) = 1/π is a good choice. We now generate random variate η obeying p(x) = 1/π in the interval −π/2 < x < +π/2 using inverse transform sampling. Values of η are obtainable from Eq. (1.2): η a p(x)d x = u. As such η = πu −π/2 where u is uniform random variable of Table 1.1 in the interval 0 < x < 1.
2.2
Evaluation of Definite Integrals Using Inverse Transform …
19
1.0
0.8
0.6
0.4
0.2
0.0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
x Fig. 2.3 Showing F(x) = Sin(x) as undashed curve and p(x) = 2x/π2 as dashed curve obtained using the command Plot[{Sin[x],(2*x)/(Pi^2)},{x,0,Pi},Frame->True, FrameLabel->{"x","F(x) = Sin(x) or p(x) "}, PlotStyle->{{Black},{Dashed,Black}}] in Mathematica 6.0.
Using Program number 2.3 and resulting Table 2.3, we now evaluate the integral with the result 2.127 rather than 2. The difference is attributable to the fact that F(x) and p(x) used are not proportional to each other. That random variate η is uniformly distributed in the interval −π/2 < x < +π/2 is evidenced by Fig. 2.6 which displays values of η against stagnant integer 1. Program number 2.3 n=654321 SeedRandom[n] Table[{i=i+1,u[i]=RandomReal[],eta[i]=Pi*u[i]-Pi/2, y[i]=Cos[eta[i]]},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","u[i]","eta[i]","y[i]"}}] Integral=Sum[Pi*y[i]/100,{i,1,100}] ListPlot[Table[{eta[i],1},{i,0,99,1}], Frame->True,FrameLabel->{"\[Eta] [i]","1"}]
20
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
N F(xi ) Table 2.2 Showing ui , ηi = π ui 1/2 , Sin(ηi )/ηi from which we get I = N1 i=1 p(xi ) = 1 N Sin(ηi ) = π 2 N Sin(ηi ) as an approximate value of the integral I = b F(x)d x = i=1 2ηi /π 2 i=1 a ηi N 2N π 2 . Using prog. 2.2 Sin(x)d x = 2.104 whereas exact value is 2. Using p(x) = 2x/π 0 i
ui
ηi = π ui 1/2
yi = Sin(ηi )/ηi
i
ui
ηi = π ui 1/2
yi = Sin(ηi )/ηi
1
0.8618
2.9165
0.0765
26
0.9276
3.0257
0.0382
2
0.4287
2.0570
0.4298
27
0.0504
0.7053
0.9191
3
0.8596
2.9127
0.0779
28
0.3510
1.8611
0.5148
4
0.0342
0.5813
0.9446
29
0.1510
1.2209
0.7694
5
0.7597
2.7382
0.1434
30
0.1761
1.3185
0.7344
6
0.4609
2.1327
0.3968
31
0.3796
1.9357
0.4826
7
0.4189
2.0333
0.4401
32
0.5239
2.2740
0.3354
8
0.4746
2.1642
0.3831
33
0.3586
1.8813
0.5061
9
0.3799
1.9364
0.4823
34
0.4202
2.0365
0.4388
10
0.1687
1.2903
0.7447
35
0.5641
2.3596
0.2986
11
0.1899
1.3692
0.7156
36
0.1372
1.1635
0.7892
12
0.0364
0.5998
0.9411
37
0.0774
0.8740
0.8775
13
0.3335
1.8142
0.5350
38
0.8924
2.9678
0.0583
14
0.4257
2.0497
0.4330
39
0.9848
3.1176
0.0077
15
0.3860
1.9520
0.4755
40
0.4854
2.1887
0.3724
16
0.2445
1.5534
0.6437
41
0.3080
1.7434
0.5651
17
0.5881
2.4093
0.2775
42
0.4945
2.2092
0.3635
18
0.2322
1.5139
0.6595
43
0.9390
3.0443
0.0319
19
0.6743
2.5798
0.2065
44
0.5923
2.4178
0.2739
20
0.9089
2.9951
0.0487
45
0.1921
1.3770
0.7126
21
0.8759
2.9402
0.0680
46
0.3830
1.9442
0.4789
22
0.3658
1.9001
0.4980
47
0.6754
2.5819
0.2057
23
0.2945
1.7050
0.5812
48
0.2944
1.7045
0.5814
24
0.6719
2.5751
0.2084
49
0.9737
3.1000
0.0134
25
0.9104
2.9975
0.0479
50
0.4501
2.1076
0.4078
51
0.7833
2.7805
0.1271
76
0.8258
2.8549
0.0991
52
0.7877
2.7883
0.1241
77
0.4101
2.0117
0.4495
53
0.7012
2.6307
0.1859
78
0.6742
2.5796
0.2066
54
0.5937
2.4208
0.2726
79
0.2088
1.4356
0.6902 (continued)
2.2
Evaluation of Definite Integrals Using Inverse Transform …
21
Table 2.2 (continued) i
ui
ηi = π ui 1/2
yi = Sin(ηi )/ηi
i
ui
ηi = π ui 1/2
yi = Sin(ηi )/ηi
55
0.6616
2.5553
0.2165
80
0.8631
2.9186
0.0758
56
0.5202
2.2658
0.3390
81
0.2042
1.4197
0.6964
57
0.9024
2.9844
0.0525
82
0.3027
1.7284
0.5714
58
0.0050
0.2219
0.9918
83
0.0504
0.7050
0.9192
59
0.2183
1.4677
0.6777
84
0.8463
2.8900
0.0861
60
0.2361
1.5264
0.6545
85
0.4134
2.0198
0.4460
61
0.4477
2.1021
0.4101
86
0.2451
1.5555
0.6428
62
0.7308
2.6856
0.1640
87
0.5565
2.3436
0.3055
63
0.3429
1.8395
0.5241
88
0.0084
0.2885
0.9862
64
0.2260
1.4935
0.6676
89
0.3532
1.8670
0.5123
65
0.1079
1.0321
0.8317
90
0.7547
2.7292
0.1468
66
0.4879
2.1944
0.3699
91
0.1188
1.0828
0.8157
67
0.1655
1.2781
0.7492
92
0.5403
2.3092
0.3203
68
0.1218
1.0966
0.8113
93
0.5235
2.2731
0.3358
69
0.3544
1.8702
0.5109
94
0.8228
2.8497
0.1010
70
0.3523
1.8647
0.5133
95
0.5526
2.3354
0.3090
71
0.3895
1.9606
0.4718
96
0.8279
2.8584
0.0977
72
0.0911
0.9484
0.8567
97
0.6770
2.5849
0.2044
73
0.9631
3.0831
0.0190
98
0.7266
2.6779
0.1670
74
0.5184
2.2619
0.3407
99
0.4398
2.0834
0.4183
75
0.6328
2.4991
0.2398
100
0.2243
1.4878
0.6698
We now take up a linear variation for p(x) given by p(x) = C x + d where C and d are parameters such that slope C = intercept d on y axis/(π/2); hence d = Cπ /2. As such p(x) +π/2 = C(x + π /2). Normalization of p(x) requires −π/2 p(x)d x = 1 or, C = 2/π2 . As such normalized probability density function is p(x) = 2x/π2 + 1/π. See Fig. 2.7 in which we have plotted F(x) = Cos(x) along with the normalized linear probability density function p(x) = 2x/π2 + 1/π. The linear probability density function p(x) = 2x/π2 +1/π is not too different from a slowly varying function of x like Cos(x). Hence it is a good choice. We now generate random variate η obeying p(x) = 2x/π2 + 1/π in the interval −π/2 < x < +π/2 using inverse transform sampling. Values of η are obtainable from Eq. (1.2): η η a p(x)d x = u or, −π/2 p(x)d x = u where u is uniform random variable of Table 1.1 in the interval 0 < x < 1. Thus we have η given by roots of the quadratic equation η2 + πη + (π2 /4−π2 u) = 0 which are η = −π/2 ± πu1/2 .
22
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
2.0
1
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
Fig. 2.4 Showing that random variate η is linearly distributed in the interval 0 to π. There are more variates for larger values of η
For η = −π/2 + πu1/2 , using Program number 2.4 and resulting Table 2.4, we now evaluate the integral with the result 2.104 rather than 2. The difference is attributable to the fact that F(x) and p(x) used are not proportional to each other. That random variate η is distributed as p(x) = 2x/π2 + 1/π in the interval −π/2 to +π/2 is evidenced by Fig. 2.8 which displays values of η against stagnant integer 1. We find that indeed there are more variates for larger values of η. Program number 2.4 n=654321 SeedRandom[n] Table[{i=i+1,u[i]=RandomReal[],eta[i]=-Pi/2+Pi*Sqrt[u[i]], y[i]=(Cos[eta[i]])/(1/Pi+2*eta[i]/Pi^2)},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","u[i]","eta[i]","y[i]"}}] Integral=(Sum[y[i],{i,1,100}])/100 ListPlot[Table[{eta[i],1},{i,0,99,1}],Frame->True,FrameLabel->{"\[Eta] [i]","1"}]
2.3
Evaluation of Definite Integrals Using Inverse Transform …
23
1.0
0.8
0.6
0.4
0.2
0.0 1.5
1.0
0.5
0.0
0.5
1.0
1.5
x Fig. 2.5 Showing F(x) = Cos(x) as undashed curve and p(x) = 1/π as dashed curve obtained using the command Plot[{Cos[x],1/Pi},{x,-Pi/2,Pi/2}, Frame->True,FrameLabel->{"x","F(x) = Cos(x) or p(x) "}, PlotStyle->{{Black},{Dashed,Black}}] in Mathematica 6.0.
2.3
Evaluation of Definite Integrals Using Inverse Transform Sampling: Example III
As Example III, we now take up the integral b I =
2 F(x)d x =
a
ex d x
(2.7)
0
where F(x) = ex . We re-write Eq. (2.7) as b I =
F(x) p(x)d x p(x)
(2.8)
a
which, as discussed in Sect. 1.6, implies that average value of F/p is the value of the integral, i.e.
24
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
N F(xi ) Table 2.3 Showing ui , ηi = π ui −π/2, Cos(ηi ) from which we get I = N1 i=1 p(xi ) = 1 N F(ηi ) = π N Cos(η ) as an approximate value of the integral I = b F(x)d x = i i=1 1/π i=1 a N N +π/2 −π/2 Cos(x)d x = 2.127 whereas exact value is 2. Using p(x) = 1/π. Using prog. 2.3 i
ui
1
0.8618
2 3
ηi = π ui −π/2
ηi = πui −π/2
yi = Cos(ηi )
i
ui
yi = Cos(ηi )
1.1367
0.4206
26
0.9276
1.3433
0.2255
0.4287
−0.2239
0.9750
27
0.0504
−1.4125
0.1577
0.8596
1.1297
0.4269
28
0.3510
−0.4682
0.8924
4
0.0342
−1.4632
0.1074
29
0.1510
−1.0963
0.4569
5
0.7597
0.8157
0.6853
30
0.1761
−1.0174
0.5256
6
0.4609
−0.1230
0.9924
31
0.3796
−0.3782
0.9293
7
0.4189
−0.2548
0.9677
32
0.5239
0.0752
0.9972
8
0.4746
−0.0799
0.9968
33
0.3586
−0.4442
0.9030
9
0.3799
−0.3773
0.9297
34
0.4202
−0.2507
0.9687
10
0.1687
−1.0408
0.5055
35
0.5641
0.2015
0.9798
11
0.1899
−0.9741
0.5619
36
0.1372
−1.1399
0.4177
12
0.0364
−1.4563
0.1142
37
0.0774
−1.3277
0.2407
13
0.3335
−0.5232
0.8662
38
0.8924
1.2329
0.3315
14
0.4257
−0.2335
0.9729
39
0.9848
1.5230
0.0478
15
0.3860
−0.3580
0.9366
40
0.4854
−0.0460
0.9989
16
0.2445
−0.8027
0.6947
41
0.3080
−0.6033
0.8235
17
0.5881
0.2769
0.9619
42
0.4945
−0.0173
0.9999
18
0.2322
−0.8413
0.6665
43
0.9390
1.3793
0.1904
19
0.6743
0.5476
0.8538
44
0.5923
0.2899
0.9583
20
0.9089
1.2847
0.2822
45
0.1921
−0.9673
0.5675
21
0.8759
1.1809
0.3801
46
0.3830
−0.3676
0.9332
22
0.3658
−0.4216
0.9125
47
0.6754
0.5510
0.8520
23
0.2945
−0.6455
0.7988
48
0.2944
−0.6460
0.7985
24
0.6719
0.5400
0.8577
49
0.9737
1.4882
0.0825
25
0.9104
1.2893
0.2778
50
0.4501
−0.1569
0.9877
51
0.7833
0.8901
0.6293
76
0.8258
1.0235
0.5204
52
0.7877
0.9040
0.6185
77
0.4101
−0.2826
0.9603
53
0.7012
0.6320
0.8068
78
0.6742
0.5473
0.8539
54
0.5937
0.2945
0.9569
79
0.2088
−0.9148
0.6099 (continued)
2.3
Evaluation of Definite Integrals Using Inverse Transform …
25
Table 2.3 (continued) i
ui
55
0.6616
56 57
ηi = π ui −π/2
ηi = πui −π/2
yi = Cos(ηi )
i
ui
0.5077
0.8739
80
0.8631
1.1407
0.4170
0.5202
0.0633
0.9980
81
0.2042
−0.9292
0.5984
0.9024
1.2643
0.3017
82
0.3027
−0.6198
0.8140
58
0.0050
−1.5551
0.0157
83
0.0504
−1.4126
0.1575
59
0.2183
−0.8851
0.6332
84
0.8463
1.0878
0.4644
60
0.2361
−0.8291
0.6755
85
0.4134
−0.2722
0.9632
61
0.4477
−0.1643
0.9865
86
0.2451
−0.8007
0.6962
62
0.7308
0.7249
0.7485
87
0.5565
0.1775
0.9843
63
0.3429
−0.4937
0.8806
88
0.0084
−1.5443
0.0265
64
0.2260
−0.8608
0.6519
89
0.3532
−0.4612
0.8955
65
0.1079
−1.2318
0.3326
90
0.7547
0.8002
0.6966
66
0.4879
−0.0380
0.9993
91
0.1188
−1.1976
0.3646
67
0.1655
−1.0509
0.4968
92
0.5403
0.1265
0.9920
68
0.1218
−1.1881
0.3735
93
0.5235
0.0739
0.9973
69
0.3544
−0.4575
0.8972
94
0.8228
1.0141
0.5284
70
0.3523
−0.4640
0.8943
95
0.5526
0.1653
0.9864
71
0.3895
−0.3472
0.9403
96
0.8279
1.0300
0.5148
72
0.0911
−1.2845
0.2824
97
0.6770
0.5561
0.8493
73
0.9631
1.4548
0.1157
98
0.7266
0.7118
0.7572
74
0.5184
0.0578
0.9983
99
0.4398
−0.1892
0.9822
75
0.6328
0.4172
0.9142
100
0.2243
−0.8662
0.6477
N 1 F(xi ) I = N p(xi )
yi = Cos(ηi )
(2.9)
i=1
where x i ’s are random values of x in the interval a < x < b obeying probability density function p(x). We now take up a linear variation for p(x) given by p(x) = C x where C is a constant. 2 Normalization of p(x) requires 0 p(x)d x = 1 or, C = 1/2. As such normalized probability density function is p(x) = x/2. See Fig. 2.9 in which we have plotted F(x) = ex along with the normalized linear probability density function p(x) = x/2. The linear probability density function p(x) = x/2 follows the function ex to some extent. Hence it is not a bad choice.
26
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
2.0
1
1.5
1.0
0.5
0.0 1.5
1.0
0.5
0.0
0.5
1.0
1.5
Fig. 2.6 Showing that random variate η is uniformly distributed in the interval −π/2 < x < +π/2
We now generate random variate η obeying p(x) = x/2 in the interval 0 < x < 2 using η inverse transform sampling. Values of η are obtainable from Eq. (1.2): a p(x)d x = u where u is random variable of Table 1.1. Thus we have η = 2 u1/2 . Using Program number 2.5 and resulting Table 2.5, we now evaluate the integral with the result 6.242 rather than 6.389. The difference is attributable to the fact that F(x) and p(x) used are not proportional to each other. That random variate η is distributed as p(x) = x/2 in the interval 0 to 2 is evidenced by Fig. 2.10 which displays values of η against stagnant integer 1. We find that indeed there are more variates for larger values of η. Program number 2.5 n=654321 SeedRandom[n] Table[{i=i+1,u[i]=RandomReal[],eta[i]=2*Sqrt[u[i]], y[i]=(Exp[eta[i]])/eta[i]},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","u[i]","eta[i]","y[i]"}}] Integral=Sum[2*y[i]/100,{i,1,100}] ListPlot[Table[{eta[i],1},{i,0,99,1}],Frame->True, FrameLabel->{"\[Eta] [i]","1"}]
2.4
Evaluation of Definite Integrals Using Inverse Transform …
27
Fig. 2.7 Showing F(x) = Cos(x) as undashed curve and p(x) = 2x/π2 + 1/π as dashed curve obtained using the command Plot[{Cos[x],2*x/Pi^2+1/Pi},{x,-Pi/2,Pi/2},Frame->True, FrameLabel->{"x","F(x) = Cos(x) or p(x) "}, PlotStyle->{{Black},{Dashed,Black}}] in Mathematica 6.0.
2.4
Evaluation of Definite Integrals Using Inverse Transform Sampling: Example IV
As Example IV, we now take up the integral b I =
5 F(x)d x =
a
loge xd x
(2.10)
1
where F(x) = loge x. We re-write Eq. (2.10) as b I =
F(x) p(x)d x p(x)
(2.11)
a
which, as discussed in Sect. 1.6, implies that average value of F/p is the value of the integral, i.e.
28
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
N F(xi ) Table 2.4 Showing ui , ηi = −π/2 + πui 1/2 , Cos(ηi )/pi from which we get I = N1 i=1 p(xi ) = b Cos(ηi ) 1 N Cos(ηi ) = 1 N i=1 i=1 2ηi /π 2 +1/π as an approximate value of the integral I = a F(x)d x = pi N N π/2 2 −π/2 Cos(x)d x = 2.104 whereas exact value is 2. Using p(x) = 2x/π + 1/π. Using prog. 2.4 Cos(ηi )/ pi
i
ui
1.3457
0.3777
26
0.9276
1.4549
0.1886
0.4862
2.1210
27
0.0504
−0.8655
4.5357
0.8596
1.3419
0.3844
28
0.3510
0.2903
2.5406
4
0.0342
−0.9895
4.6615
29
0.1510
−0.3499
3.7970
5
0.7597
1.1674
0.7075
30
0.1761
−0.2523
3.6242
6
0.4609
0.5619
1.9581
31
0.3796
0.3649
2.3816
7
0.4189
0.4625
2.1719
32
0.5239
0.7032
1.6553
8
0.4746
0.5934
1.8904
33
0.3586
0.3105
2.4976
9
0.3799
0.3656
2.3801
34
0.4202
0.4657
2.1652
10
0.1687
−0.2805
3.6750
35
0.5641
0.7888
1.4738
11
0.1899
−0.2016
3.5312
36
0.1372
−0.4073
3.8943
12
0.0364
−0.9710
4.6442
37
0.0774
−0.6968
4.3301
13
0.3335
0.2434
2.6400
38
0.8924
1.3970
0.2875
14
0.4257
0.4789
2.1367
39
0.9848
1.5468
0.0380
15
0.3860
0.3812
2.3467
40
0.4854
0.6179
1.8378
16
0.2445
−0.0174
3.1764
41
0.3080
0.1726
2.7885
17
0.5881
0.8385
1.3694
42
0.4945
0.6384
1.7939
18
0.2322
−0.0569
3.2544
43
0.9390
1.4735
0.1574
19
0.6743
1.0090
1.0191
44
0.5923
0.8470
1.3517
20
0.9089
1.4243
0.2405
45
0.1921
−0.1938
3.5167
21
0.8759
1.3694
0.3358
46
0.3830
0.3734
2.3632
22
0.3658
0.3293
2.4575
47
0.6754
1.0111
1.0149
23
0.2945
0.1342
2.8684
48
0.2944
0.1337
2.8692
24
0.6719
1.0043
1.0284
49
0.9737
1.5292
0.0661
25
0.9104
1.4267
0.2363
50
0.4501
0.5368
2.0122
51
0.7833
1.2097
0.6270
76
0.8258
1.2841
0.4889
52
0.7877
1.2175
0.6123
77
0.4101
0.4409
2.2184
53
0.7012
1.0599
0.9173
78
0.6742
1.0088
1.0194
54
0.5937
0.8500
1.3455
79
0.2088
−0.1352
3.4062
i
ui
1
0.8618
2
0.4287
3
ηi = − π/2+πui 1/2
ηi = − π/2+πui 1/2
Cos(ηi )/ pi
(continued)
2.4
Evaluation of Definite Integrals Using Inverse Transform …
29
Table 2.4 (continued) Cos(ηi )/ pi
i
ui
0.9845
1.0684
80
0.8631
1.3478
0.3739
0.6950
1.6729
81
0.2042
−0.1511
3.4364
0.9024
1.4136
0.2589
82
0.3027
0.1576
2.8197
58
0.0050
−1.3489
4.8944
83
0.0504
−0.8658
4.5361
59
0.2183
−0.1031
3.3444
84
0.8463
1.3193
0.4250
60
0.2361
−0.0444
3.2297
85
0.4134
0.4490
2.2010
61
0.4477
0.5313
2.0240
86
0.2451
−0.0153
3.1722
62
0.7308
1.1148
0.8092
87
0.5565
0.7728
1.5075
63
0.3429
0.2687
2.5864
88
0.0084
−1.2823
4.8666
64
0.2260
−0.0773
3.2943
89
0.3532
0.2962
2.5280
65
0.1079
−0.5387
4.1043
90
0.7547
1.1584
0.7247
66
0.4879
0.6236
1.8255
91
0.1188
−0.4880
4.0255
67
0.1655
−0.2927
3.6969
92
0.5403
0.7384
1.5805
68
0.1218
−0.4742
4.0036
93
0.5235
0.7023
1.6572
69
0.3544
0.2994
2.5213
94
0.8228
1.2789
0.4984
70
0.3523
0.2939
2.5329
95
0.5526
0.7646
1.5249
71
0.3895
0.3898
2.3282
96
0.8279
1.2877
0.4823
72
0.0911
−0.6224
4.2276
97
0.6770
1.0142
1.0086
73
0.9631
1.5123
0.0936
98
0.7266
1.1071
0.8243
74
0.5184
0.6911
1.6811
99
0.4398
0.5126
2.0642
75
0.6328
0.9283
1.1832
100
0.2243
−0.0830
3.3055
i
ui
55
0.6616
56
0.5202
57
ηi = − π/2+πui 1/2
I =
N 1 F(xi ) N p(xi )
ηi = − π/2+πui 1/2
Cos(ηi )/ pi
(2.12)
i=1
where x i ’s are random values of x in the interval a < x < b obeying probability density function p(x). We now take up an exponential variation for p(x) given by p(x) = C ex where C 5 is a constant. Normalization of p(x) requires 1 p(x)d x = 1 or, C = 1/145. As such normalized probability density function is p(x) = ex /145. See Fig. 2.11 in which we have plotted F(x) = loge x along with the normalized probability density function p(x) = ex /145. The probability density function p(x) = ex /145 follows the function loge x well. Hence it is a good choice.
30
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
2.0
1
1.5
1.0
0.5
0.0 1.0
0.5
0.0
0.5
1.0
1.5
Fig. 2.8 Showing that random variate η is linearly distributed in the interval −π/2 to +π/2. There are more variates for larger values of η
We now generate random variate η obeying p(x) = ex /145 in the interval 1 < x < 5 using η inverse transform sampling. Values of η are obtainable from Eq. (1.2): a p(x)d x = u η or, 1 p(x)d x = u. As such η = loge (145u + 2.72). Here u is uniform random variable of Table 1.1 in the interval 0–1. For ηi = loge (145ui + 2.72), using Program number 2.6 and resulting Table 2.6, we now evaluate the integral with the result 4.023 rather than 4.047. The difference is attributable to the fact that F(x) and p(x) used are not proportional to each other. That random variate η is non-uniformly distributed like p(x) = ex /145 in the interval 1–5 is evidenced by Fig. 2.12 which displays values of η against stagnant integer 1. Program number 2.6 n=654321 SeedRandom[n] Table[{i=i+1,u[i]=RandomReal[],eta[i]=Log[145*u[i]+2.72], y[i]=(Log[eta[i]])/(Exp[eta[i]])},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2},
2.5
Evaluation of Definite Integrals Using Inverse Transform …
31
7 6 5 4 3 2 1 0 0.0
0.5
1.0
1.5
2.0
x Fig. 2.9 Showing F(x) = ex as undashed curve and p(x) = x/2 as dashed curve obtained using the command Plot[{Exp[x],x/2},{x,0,2},Frame->True, FrameLabel->{"x","F(x) = e^x or p(x)"}, PlotStyle->{{Black},{Dashed, Black}}] in Mathematica 6.0.
TableHeadings->{None,{"i","u[i]","eta[i]","y[i]"}}] Integral=Sum[145*y[i]/100,{i,1,100}] ListPlot[Table[{eta[i],1},{i,0,99,1}], Frame->True,FrameLabel->{"\[Eta] [i]","1"}]
2.5
Evaluation of Definite Integrals Using Inverse Transform Sampling: Example V
As Example V, we now take up the integral b I =
3 F(x)d x =
a
0
1 dx 1 + x2
(2.13)
32
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
N F(xi ) Table 2.5 Showing tabulated values of ui , ηi = 2ui 1/2 , eηi /ηi from which we get I = N1 i=1 p(xi ) η η i i b 2 N N = N1 i=1 ηe /2 = N2 i=1 eηi as an approximate value of the integral I = a F(x)d x = 0 e x d x i
= 6.242 whereas exact value is 6.389. Using p(x) = x/2. Using prog. 2.5 i
ui
ηi = 2ui 1/2
eηi /ηi
i
ui
ηi = 2ui 1/2
eηi /ηi
1
0.8618
1.8567
3.4484
26
0.9276
1.9262
3.5632
2
0.4287
1.3095
2.8288
27
0.0504
0.4490
3.4894
3
0.8596
1.8543
3.4445
28
0.3510
1.1848
2.7600
4
0.0342
0.3701
3.9122
29
0.1510
0.7773
2.7989
5
0.7597
1.7432
3.2788
30
0.1761
0.8394
2.7579
6
0.4609
1.3577
2.8631
31
0.3796
1.2323
2.7827
7
0.4189
1.2945
2.8190
32
0.5239
1.4477
2.9380
8
0.4746
1.3778
2.8786
33
0.3586
1.1977
2.7657
9
0.3799
1.2327
2.7829
34
0.4202
1.2965
2.8202
10
0.1687
0.8215
2.7680
35
0.5641
1.5022
2.9900
11
0.1899
0.8716
2.7429
36
0.1372
0.7407
2.8316
12
0.0364
0.3818
3.8368
37
0.0774
0.5564
3.1352
13
0.3335
1.1549
2.7481
38
0.8924
1.8894
3.5013
14
0.4257
1.3049
2.8258
39
0.9848
1.9847
3.6665
15
0.3860
1.2427
2.7882
40
0.4854
1.3934
2.8911
16
0.2445
0.9889
2.7185
41
0.3080
1.1099
2.7336
17
0.5881
1.5338
3.0224
42
0.4945
1.4064
2.9019
18
0.2322
0.9638
2.7201
43
0.9390
1.9381
3.5836
19
0.6743
1.6423
3.1463
44
0.5923
1.5392
3.0281
20
0.9089
1.9068
3.5302
45
0.1921
0.8766
2.7410
21
0.8759
1.8718
3.4725
46
0.3830
1.2377
2.7856
22
0.3658
1.2097
2.7713
47
0.6754
1.6437
3.1479
23
0.2945
1.0854
2.7277
48
0.2944
1.0852
2.7276
24
0.6719
1.6394
3.1426
49
0.9737
1.9735
3.6463
25
0.9104
1.9083
3.5328
50
0.4501
1.3417
2.8513
51
0.7833
1.7701
3.3171
76
0.8258
1.8175
3.3872
52
0.7877
1.7751
3.3242
77
0.4101
1.2807
2.8103
53
0.7012
1.6747
3.1870
78
0.6742
1.6422
3.1461
54
0.5937
1.5411
3.0301
79
0.2088
0.9139
2.7290
55
0.6616
1.6268
3.1273
80
0.8631
1.8581
3.4505 (continued)
2.5
Evaluation of Definite Integrals Using Inverse Transform …
33
Table 2.5 (continued) i
ui
ηi = 2ui 1/2
eηi /ηi
i
ui
ηi = 2ui 1/2
eηi /ηi
56
0.5202
1.4424
2.9332
81
0.2042
0.9038
2.7318
57
0.9024
1.8999
3.5188
82
0.3027
1.1004
2.7312
58
0.0050
0.1413
8.1521
83
0.0504
0.4488
3.4903
59
0.2183
0.9344
2.7244
84
0.8463
1.8399
3.4218
60
0.2361
0.9718
2.7194
85
0.4134
1.2859
2.8135
61
0.4477
1.3382
2.8488
86
0.2451
0.9902
2.7184
62
0.7308
1.7097
3.2329
87
0.5565
1.4920
2.9799
63
0.3429
1.1711
2.7543
88
0.0084
0.1837
6.5418
64
0.2260
0.9508
2.7217
89
0.3532
1.1886
2.7616
65
0.1079
0.6570
2.9360
90
0.7547
1.7375
3.2708
66
0.4879
1.3970
2.8941
91
0.1188
0.6893
2.8903
67
0.1655
0.8136
2.7729
92
0.5403
1.4701
2.9587
68
0.1218
0.6981
2.8792
93
0.5235
1.4471
2.9374
69
0.3544
1.1906
2.7625
94
0.8228
1.8142
3.3822
70
0.3523
1.1871
2.7610
95
0.5526
1.4868
2.9748
71
0.3895
1.2482
2.7912
96
0.8279
1.8197
3.3907
72
0.0911
0.6038
3.0293
97
0.6770
1.6456
3.1503
73
0.9631
1.9627
3.6270
98
0.7266
1.7048
3.2263
74
0.5184
1.4400
2.9310
99
0.4398
1.3263
2.8403
75
0.6328
1.5910
3.0852
100
0.2243
0.9471
2.7222
where F(x) =
1 . 1+x 2
We re-write Eq. (2.13) as b I =
F(x) p(x)d x p(x)
(2.14)
a
which, as discussed in Sect. 1.6, implies that average value of F/p is the value of the integral, i.e. N 1 F(xi ) I = N p(xi )
(2.15)
i=1
where x i ’s are random values of x in the interval a < x < b obeying probability density function p(x).
34
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
2.0
1
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
Fig. 2.10 Showing that random variate η is linearly distributed in the interval 0–2. There are more variates for larger values of η
We now take up an exponential variation for p(x) given by p(x) = C e−x where C is a 3 constant. Normalization of p(x) requires 0 p(x)d x = 1 or, C = 1.05. As such normalized probability density function is p(x) = 1.05e−x . See Fig. 2.13 in which we have plotted 1 −x . The F(x) = 1+x 2 along with the normalized probability density function p(x) = 1.05 e
1 probability density function p(x) = 1.05e−x follows the function 1+x 2 quite well. Hence it is a very good choice. We now generate random variate η obeying p(x) = 1.05 e−x in the interval 0 < x < 3 η using inverse transform sampling. Values of η are obtainable from Eq. (1.2): a p(x)d x = η u or, 0 p(x)d x = u. As such η = −loge (1 − 0.95u) where u is uniform random variable of Table 1.1 in the interval 0 < x < 1. Using η = −loge (1 − 0.95u) and Program number 2.7 and resulting Table 2.7, we now evaluate the integral with the result 1.238 rather than 1.249. The absence of difference is attributable to the fact that F(x) and p(x) used are almost proportional to each other. That random variate η is non-uniformly distributed in the interval 0–3 is evidenced by Fig. 2.14 which displays values of η against stagnant integer 1.
2.5
Evaluation of Definite Integrals Using Inverse Transform …
35
1.5
1.0
0.5
0.0 1
2
3
4
5
x Fig. 2.11 Showing F(x) = loge x as undashed curve and p(x) = ex /145 as dashed curve obtained using the command a=1/145; p1=Plot[Log[E,x],{x,1,5},Frame->True, FrameLabel->{"x","F(x) = Log(x) or p(x)"}, PlotStyle->{Black}]; p2=Plot[a*Exp[x],{x,1,5},PlotStyle->{Dashed,Black}, Frame->True]; Show[p1,p2] in Mathematica 6.0.
Program number 2.7 n=654321 SeedRandom[n] Table[{i=i+1,u[i]=RandomReal[],eta[i]=-Log[1-0.95*u[i]], y[i]=(Exp[eta[i]])/(1+eta[i]^2)},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","u[i]","eta[i]","y[i]"}}] Integral=Sum[0.95*y[i]/100,{i,1,100}] ListPlot[Table[{eta[i],1},{i,0,99,1}],Frame->True, FrameLabel->{"\[Eta] [i]","1"}]
36
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
N F(xi ) Table 2.6 Showing ui , ηi = loge (145ui + 2.72), Loge ηi /eηi from which we get I = N1 i=1 p(xi ) η η Log Log b N N 145 e i e i = N1 i=1 as an approximate value of the integral I = a F(x)d x = i=1 eηi eηi /145 = N 5 x 1 Loge x d x = 4.023 whereas exact value is 4.047. Using p(x) = e /145. Using prog. 2.6 i
ui
ηi
Loge ηi eηi
i
ui
ηi
Loge ηi eηi
1
0.8618
4.8496
0.0124
26
0.9276
4.9216
0.0116
2
0.4287
4.1726
0.0220
27
0.0504
2.3054
0.0833
3
0.8596
4.8470
0.0124
28
0.3510
3.9817
0.0258
4
0.0342
2.0393
0.0927
29
0.1510
3.2036
0.0473
5
0.7597
4.7262
0.0138
30
0.1761
3.3415
0.0427
6
0.4609
4.2420
0.0208
31
0.3796
4.0564
0.0242
7
0.4189
4.1504
0.0224
32
0.5239
4.3655
0.0187
8
0.4746
4.2702
0.0203
33
0.3586
4.0022
0.0253
9
0.3799
4.0571
0.0242
34
0.4202
4.1534
0.0224
10
0.1687
3.3025
0.0440
35
0.5641
4.4370
0.0176
11
0.1899
3.4099
0.0405
36
0.1372
3.1184
0.0503
12
0.0364
2.0800
0.0915
37
0.0774
2.6349
0.0695
13
0.3335
3.9333
0.0268
38
0.8924
4.8837
0.0120
14
0.4257
4.1658
0.0221
39
0.9848
4.9803
0.0110
15
0.3860
4.0724
0.0239
40
0.4854
4.2918
0.0199
16
0.2445
3.6421
0.0339
41
0.3080
3.8581
0.0285
17
0.5881
4.4774
0.0170
42
0.4945
4.3097
0.0196
18
0.2322
3.5943
0.0352
43
0.9390
4.9336
0.0115
19
0.6743
4.6101
0.0152
44
0.5923
4.4842
0.0169
20
0.9089
4.9017
0.0118
45
0.1921
3.4202
0.0402
21
0.8759
4.8654
0.0122
46
0.3830
4.0648
0.0241
22
0.3658
4.0211
0.0250
47
0.6754
4.6117
0.0152
23
0.2945
3.8161
0.0295
48
0.2944
3.8156
0.0295
24
0.6719
4.6066
0.0153
49
0.9737
4.9692
0.0111 (continued)
2.5
Evaluation of Definite Integrals Using Inverse Transform …
37
Table 2.6 (continued) i
ui
ηi
Loge ηi eηi
i
ui
ηi
Loge ηi eηi
25
0.9104
4.9033
0.0118
50
0.4501
4.2192
0.0212
51
0.7833
4.7562
0.0134
76
0.8258
4.8078
0.0128
52
0.7877
4.7617
0.0133
77
0.4101
4.1300
0.0228
53
0.7012
4.6482
0.0147
78
0.6742
4.6100
0.0152
54
0.5937
4.4865
0.0169
79
0.2088
3.4964
0.0379
55
0.6616
4.5916
0.0155
80
0.8631
4.8510
0.0124
56
0.5202
4.3585
0.0188
81
0.2042
3.4760
0.0385
57
0.9024
4.8946
0.0119
82
0.3027
3.8418
0.0289
58
0.0050
1.2365
0.0616
83
0.0504
2.3047
0.0833
59
0.2183
3.5371
0.0368
84
0.8463
4.8317
0.0126
60
0.2361
3.6096
0.0347
85
0.4134
4.1377
0.0227
61
0.4477
4.2142
0.0213
86
0.2451
3.6446
0.0338
62
0.7308
4.6884
0.0142
87
0.5565
4.4238
0.0178
63
0.3429
3.9596
0.0262
88
0.0084
1.3720
0.0802
64
0.2260
3.5693
0.0359
89
0.3532
3.9877
0.0256
65
0.1079
2.9106
0.0582
90
0.7547
4.7199
0.0138
66
0.4879
4.2968
0.0198
91
0.1188
2.9930
0.0550
67
0.1655
3.2853
0.0445
92
0.5403
4.3952
0.0183
68
0.1218
3.0148
0.0541
93
0.5235
4.3648
0.0187
69
0.3544
3.9910
0.0256
94
0.8228
4.8042
0.0129
70
0.3523
3.9854
0.0257
95
0.5526
4.4170
0.0179
71
0.3895
4.0808
0.0238
96
0.8279
4.8102
0.0128
72
0.0911
2.7685
0.0639
97
0.6770
4.6140
0.0152
73
0.9631
4.9584
0.0112
98
0.7266
4.6828
0.0143
74
0.5184
4.3553
0.0189
99
0.4398
4.1970
0.0216
75
0.6328
4.5484
0.0160
100
0.2243
3.5621
0.0361
38
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
2.0
1
1.5
1.0
0.5
0.0 1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
Fig. 2.12 Showing that random variate η is non-uniformly distributed like p(x) = ex /145 in the interval 1–5. There are more variates for larger values of η 1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
x
Fig. 2.13 Showing F(x) = 1 2 as undashed curve and p(x) = 1.05 e−x as dashed curve obtained 1+x using the command p1=Plot[1/(1+x^2),{x,0,3},PlotRange->{0,1},Frame->True, FrameLabel->{"x","F(x) = 1/(1+x^2) or p(x)"}, PlotStyle->{Black}]; p2=Plot[1.05*Exp[-x],{x,0,3},PlotStyle->{Dashed,Black}]; Show[p1,p2] in Mathematica 6.0.
2.5
Evaluation of Definite Integrals Using Inverse Transform …
39
Table 2.7 Showing tabulated values of ui , ηi = −loge (1 − 0.95ui ), eηi /(1 + ηi2 ) from which we 2 N F(xi ) 1 N 1/(1+ηi ) = 0.95 N eηi get I = N1 i=1 i=1 1.05e−ηi i=1 (1+η2 ) as an approximate value of the N p(xi ) = N i integral I = ab F(x)d x = 03 1 2 d x = 1.238 whereas exact value is 1.249. Using p(x) = 1.05
1+x
e−x . Using prog. 2.7 i
ui
ηi
eηi (1+ηi2 )
i
ui
ηi
eηi (1+ηi2 )
1
0.8618
1.7078
1.4086
26
0.9276
2.1304
1.5199
2
0.4287
0.5230
1.3247
27
0.0504
0.0491
1.0478
3
0.8596
1.6962
1.4065
28
0.3510
0.4056
1.2883
4
0.0342
0.0331
1.0325
29
0.1510
0.1549
1.1402
5
0.7597
1.2790
1.3632
30
0.1761
0.1831
1.1620
6
0.4609
0.5759
1.3357
31
0.3796
0.4473
1.3033
7
0.4189
0.5074
1.3209
32
0.5239
0.6887
1.3506
8
0.4746
0.5994
1.3397
33
0.3586
0.4166
1.2925
9
0.3799
0.4477
1.3035
34
0.4202
0.5095
1.3214
10
0.1687
0.1747
1.1556
35
0.5641
0.7677
1.3558
11
0.1899
0.1990
1.1737
36
0.1372
0.1396
1.1279
12
0.0364
0.0352
1.0346
37
0.0774
0.0764
1.0731
13
0.3335
0.3810
1.2782
38
0.8924
1.8827
1.4459
14
0.4257
0.5182
1.3236
39
0.9848
2.7418
1.8215
15
0.3860
0.4569
1.3064
40
0.4854
0.6182
1.3425
16
0.2445
0.2643
1.2175
41
0.3080
0.3461
1.2623
17
0.5881
0.8181
1.3576
42
0.4945
0.6344
1.3447
18
0.2322
0.2492
1.2080
43
0.9390
2.2264
1.5556
19
0.6743
1.0233
1.3591
44
0.5923
0.8271
1.3578
20
0.9089
1.9913
1.4753
45
0.1921
0.2015
1.1755
21
0.8759
1.7843
1.4235
46
0.3830
0.4523
1.3050
22
0.3658
0.4270
1.2963
47
0.6754
1.0262
1.3591
23
0.2945
0.3282
1.2535
48
0.2944
0.3280
1.2534
24
0.6719
1.0169
1.3591
49
0.9737
2.5907
1.7297 (continued)
40
2 Evaluation of Definite Integrals Using Inverse Transform Sampling …
Table 2.7 (continued) i
ui
ηi
eηi (1+ηi2 )
i
ui
ηi
eηi (1+ηi2 )
25
0.9104
2.0015
1.4783
50
0.4501
0.5578
1.3323
51
0.7833
1.3632
1.3675
76
0.8258
1.5348
1.3829
52
0.7877
1.3797
1.3686
77
0.4101
0.4936
1.3173
53
0.7012
1.0970
1.3593
78
0.6742
1.0231
1.3591
54
0.5937
0.8303
1.3579
79
0.2088
0.2211
1.1893
55
0.6616
0.9903
1.3591
80
0.8631
1.7145
1.4098
56
0.5202
0.6815
1.3499
81
0.2042
0.2157
1.1856
57
0.9024
1.9471
1.4627
82
0.3027
0.3391
1.2589
58
0.0050
0.0048
1.0047
83
0.0504
0.0490
1.0477
59
0.2183
0.2324
1.1970
84
0.8463
1.6294
1.3956
60
0.2361
0.2540
1.2110
85
0.4134
0.4987
1.3186
61
0.4477
0.5540
1.3315
86
0.2451
0.2651
1.2180
62
0.7308
1.1849
1.3604
87
0.5565
0.7522
1.3550
63
0.3429
0.3941
1.2837
88
0.0084
0.0080
1.0080
64
0.2260
0.2417
1.2031
89
0.3532
0.4088
1.2895
65
0.1079
0.1082
1.1014
90
0.7547
1.2622
1.3625
66
0.4879
0.6227
1.3431
91
0.1188
0.1197
1.1113
67
0.1655
0.1711
1.1528
92
0.5403
0.7200
1.3530
68
0.1218
0.1230
1.1140
93
0.5235
0.6879
1.3505
69
0.3544
0.4105
1.2902
94
0.8228
1.5216
1.3814
70
0.3523
0.4075
1.2890
95
0.5526
0.7444
1.3546
71
0.3895
0.4620
1.3081
96
0.8279
1.5440
1.3840
72
0.0911
0.0906
1.0859
97
0.6770
1.0305
1.3592
73
0.9631
2.4643
1.6620
98
0.7266
1.1719
1.3602
74
0.5184
0.6782
1.3496
99
0.4398
0.5409
1.3288
75
0.6328
0.9192
1.3590
100
0.2243
0.2396
1.2018
2.5
Evaluation of Definite Integrals Using Inverse Transform …
41
2.0
1
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
2.5
Fig. 2.14 Showing that random variate η is non-uniformly distributed in the interval 0 to 3. There are fewer variates for larger values of η as expected from p(x) = 1.05 e−x
Part II Monte Carlo Methods Using Acceptance-Rejection Sampling Utilizing Mathematica
3
Introduction for Part II
This chapter provides us with background for Chaps. 4 and 5. It introduces readers to continuous, uniform and Gaussian random variables. Acceptance-rejection method of generating random variate for given probability distribution function, and concept of importance sampling are also introduced. This chapter repeats much of Chap. 1; the purpose is to enable readers to read part I and part II of the book almost independently.
3.1
Random Variable
In the context of Monte Carlo methods, to define a random variable, we must indicate the values that the variable can assume and probabilities of occurrence of these values. We do not know value of the variable in any given case; but we do know the values that the variable can assume and the probabilities of occurrence of these values.
3.2
Continuous Random Variable
A random variable x is called continuous if it can assume any fractional value in a certain interval, say a to b. Besides specifying the interval containing all its possible values, we need to state a function p(x) called probability density function or probability distribution function. We have (1) p(x) ≥ 0 for a ≤ x ≤ b, (2) product p(x)dx is probability that x ≤ x ≤ x + dx,
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Chowdhury, Monte Carlo Methods Utilizing Mathematica®, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-23294-7_3
45
46
3 Introduction for Part II
b (3) a p(x)d x is probability that a ≤ x ≤ b where a < a and b < b, b (4) a p(x)d x = 1 indicating the surety that x lies in the interval a to b. As such, average or so called expectation value of x is given by A x = b and average value of any function of x, say f (x), is A f = a f (x) p(x)d x.
3.3
b a
x p(x)d x,
Uniform Random Variable
A random variable u in the interval 0 ≤ x ≤ 1 having constant, say C, probability density function p(x) = C is said to be a uniform random variable. The requirement that 1 1 0 p(x)d x = 1 gives 0 Cd x = 1 which gives C = 1. Thus probability density function of uniform random variable in the interval 0–1 is p(x) = 1. Program number 3.1 n=654321 SeedRandom[n] Table[{i=i+1,u[i]=RandomReal[]},{i,0,99,1}]; TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","u[i]"}}] ListPlot[Table[{u[i],1},{i,0,99,1}], Frame->True,FrameLabel->{"u [i]","1"}] Program number 3.1 in Mathematica 6.0 can provide us 100 uniform random numbers ui ’s in the interval 0–1 and the result is shown in Table 3.1 by rounding the numbers ui ’s to 4 decimal place in Microsoft Excel. In the program, value of n has been chosen by trial and error so that the random numbers ui ’s are almost uniformly distributed in the interval 0–1 as vindicated by Fig. 3.1 which displays the numbers ui ’s against a stagnant integer 1. This is a quality check of random numbers ui ’s. A benefit of using the command SeedRandom[n] in Program number 3.1 is that it enables us to get the values of ui ’s reproducibly.
3.4
Normal or Gaussian Random Variable
Probability density function p(x) for normal random variable x is given by 1 1 x −a 2 p(x) = √ exp − 2 σ σ 2π
(3.1)
3.4
Normal or Gaussian Random Variable
47
Table 3.1 Showing 100 uniform random numbers ui ’s (in the interval 0 to 1) obtained using program number 3.1 in Mathematica i
ui
i
ui
i
ui
i
ui
1
0.8618
26
0.9276
51
0.7833
76
0.8258
2
0.4287
27
0.0504
52
0.7877
77
0.4101
3
0.8596
28
0.3510
53
0.7012
78
0.6742
4
0.0342
29
0.1510
54
0.5937
79
0.2088
5
0.7597
30
0.1761
55
0.6616
80
0.8631
6
0.4609
31
0.3796
56
0.5202
81
0.2042
7
0.4189
32
0.5239
57
0.9024
82
0.3027
8
0.4746
33
0.3586
58
0.0050
83
0.0504
9
0.3799
34
0.4202
59
0.2183
84
0.8463
10
0.1687
35
0.5641
60
0.2361
85
0.4134
11
0.1899
36
0.1372
61
0.4477
86
0.2451
12
0.0364
37
0.0774
62
0.7308
87
0.5565
13
0.3335
38
0.8924
63
0.3429
88
0.0084
14
0.4257
39
0.9848
64
0.2260
89
0.3532
15
0.3860
40
0.4854
65
0.1079
90
0.7547
16
0.2445
41
0.3080
66
0.4879
91
0.1188
17
0.5881
42
0.4945
67
0.1655
92
0.5403
18
0.2322
43
0.9390
68
0.1218
93
0.5235
19
0.6743
44
0.5923
69
0.3544
94
0.8228
20
0.9089
45
0.1921
70
0.3523
95
0.5526
21
0.8759
46
0.3830
71
0.3895
96
0.8279
22
0.3658
47
0.6754
72
0.0911
97
0.6770
23
0.2945
48
0.2944
73
0.9631
98
0.7266
24
0.6719
49
0.9737
74
0.5184
99
0.4398
25
0.9104
50
0.4501
75
0.6328
100
0.2243
defined for −∝ < x < +∝ . Average value of x is a and variance of x is σ 2 . According a+3σ to so called rule of 3 sigma, a−3σ p(x)d x ≈ 0.997. Program number 3.2 n=654321 SeedRandom[n] Table[{i=i+1,g[i] = RandomReal[NormalDistribution[0,1]]}, {i,0,99,1}];
48
3 Introduction for Part II
2.0
1
1.5
1.0
0.5
0.0 0.0
0.2
0.4
0.6
0.8
1.0
Fig. 3.1 Uniform random numbers ui ’s of Table 3.1 plotted against a stagnant integer 1 to vindicate that ui ’s are indeed almost uniformly distributed in the interval 0–1
TableForm[%,TableSpacing->{2,2}, TableHeadings->{None,{"i","g[i]"}}] ListPlot[Table[{g[i],1},{i,0,99,1}], Frame- >True,FrameLabel->{"g [i]","1"}] Program number 3.2 in Mathematica 6.0 can provide us 100 Gaussian random numbers gi ’s in the interval −3 to +3 and the result is shown in Table 3.2 by rounding the numbers gi ’s to 4 decimal place in Microsoft Excel. In the program, value of n has been chosen by trial and error so that the random numbers gi ’s are so distributed in the interval −3 to + 3 that most of the numbers are close to the average value x = 0 as vindicated by Fig. 3.2 which displays the numbers gi ’s against a stagnant integer 1. This is a quality check of random numbers gi ’s. A benefit of using the command SeedRandom[n] in Program number 3.2 is that it enables us to get the values of gi ’s reproducibly.
3.5
Generating Random Variates by Acceptance-Rejection Method
49
Table 3.2 Showing 100 Gaussian random numbers gi ’s (in the interval − 3 to + 3) obtained using program number 3.2 in Mathematica. For a = 0 and σ = 1 i
gi
1
−0.4915
26
0.1642
51
0.0250
76
0.2147
2
0.5374
27
−0.7006
52
0.0436
77
−0.1304
3
−0.7192
28
−0.9017
53
0.6939
78
1.2606
4
−1.3024
29
0.4529
54
−0.5178
79
1.8331
5
0.6802
30
0.1524
55
0.9525
80
1.2424
6
1.7751
31
−0.1529
56
−1.5620
81
−0.6542
7
−1.3234
32
0.2197
57
0.7447
82
−0.6587
8
0.0478
33
−2.1041
58
0.5326
83
−0.7827
9
0.1149
34
1.3676
59
0.3137
84
1.1034
10
0.7463
35
−0.8635
60
0.5035
85
−0.1103
11
−0.3424
36
1.1543
61
−1.0072
86
−0.1061
12
−0.7370
37
−0.2724
62
1.6373
87
−0.2752
13
0.3892
38
0.4385
63
−0.4160
88
0.6360
14
−1.4486
39
−0.6121
64
−0.2576
89
0.6611
15
0.8703
40
1.1545
65
2.3177
90
0.6861
16
−1.3761
41
−0.5795
66
0.0741
91
1.1652
17
−1.2559
42
1.3903
67
−0.5673
92
2.6803
18
0.6966
43
0.0406
68
−0.0757
93
−0.8103
19
1.7650
44
1.0811
69
0.1267
94
−1.3781
20
−0.1744
45
0.0427
70
−1.0962
95
−0.2635
21
−1.5339
46
−1.9984
71
−0.2686
96
−0.3182
22
−0.2967
47
0.5023
72
0.7321
97
0.7618
23
−1.3473
48
0.5118
73
2.0250
98
0.9291
24
−0.2439
49
−0.1296
74
−0.5563
99
−0.9932
25
−0.2195
50
0.2063
75
0.7393
3.5
i
gi
i
gi
i
100
gi
0.4787
Generating Random Variates by Acceptance-Rejection Method
See Fig. 3.3. Suppose, we wish to generate random variates x corresponding to a given probability distribution function p(x) defined for the interval x = a to b. Suppose, non-zero values of p(x) span from 0 to c. In acceptance-rejection method, 2 sets of uniform random numbers u1 ’s and u2 ’s in the interval 0–1 are first generated. Thereafter, linear transformations are carried out using
50
3 Introduction for Part II
2.0
1
1.5
1.0
0.5
0.0 2
1
0
1
2
Fig. 3.2 Gaussian random numbers gi ’s of Table 3.2 plotted against a stagnant integer 1 to vindicate that gi ’s are so distributed in the interval −3 to +3 that most of the numbers are close to the average value x = 0. For a = 0 and σ = 1 Fig. 3.3 Figure to help explain acceptance-rejection method of obtaining random variates for a given probability distribution function p(x)
p(x) c
(0, 0)
a
b
x
x
x + dx
U 1 = a + (b − a)u1 and U 2 = 0 + (c − 0)u2 known from sampling by inverse transform method, to get 2 sets of uniform random numbers U 1 ’s and U 2 ’s in the interval a to b and 0 to c respectively. The next steps are to calculate p(x = U 1 ) for each value of U 1 and if U 2 < p(x = U 1 ), that U 1 is accepted as a valid random variate for the probability distribution function p(x); otherwise that value of U 1 is rejected.
3.6 Variance Reduction and Importance Sampling
51
Validity of the method narrated above can be understood as follows. Since U 1 is a uniform random variable in the interval a to b, probability that U 1 falls in the interval x to x + dx is proportional to the width of the interval which is dx. Again, since U 2 is a uniform random variable in the interval 0 to c, probability that U 2 falls below p(x = U 1 ) is proportional to the interval p(x = U 1 ) − 0 = p (x = U 1 ). Thus the probability that U 1 is within x to x + dx and U 2 is below p(x = U 1 ) is the product of p(x = U 1 ) and dx which is p(x)dx.
3.6
Variance Reduction and Importance Sampling
Let us consider a definite integral b I =
F(x)d x
(3.2)
a
Let us re-write Eq. (3.2) as b I=
F(x) p(x)d x = p(x)
a
b f (x) p(x)d x
(3.3)
a
where f (x) = F(x)/ p(x). As discussed in Sect. 3.2, average value of any function of x, say f (x), is given by b A f = a f (x) p(x)d x where p(x) is a normalized probability density function. Thus according to Eq. (3.3), value I of the integral is average value of F/p. We can choose any functional form for p(x), but as proved below, choosing p(x) as proportional to F(x) ensures that variance of f (x) = F/p is small. This will ensure a b better value of the average and hence a better value of the integral I = a F(x)d x. Variance of f (x) is square of standard deviation. Variance of f (x) is given by V [ f (x)] = A[ ( f − A f )2 ]
(3.4)
where A stands for average value. Thus
or,
V f (x) = A[ f 2 − 2 f A f + A2f ] = A f 2 − 2 A f A f + A2f = A f 2 − A2f ⎛ b ⎞2 b V f (x) = f 2 (x) p(x)d x − ⎝ f (x) p(x)d x ⎠ a
b = a
⎛ F2 pd x − ⎝ p2
a
b a
⎞2 F p(x)d x ⎠ p
52
3 Introduction for Part II
b =
⎛ F2 p
a
dx − ⎝
b
⎞2 F dx⎠
a
Thus the variance is b V f (x) =
F2 dx − I2 p
(3.5)
a
If p(x) is taken as
p(x) =
|F(x)| b
(3.6)
|F(x)|d x
a
the variance given by Eq. (3.5) becomes V f (x) = ⎛ or, V f (x) = ⎝
b a
b a
F2
|F(x)| b a |F(x)|d x
dx − I2
⎛ b ⎞⎛ b ⎞ ⎞2 |F(x)|d x ⎠⎝ |F(x)|d x ⎠ − I 2 = ⎝ |F(x)|d x ⎠ − I 2 a
(3.7)
a
b Equation (3.7) in conjunction with Eq. (3.2): I = a F(x)d x reveals that the variance vanishes if the integrand F(x) does not change sign. If it does, the variance will be small if condition laid by Eq. (3.6) is met. Thus the probability density function p(x) should be proportional to the integrand |F(x)|. This is the so called importance sampling. Looking back at Eq. (3.3), we find that value of the integral is approximately N F(xi ) determined by the value of the average I = N1 i=1 p(xi ) . As recommended by importance sampling, we need to take p(x) = C F(x) where C is a constant; we need to b b . Thus normalize p(x) first; a p(x)d x = 1 gives a C F(x)d x = 1 or, C = b 1 F(x)d x ba F(xi ) F(x) 1 N F(xi ) 1 N p(x) = b . The sum N i=1 p(xi ) becomes N i=1 F(xi ) = a F(x)d x. Thus a
F(x)d x
N
F(xi ) i=1 p(xi )
b a F(x)d x
the sum N1 equals the integral if probability density function p(x) is taken proportional to the integrand |F(x)|. Slight variation of p(x) from proportionality with N F(xi ) |F(x)| will result in slight difference of N1 i=1 p(xi ) from actual value of the integral b I = a F(x)d x.
4
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling Utilizing Mathematica
After getting required background in Chaps. 3, this chapter provides practical demonstrations of how to carry out definite integrals using Monte Carlo method using random variates obtained by acceptance-rejection method. Random variates have been obtained corresponding to probability distribution functions linear, Gaussian and exponential in the variable. Programs written in Mathematica have been used in this regard.
4.1
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example I
As example I, we now take up the integral b I =
π F(x)d x =
a
Sin(x)d x
(4.1)
0
where F(x) = Sin(x). We re-write Eq. (4.1) as b I =
F(x) p(x)d x p(x)
(4.2)
a
which, as discussed in Sect. 3.6, implies that average value of F/p is the value of the integral, i.e. I =
N 1 F(xi ) N p(xi )
(4.3)
i=1
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Chowdhury, Monte Carlo Methods Utilizing Mathematica®, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-23294-7_4
53
54
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
where x i ’s are random values of x in the interval a < x < b obeying probability density function p(x). We take up a linear variation for p(x) given by p(x) = C x where C is a constant. π Normalization requires that 0 C x d x = 1 which gives C = 2/π2 . Thus normalized p(x) = 2x/π2 . See Fig. 4.1 in which we have plotted F(x) = Sin(x) along with the normalized linear probability density function p(x) = 2x/π2 . The linear probability density function p(x) = 2x/π2 is not too different from a slowly varying function of x like Sin(x). Hence it is a good choice. We now generate random variate η obeying p(x) = 2x/π2 in the interval 0 < x < π using acceptance-rejection sampling, using Program number 4.1 and resulting Table 4.1. From Table 4.1, we manually gather accepted random variates ηa which we use in Program number 4.2 and resulting Table 4.2 to evaluate the integral with the result 2.095 rather than the exact value 2.
1.0
0.8
0.6
0.4
0.2
0.0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
x Fig. 4.1 Showing F(x) = Sin(x) as undashed curve and p(x) = 2x/π2 as dashed curve obtained using the command Plot[{Sin[x],(2*x)/(Pi^2)},{x,0,Pi},Frame-> True, FrameLabel-> {"x", "F(x) = Sin(x) or p(x) "}, PlotStyle-> {{Black},{Dashed,Black}}] in Mathematica 6.0
4.1
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
55
Table 4.1 Showing values of u1[i], u2[i], U1[i], U2[i] and eta[i]. Using prog. 4.1 i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
1
0.8618
0.4287
2.7075
0.2787
2.7075
1
2
0.8596
0.0342
2.7005
0.0223
2.7005
2
3
0.7597
0.4609
2.3865
0.2996
2.3865
3
4
0.4189
0.4746
1.3160
0.3085
Null
4
5
0.3799
0.1687
1.1935
0.1097
1.1935
5
6
0.1899
0.0364
0.5967
0.0237
0.5967
6
7
0.3335
0.4257
1.0476
0.2767
Null
7
8
0.3860
0.2445
1.2128
0.1589
1.2128
8
9
0.5881
0.2322
1.8477
0.1509
1.8477
9
10
0.6743
0.9089
2.1184
0.5908
Null
10
11
0.8759
0.3658
2.7517
0.2378
2.7517
11
12
0.2945
0.6719
0.9253
0.4367
Null
12
13
0.9104
0.9276
2.8601
0.6029
Null
13
14
0.0504
0.3510
0.1583
0.2281
Null
14
15
0.1510
0.1761
0.4745
0.1145
Null
15
16
0.3796
0.5239
1.1926
0.3406
Null
16
17
0.3586
0.4202
1.1266
0.2731
Null
17
18
0.5641
0.1372
1.7723
0.0892
1.7723
18
19
0.0774
0.8924
0.2431
0.5801
Null
19
20
0.9848
0.4854
3.0938
0.3155
3.0938
20
21
0.3080
0.4945
0.9675
0.3214
Null
21
22
0.9390
0.5923
2.9501
0.3850
2.9501
22
23
0.1921
0.3830
0.6035
0.2490
Null
23
24
0.6754
0.2944
2.1218
0.1914
2.1218
24
25
0.9737
0.4501
3.0590
0.2925
3.0590
25
26
0.7833
0.7877
2.4609
0.5120
Null
26
27
0.7012
0.5937
2.2028
0.3859
2.2028
27
28
0.6616
0.5202
2.0785
0.3381
2.0785
28
29
0.9024
0.0050
2.8351
0.0032
2.8351
29
30
0.2183
0.2361
0.6857
0.1534
Null
30
31
0.4477
0.7308
1.4065
0.4750
Null
31
32
0.3429
0.2260
1.0771
0.1469
1.0771
32 (continued)
56
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.1 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
33
0.1079
0.4879
0.3390
0.3171
Null
33
34
0.1655
0.1218
0.5199
0.0792
0.5199
34
35
0.3544
0.3523
1.1133
0.2290
Null
35
36
0.3895
0.0911
1.2236
0.0592
1.2236
36
37
0.9631
0.5184
3.0256
0.3370
3.0256
37
38
0.6328
0.8258
1.9880
0.5368
Null
38
39
0.4101
0.6742
1.2882
0.4382
Null
39
40
0.2088
0.8631
0.6560
0.5610
Null
40
41
0.2042
0.3027
0.6416
0.1968
Null
41
42
0.0504
0.8463
0.1582
0.5501
Null
42
43
0.4134
0.2451
1.2986
0.1593
1.2986
43
44
0.5565
0.0084
1.7483
0.0055
1.7483
44
45
0.3532
0.7547
1.1096
0.4906
Null
45
46
0.1188
0.5403
0.3732
0.3512
Null
46
47
0.5235
0.8228
1.6447
0.5348
Null
47
48
0.5526
0.8279
1.7361
0.5381
Null
48
49
0.6770
0.7266
2.1269
0.4723
Null
49
50
0.4398
0.2243
1.3816
0.1458
1.3816
50
51
0.9524
0.7628
2.9922
0.4958
2.9922
51
52
0.9227
0.7673
2.8988
0.4988
2.8988
52
53
0.7744
0.0388
2.4329
0.0253
2.4329
53
54
0.4234
0.3146
1.3302
0.2045
1.3302
54
55
0.2849
0.8526
0.8952
0.5542
Null
55
56
0.2712
0.4589
0.8521
0.2983
Null
56
57
0.5662
0.1270
1.7787
0.0826
1.7787
57
58
0.3261
0.1921
1.0244
0.1249
1.0244
58
59
0.9294
0.0971
2.9197
0.0631
2.9197
59
60
0.8678
0.0527
2.7264
0.0342
2.7264
60
61
0.6010
0.5101
1.8880
0.3315
1.8880
61
62
0.0491
0.8662
0.1543
0.5630
Null
62
63
0.8311
0.3699
2.6111
0.2404
2.6111
63
64
0.6055
0.7086
1.9021
0.4606
Null
64 (continued)
4.1
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
57
Table 4.1 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
65
0.0621
0.0293
0.1951
0.0190
0.1951
65
66
0.9600
0.7918
3.0158
0.5146
3.0158
66
67
0.2790
0.3078
0.8764
0.2000
Null
67
68
0.9336
0.7173
2.9329
0.4662
2.9329
68
69
0.9645
0.1720
3.0300
0.1118
3.0300
69
70
0.1165
0.6613
0.3660
0.4299
Null
70
71
0.9249
0.6311
2.9056
0.4102
2.9056
71
72
0.6276
0.1130
1.9716
0.0734
1.9716
72
73
0.1215
0.9735
0.3816
0.6328
Null
73
74
0.0148
0.2807
0.0465
0.1824
Null
74
75
0.1744
0.8147
0.5479
0.5296
Null
75
76
0.9503
0.8674
2.9854
0.5638
2.9854
76
77
0.3100
0.7364
0.9738
0.4787
Null
77
78
0.2646
0.1094
0.8312
0.0711
0.8312
78
79
0.1454
0.9416
0.4569
0.6121
Null
79
80
0.3930
0.0684
1.2345
0.0445
1.2345
80
81
0.6325
0.3698
1.9870
0.2404
1.9870
81
82
0.3261
0.3225
1.0246
0.2096
Null
82
83
0.7350
0.4885
2.3089
0.3175
2.3089
83
84
0.4781
0.0687
1.5020
0.0447
1.5020
84
85
0.7332
0.7276
2.3035
0.4730
Null
85
86
0.2226
0.7403
0.6995
0.4812
Null
86
87
0.6044
0.2942
1.8987
0.1912
1.8987
87
88
0.3337
0.8206
1.0484
0.5334
Null
88
89
0.6196
0.1319
1.9464
0.0858
1.9464
89
90
0.4761
0.1548
1.4957
0.1006
1.4957
90
91
0.1495
0.1481
0.4696
0.0962
Null
91
92
0.0263
0.9820
0.0826
0.6383
Null
92
93
0.6774
0.4351
2.1281
0.2828
2.1281
93
94
0.3336
0.4401
1.0481
0.2861
Null
94
95
0.3045
0.2773
0.9565
0.1803
0.9565
95
96
0.9500
0.5186
2.9844
0.3371
2.9844
96 (continued)
58
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.1 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
97
0.7374
0.9651
2.3167
0.6273
Null
97
98
0.6385
0.9689
2.0060
0.6298
Null
98
99
0.5705
0.4434
1.7922
0.2882
1.7922
99
100
0.5668
0.8242
1.7808
0.5357
Null
100
Table 4.2 Showing etaa[i] and y[i] = (Sin(etaa[i]))/etaa[i]. Using prog. 4.2 i
etaa[i]
y[i]
i
etaa[i]
y[i]
1
2.7075
0.1553
26
2.4329
0.2675
2
2.7005
0.1581
27
1.3302
0.7301
3
2.3865
0.2872
28
1.7787
0.5501
4
1.1935
0.7789
29
1.0244
0.8341
5
0.5967
0.9417
30
2.9197
0.0754
6
1.2128
0.7723
31
2.7264
0.1480
7
1.8477
0.5206
32
1.8880
0.5032
8
2.7517
0.1381
33
2.6111
0.1938
9
1.7723
0.5528
34
0.1951
0.9937
10
3.0938
0.0154
35
3.0158
0.0416
11
2.9501
0.0645
36
2.9329
0.0706
12
2.1218
0.4015
37
3.0300
0.0368
13
3.0590
0.0270
38
2.9056
0.0805
14
2.2028
0.3663
39
1.9716
0.4670
15
2.0785
0.4204
40
2.9854
0.0521
16
2.8351
0.1064
41
0.8312
0.8888
17
1.0771
0.8176
42
1.2345
0.7647
18
0.5199
0.9556
43
1.9870
0.4603
19
1.2236
0.7685
44
2.3089
0.3204
20
3.0256
0.0382
45
1.5020
0.6642
21
1.2986
0.7417
46
1.8987
0.4986
22
1.7483
0.5630
47
1.9464
0.4779
23
1.3816
0.7109
48
1.4957
0.6667
24
2.9922
0.0497
49
2.1281
0.3988
25
2.8988
0.0829
50
0.9565
0.8544
51
2.9844
0.0524
52
1.7922
0.5443
4.1
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
Program number 4.1 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[], u2[i] = RandomReal[], U1[i] = 0 + Pi*u1[i],U2[i] = 0.65*u2[i], If[U2[i] < 2*U1[i]/Pi^2,eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]", "U1[i]","U2[i]","eta[i]"}}] Program number 4.2 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + Pi*u1[i],U2[i] = 0.65*u2[i], If[U2[i] < 2*U1[i]/Pi^2,eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","eta[i]"}}] etaa[1] = eta[1];etaa[2] = eta[2];etaa[3] = eta[3];etaa[4] = eta[5];etaa[5] = eta[6]; etaa[6] = eta[8];etaa[7] = eta[9];etaa[8] = eta[11];etaa[9] = eta[18]; etaa[10] = eta[20];etaa[11] = eta[22];etaa[12] = eta[24];etaa[13] = eta[25]; etaa[14] = eta[27];etaa[15] = eta[28];etaa[16] = eta[29];etaa[17] = eta[32]; etaa[18] = eta[34];etaa[19] = eta[36];etaa[20] = eta[37];etaa[21] = eta[43]; etaa[22] = eta[44];etaa[23] = eta[50];etaa[24] = eta[51];etaa[25] = eta[52]; etaa[26] = eta[53];etaa[27] = eta[54];etaa[28] = eta[57];etaa[29] = eta[58]; etaa[30] = eta[59];etaa[31] = eta[60];etaa[32] = eta[61];etaa[33] = eta[63]; etaa[34] = eta[65];etaa[35] = eta[66];etaa[36] = eta[68];etaa[37] = eta[69]; etaa[38] = eta[71];etaa[39] = eta[72];etaa[40] = eta[76];etaa[41] = eta[78]; etaa[42] = eta[80];etaa[43] = eta[81];etaa[44] = eta[83];etaa[45] = eta[84]; etaa[46] = eta[87];etaa[47] = eta[89];etaa[48] = eta[90];etaa[49] = eta[93]; etaa[50] = eta[95];etaa[51] = eta[96];etaa[52] = eta[99]; Table[{i = i + 1,etaa[i],y[i] = (Sin[etaa[i]])/etaa[i]},{i,0,51,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","etaa[i]","y[i]"}}] Integral = Sum[Pi^2*y[i]/(2*52),{i,1,52}] ListPlot[Table[{etaa[i],1},{i,0,51,1}], Frame-> True,FrameLabel-> {"\[Eta]a [i]","1"}]
59
60
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
2.0
1
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
Fig. 4.2 Showing that random variates ηa are linearly distributed in the interval 0 to π. There are more variates for larger values of ηa
That random variate ηa is distributed as p(x) = 2x/π2 in the interval 0 to π is evidenced by Fig. 4.2 which displays values of ηa against stagnant integer 1. We find that indeed there are more variates for larger values of ηa. We now up a normalized Gaussian variation for p(x) given by p(x) = take 2 √1 exp − 1 x−a where a is average and σ is variance of x. See Fig. 4.3 in which 2 σ σ 2π we have plotted F(x) = Sin(x) along with the normalized Gaussian probability density function p(x). As Fig. 4.3 shows, the Gaussian probability density function p(x) with a = π/2 and σ = 0.5 is not too different from the function Sin(x). Hence it is ideal for use as p(x). 2 We now generate random variate g obeying p(g) = √1 exp − 21 g−a with a σ σ 2π = π/2 and σ = 0.5 in the interval 0 < g < π using acceptance-rejection sampling, using Program number 4.3 and resulting Table 4.3. From Table 4.3, we manually gather accepted random variates ga which we use in Program number 4.4 and resulting Table 4.4 to evaluate the integral with the result 2.132 rather than the exact value 2.
4.1
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
61
1.0
0.8
0.6
0.4
0.2
0.0 0.0
0.5
1.0
1.5
2.0
2.5
3.0
x 2 Fig. 4.3 Showing F(x) = Sin(x) as undashed curve and Gaussian p(x) = √1 exp − 21 x−a σ σ 2π as dashed curve obtained using the command sig = 0.5;a = Pi/2; p1 = Plot[Sin[x],{x,0,Pi},PlotStyle-> {Black},Frame-> True, FrameLabel-> {"x","F(x) = Sin (x) or Gaussian (x)"}]; p2 = Plot[(1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((x-a)/sig)^2],{x,0,Pi}, PlotStyle-> {Dashed,Black}];Show[p1,p2] in Mathematica 6.0
Program number 4.3 n = 654321 SeedRandom[n] sig = 0.5;a = Pi/2; Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + Pi*u1[i],U2[i] = 0 + 0.8*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))* Exp[-0.5*((U1[i]-a)/sig)^2],g[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]", "U1[i]","U2[i]","g[i]"}}]
62
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.3 Showing values of u1[i], u2[i], U1[i], U2[i] and g[i]. Using prog. 4.3 i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
1
0.8618
0.4287
2.7075
0.3430
Null
1
2
0.8596
0.0342
2.7005
0.0274
2.7005
2
3
0.7597
0.4609
2.3865
0.3687
Null
3
4
0.4189
0.4746
1.3160
0.3797
1.3160
4
5
0.3799
0.1687
1.1935
0.1350
1.1935
5
6
0.1899
0.0364
0.5967
0.0292
0.5967
6
7
0.3335
0.4257
1.0476
0.3406
1.0476
7
8
0.3860
0.2445
1.2128
0.1956
1.2128
8
9
0.5881
0.2322
1.8477
0.1858
1.8477
9
10
0.6743
0.9089
2.1184
0.7271
Null
10
11
0.8759
0.3658
2.7517
0.2927
Null
11
12
0.2945
0.6719
0.9253
0.5375
Null
12
13
0.9104
0.9276
2.8601
0.7421
Null
13
14
0.0504
0.3510
0.1583
0.2808
Null
14
15
0.1510
0.1761
0.4745
0.1409
Null
15
16
0.3796
0.5239
1.1926
0.4192
1.1926
16
17
0.3586
0.4202
1.1266
0.3362
1.1266
17
18
0.5641
0.1372
1.7723
0.1097
1.7723
18
19
0.0774
0.8924
0.2431
0.7140
Null
19
20
0.9848
0.4854
3.0938
0.3883
Null
20
21
0.3080
0.4945
0.9675
0.3956
Null
21
22
0.9390
0.5923
2.9501
0.4738
Null
22
23
0.1921
0.3830
0.6035
0.3064
Null
23
24
0.6754
0.2944
2.1218
0.2355
2.1218
24
25
0.9737
0.4501
3.0590
0.3600
Null
25
26
0.7833
0.7877
2.4609
0.6302
Null
26
27
0.7012
0.5937
2.2028
0.4750
Null
27
28
0.6616
0.5202
2.0785
0.4161
2.0785
28
29
0.9024
0.0050
2.8351
0.0040
2.8351
29
30
0.2183
0.2361
0.6857
0.1889
Null
30
31
0.4477
0.7308
1.4065
0.5846
1.4065
31
32
0.3429
0.2260
1.0771
0.1808
1.0771
32 (continued)
4.1
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
63
Table 4.3 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
33
0.1079
0.4879
0.3390
0.3903
Null
33
34
0.1655
0.1218
0.5199
0.0975
Null
34
35
0.3544
0.3523
1.1133
0.2819
1.1133
35
36
0.3895
0.0911
1.2236
0.0729
1.2236
36
37
0.9631
0.5184
3.0256
0.4147
Null
37
38
0.6328
0.8258
1.9880
0.6606
Null
38
39
0.4101
0.6742
1.2882
0.5394
1.2882
39
40
0.2088
0.8631
0.6560
0.6905
Null
40
41
0.2042
0.3027
0.6416
0.2422
Null
41
42
0.0504
0.8463
0.1582
0.6770
Null
42
43
0.4134
0.2451
1.2986
0.1961
1.2986
43
44
0.5565
0.0084
1.7483
0.0067
1.7483
44
45
0.3532
0.7547
1.1096
0.6038
Null
45
46
0.1188
0.5403
0.3732
0.4322
Null
46
47
0.5235
0.8228
1.6447
0.6582
1.6447
47
48
0.5526
0.8279
1.7361
0.6623
1.7361
48
49
0.6770
0.7266
2.1269
0.5812
Null
49
50
0.4398
0.2243
1.3816
0.1794
1.3816
50
51
0.9524
0.7628
2.9922
0.6102
Null
51
52
0.9227
0.7673
2.8988
0.6139
Null
52
53
0.7744
0.0388
2.4329
0.0311
2.4329
53
54
0.4234
0.3146
1.3302
0.2517
1.3302
54
55
0.2849
0.8526
0.8952
0.6821
Null
55
56
0.2712
0.4589
0.8521
0.3672
Null
56
57
0.5662
0.1270
1.7787
0.1016
1.7787
57
58
0.3261
0.1921
1.0244
0.1537
1.0244
58
59
0.9294
0.0971
2.9197
0.0777
Null
59
60
0.8678
0.0527
2.7264
0.0421
2.7264
60
61
0.6010
0.5101
1.8880
0.4081
1.8880
61
62
0.0491
0.8662
0.1543
0.6930
Null
62
63
0.8311
0.3699
2.6111
0.2959
Null
63
64
0.6055
0.7086
1.9021
0.5669
1.9021
64 (continued)
64
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.3 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
65
0.0621
0.0293
0.1951
0.0234
Null
65
66
0.9600
0.7918
3.0158
0.6334
Null
66
67
0.2790
0.3078
0.8764
0.2462
0.8764
67
68
0.9336
0.7173
2.9329
0.5738
Null
68
69
0.9645
0.1720
3.0300
0.1376
Null
69
70
0.1165
0.6613
0.3660
0.5291
Null
70
71
0.9249
0.6311
2.9056
0.5049
Null
71
72
0.6276
0.1130
1.9716
0.0904
1.9716
72
73
0.1215
0.9735
0.3816
0.7788
Null
73
74
0.0148
0.2807
0.0465
0.2246
Null
74
75
0.1744
0.8147
0.5479
0.6518
Null
75
76
0.9503
0.8674
2.9854
0.6939
Null
76
77
0.3100
0.7364
0.9738
0.5891
Null
77
78
0.2646
0.1094
0.8312
0.0875
0.8312
78
79
0.1454
0.9416
0.4569
0.7533
Null
79
80
0.3930
0.0684
1.2345
0.0547
1.2345
80
81
0.6325
0.3698
1.9870
0.2958
1.9870
81
82
0.3261
0.3225
1.0246
0.2580
1.0246
82
83
0.7350
0.4885
2.3089
0.3908
Null
83
84
0.4781
0.0687
1.5020
0.0550
1.5020
84
85
0.7332
0.7276
2.3035
0.5821
Null
85
86
0.2226
0.7403
0.6995
0.5922
Null
86
87
0.6044
0.2942
1.8987
0.2354
1.8987
87
88
0.3337
0.8206
1.0484
0.6565
Null
88
89
0.6196
0.1319
1.9464
0.1056
1.9464
89
90
0.4761
0.1548
1.4957
0.1238
1.4957
90
91
0.1495
0.1481
0.4696
0.1184
Null
91
92
0.0263
0.9820
0.0826
0.7856
Null
92
93
0.6774
0.4351
2.1281
0.3481
2.1281
93
94
0.3336
0.4401
1.0481
0.3521
1.0481
94
95
0.3045
0.2773
0.9565
0.2219
0.9565
95
96
0.9500
0.5186
2.9844
0.4149
Null
96 (continued)
4.1
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
65
Table 4.3 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
97
0.7374
0.9651
2.3167
0.7721
Null
97
98
0.6385
0.9689
2.0060
0.7751
Null
98
99
0.5705
0.4434
1.7922
0.3547
1.7922
99
100
0.5668
0.8242
1.7808
0.6593
1.7808
100
Table 4.4 Showing values of ga[i], p[i] and y[i] = (Sin(ga[i]))/p[i]. Using prog. 4.4 i
ga[i]
p[i]
y[i]
i
ga[i]
p[i]
y[i]
1
2.7005
0.0621
6.8699
26
1.7787
0.7318
1.3370
2
1.3160
0.7008
1.3810
27
1.0244
0.4391
1.9457
3
1.1935
0.6002
1.5489
28
2.7264
0.0552
7.3058
4
0.5967
0.1196
4.6977
29
1.8880
0.6525
1.4562
5
1.0476
0.4615
1.8769
30
1.9021
0.6406
1.4761
6
1.2128
0.6175
1.5168
31
0.8764
0.3041
2.5265
7
1.8477
0.6844
1.4054
32
1.9716
0.5786
1.5913
8
1.1926
0.5994
1.5504
33
0.8312
0.2672
2.7647
9
1.1266
0.5378
1.6791
34
1.2345
0.6364
1.4833
10
1.7723
0.7357
1.3318
35
1.9870
0.5642
1.6210
11
2.1218
0.4347
1.9599
36
1.0246
0.4393
1.9451
12
2.0785
0.4765
1.8339
37
1.5020
0.7904
1.2622
13
2.8351
0.0326
9.2479
38
1.8987
0.6435
1.4712
14
1.4065
0.7560
1.3050
39
1.9464
0.6017
1.5461
15
1.0771
0.4900
1.7970
40
1.4957
0.7889
1.2640
16
1.1133
0.5250
1.7089
41
2.1281
0.4287
1.9797
17
1.2236
0.6269
1.4999
42
1.0481
0.4620
1.8755
18
1.2882
0.6801
1.4120
43
0.9565
0.3751
2.1786
19
1.2986
0.6880
1.4000
44
1.7922
0.7234
1.3487
20
1.7483
0.7491
1.3139
45
1.7808
0.7306
1.3388
21
1.6447
0.7892
1.2636
22
1.7361
0.7554
1.3057
23
1.3816
0.7428
1.3223
24
2.4329
0.1805
3.6062
25
1.3302
0.7106
1.3667
66
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Program number 4.4 n = 654321 SeedRandom[n] sig = 0.5;a = Pi/2; Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + Pi*u1[i],U2[i] = 0 + 0.8*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))* Exp[-0.5*((U1[i]-a)/sig)^2],g[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","g[i]"}}] ga[1] = g[2];ga[2] = g[4];ga[3] = g[5];ga[4] = g[6];ga[5] = g[7];ga[6] = g[8]; ga[7] = g[9];ga[8] = g[16];ga[9] = g[17];ga[10] = g[18];ga[11] = g[24]; ga[12] = g[28];ga[13] = g[29];ga[14] = g[31];ga[15] = g[32];ga[16] = g[35]; ga[17] = g[36];ga[18] = g[39];ga[19] = g[43];ga[20] = g[44];ga[21] = g[47]; ga[22] = g[48];ga[23] = g[50];ga[24] = g[53];ga[25] = g[54];ga[26] = g[57]; ga[27] = g[58];ga[28] = g[60];ga[29] = g[61];ga[30] = g[64];ga[31] = g[67]; ga[32] = g[72];ga[33] = g[78];ga[34] = g[80];ga[35] = g[81];ga[36] = g[82]; ga[37] = g[84];ga[38] = g[87];ga[39] = g[89];ga[40] = g[90];ga[41] = g[93]; ga[42] = g[94];ga[43] = g[95];ga[44] = g[99];ga[45] = g[100]; Table[{i = i + 1,ga[i], p[i] = (1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((ga[i]-a)/sig)^2], y[i] = (Sin[ga[i]])/p[i]},{i,0,44,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","ga[i]", "p[i]","y[i]"}}] Integral = Sum[y[i]/(45),{i,1,45}] ListPlot[Table[{ga[i],1},{i,0,44,1}],Frame-> True, FrameLabel-> {"ga [i]","1"}] 2 That random variate ga is distributed as p(x) = √1 exp − 21 x−a in the interval σ σ 2π 0 to π is evidenced by Fig. 4.4 which displays values of ga against stagnant integer 1. We find that indeed there are more variates near ga = π /2.
4.2
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
67
2.0
1
1.5
1.0
0.5
0.0 1.0
1.5
2.0
2.5
ga Fig. 4.4 Showing random variates ga are distributed in the interval 0 to π as p(x) that 2 1 1 x−a √ exp − 2 σ . There are more variates near ga = π/2
=
σ 2π
4.2
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example II
As example II, we now take up the integral +π/2
b I =
F(x)d x =
Cos(x)d x
(4.4)
−π/2
a
where F(x) = Cos(x). We re-write Eq. (4.4) as b I =
F(x) p(x)d x p(x)
(4.5)
a
which, as discussed in Sect. 3.6, implies that average value of F/p is the value of the integral, i.e. I =
N 1 F(xi ) N p(xi ) i=1
(4.6)
68
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
where x i ’s are random values of x in the interval a < x < b obeying probability density function p(x). We now take up a linear variation for p(x) given by p(x) = C x + d where C and d are parameters such that slope C = intercept d on y axis /(π/2); hence d = Cπ /2. As +π/2 such p(x) = C(x + π /2). Normalization of p(x) requires −π/2 p(x)d x = 1 or, C = 2/π2 . As such normalized probability density function is p(x) = 2x/π2 + 1/π. See Fig. 4.5 in which we have plotted F(x) = Cos(x) along with the normalized linear probability density function p(x) = 2x/π2 + 1/π. The linear probability density function p(x) = 2x/π2 + 1/π is not too different from a slowly varying function of x like Cos(x). Hence it is a good choice. We now generate random variate η obeying p(x) = 2x/π2 + 1/π in the interval −π/2 < x < +π/2 using acceptance-rejection sampling, using Program number 4.5 and resulting Table 4.5. From Table 4.5, we manually gather accepted random variates ηa which we
1.0
0.8
0.6
0.4
0.2
0.0 1.5
1.0
0.5
0.0
0.5
1.0
1.5
x Fig. 4.5 Showing F(x) = Cos(x) as undashed curve and p(x) = 2x/π2 + 1/π as dashed curve obtained using the command Plot[{Cos[x],2*x/Pi^2 + 1/Pi},{x,-Pi/2,Pi/2},Frame-> True, FrameLabel-> {"x","F(x) = Cos(x) or p(x)"}, PlotStyle-> {{Black},{Dashed,Black}}] in Mathematica 6.0
4.2
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
69
use in Program number 4.6 and resulting Table 4.6 to evaluate the integral with the result 2.095 rather than the exact value 2. Program number 4.5 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = -Pi/2 + Pi*u1[i],U2[i] = 0 + 0.65*u2[i], If[U2[i] < 2*U1[i]/Pi^2 + 1/Pi,eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]", "U1[i]","U2[i]","eta[i]"}}] Program number 4.6 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = -Pi/2 + Pi*u1[i],U2[i] = 0 + 0.65*u2[i], If[U2[i] < 2*U1[i]/Pi^2 + 1/Pi,eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","eta[i]"}}] etaa[1] = eta[1];etaa[2] = eta[2];etaa[3] = eta[3];etaa[4] = eta[5]; etaa[5] = eta[6];etaa[6] = eta[8];etaa[7] = eta[9];etaa[8] = eta[11]; etaa[9] = eta[18];etaa[10] = eta[20];etaa[11] = eta[22];etaa[12] = eta[24]; etaa[13] = eta[25];etaa[14] = eta[27];etaa[15] = eta[28];etaa[16] = eta[29]; etaa[17] = eta[32];etaa[18] = eta[34];etaa[19] = eta[36];etaa[20] = eta[37]; etaa[21] = eta[43];etaa[22] = eta[44];etaa[23] = eta[50];etaa[24] = eta[51]; etaa[25] = eta[52];etaa[26] = eta[53];etaa[27] = eta[54];etaa[28] = eta[57]; etaa[29] = eta[58];etaa[30] = eta[59];etaa[31] = eta[60];etaa[32] = eta[61]; etaa[33] = eta[63];etaa[34] = eta[65];etaa[35] = eta[66];etaa[36] = eta[68]; etaa[37] = eta[69];etaa[38] = eta[71];etaa[39] = eta[72];etaa[40] = eta[76]; etaa[41] = eta[78];etaa[42] = eta[80];etaa[43] = eta[81];etaa[44] = eta[83]; etaa[45] = eta[84];etaa[46] = eta[87];etaa[47] = eta[89];etaa[48] = eta[90]; etaa[49] = eta[93];etaa[50] = eta[95];etaa[51] = eta[96];etaa[52] = eta[99]; Table[{i = i + 1,etaa[i],p[i] = 2*etaa[i]/Pi^2 + 1/Pi, y[i] = (Cos[etaa[i]])/p[i]},{i,0,51,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","etaa[i]","p[i]", "(Cos[etaa[i]])/p[i]"}}] Integral = Sum[y[i]/(52),{i,1,52}]
70
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.5 Showing u1[i], u2[i], U1[i], U2[i] and eta[i]. Using prog. 4.5 i
u1[i]
u2[i]
1
0.8618
0.4287
2
0.8596
3
0.7597
4
U1[i]
U2[i]
eta[i]
i
1.1367
0.2787
1.1367
1
0.0342
1.1297
0.0223
1.1297
2
0.4609
0.8157
0.2996
0.8157
3
0.4189
0.4746
−0.2548
0.3085
Null
4
5
0.3799
0.1687
−0.3773
0.1097
−0.3773
5
6
0.1899
0.0364
−0.9741
0.0237
−0.9741
6
7
0.3335
0.4257
−0.5232
0.2767
Null
7
8
0.3860
0.2445
−0.3580
0.1589
−0.3580
8
9
0.5881
0.2322
0.2769
0.1509
0.2769
9
10
0.6743
0.9089
0.5476
0.5908
Null
10
11
0.8759
0.3658
1.1809
0.2378
1.1809
11
12
0.2945
0.6719
−0.6455
0.4367
Null
12
13
0.9104
0.9276
1.2893
0.6029
Null
13
14
0.0504
0.3510
−1.4125
0.2281
Null
14
15
0.1510
0.1761
−1.0963
0.1145
Null
15
16
0.3796
0.5239
−0.3782
0.3406
Null
16
17
0.3586
0.4202
−0.4442
0.2731
Null
17
18
0.5641
0.1372
0.2015
0.0892
0.2015
18
19
0.0774
0.8924
−1.3277
0.5801
Null
19
20
0.9848
0.4854
1.5230
0.3155
1.5230
20
21
0.3080
0.4945
−0.6033
0.3214
Null
21
22
0.9390
0.5923
1.3793
0.3850
1.3793
22
23
0.1921
0.3830
−0.9673
0.2490
Null
23
24
0.6754
0.2944
0.5510
0.1914
0.5510
24
25
0.9737
0.4501
1.4882
0.2925
1.4882
25
26
0.7833
0.7877
0.8901
0.5120
Null
26
27
0.7012
0.5937
0.6320
0.3859
0.6320
27
28
0.6616
0.5202
0.5077
0.3381
0.5077
28
29
0.9024
0.0050
1.2643
0.0032
1.2643
29
30
0.2183
0.2361
−0.8851
0.1534
Null
30
31
0.4477
0.7308
−0.1643
0.4750
Null
31
32
0.3429
0.2260
−0.4937
0.1469
−0.4937
32 (continued)
4.2
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
71
Table 4.5 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
33
0.1079
0.4879
−1.2318
0.3171
Null
33
34
0.1655
0.1218
−1.0509
0.0792
−1.0509
34
35
0.3544
0.3523
−0.4575
0.2290
Null
35
36
0.3895
0.0911
−0.3472
0.0592
−0.3472
36
37
0.9631
0.5184
1.4548
0.3370
1.4548
37
38
0.6328
0.8258
0.4172
0.5368
Null
38
39
0.4101
0.6742
−0.2826
0.4382
Null
39
40
0.2088
0.8631
−0.9148
0.5610
Null
40
41
0.2042
0.3027
−0.9292
0.1968
Null
41
42
0.0504
0.8463
−1.4126
0.5501
Null
42
43
0.4134
0.2451
−0.2722
0.1593
−0.2722
43
44
0.5565
0.0084
0.1775
0.0055
0.1775
44
45
0.3532
0.7547
−0.4612
0.4906
Null
45
46
0.1188
0.5403
−1.1976
0.3512
Null
46
47
0.5235
0.8228
0.0739
0.5348
Null
47
48
0.5526
0.8279
0.1653
0.5381
Null
48
49
0.6770
0.7266
0.5561
0.4723
Null
49
50
0.4398
0.2243
−0.1892
0.1458
−0.1892
50
51
0.9524
0.7628
1.4214
0.4958
1.4214
51
52
0.9227
0.7673
1.3281
0.4988
1.3281
52
53
0.7744
0.0388
0.8621
0.0253
0.8621
53
54
0.4234
0.3146
−0.2406
0.2045
−0.2406
54
55
0.2849
0.8526
−0.6756
0.5542
Null
55
56
0.2712
0.4589
−0.7187
0.2983
Null
56
57
0.5662
0.1270
0.2079
0.0826
0.2079
57
58
0.3261
0.1921
−0.5464
0.1249
−0.5464
58
59
0.9294
0.0971
1.3489
0.0631
1.3489
59
60
0.8678
0.0527
1.1556
0.0342
1.1556
60
61
0.6010
0.5101
0.3172
0.3315
0.3172
61
62
0.0491
0.8662
−1.4165
0.5630
Null
62
63
0.8311
0.3699
1.0403
0.2404
1.0403
63
64
0.6055
0.7086
0.3313
0.4606
Null
64 (continued)
72
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.5 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
65
0.0621
0.0293
−1.3757
0.0190
−1.3757
65
66
0.9600
0.7918
1.4450
0.5146
1.4450
66
67
0.2790
0.3078
−0.6944
0.2000
Null
67
68
0.9336
0.7173
1.3621
0.4662
1.3621
68
69
0.9645
0.1720
1.4592
0.1118
1.4592
69
70
0.1165
0.6613
−1.2048
0.4299
Null
70
71
0.9249
0.6311
1.3348
0.4102
1.3348
71
72
0.6276
0.1130
0.4008
0.0734
0.4008
72
73
0.1215
0.9735
−1.1892
0.6328
Null
73
74
0.0148
0.2807
−1.5243
0.1824
Null
74
75
0.1744
0.8147
−1.0229
0.5296
Null
75
76
0.9503
0.8674
1.4146
0.5638
1.4146
76
77
0.3100
0.7364
−0.5970
0.4787
Null
77
78
0.2646
0.1094
−0.7396
0.0711
−0.7396
78
79
0.1454
0.9416
−1.1139
0.6121
Null
79
80
0.3930
0.0684
−0.3363
0.0445
−0.3363
80
81
0.6325
0.3698
0.4162
0.2404
0.4162
81
82
0.3261
0.3225
−0.5462
0.2096
Null
82
83
0.7350
0.4885
0.7381
0.3175
0.7381
83
84
0.4781
0.0687
−0.0688
0.0447
−0.0688
84
85
0.7332
0.7276
0.7327
0.4730
Null
85
86
0.2226
0.7403
−0.8713
0.4812
Null
86
87
0.6044
0.2942
0.3279
0.1912
0.3279
87
88
0.3337
0.8206
−0.5224
0.5334
Null
88
89
0.6196
0.1319
0.3756
0.0858
0.3756
89
90
0.4761
0.1548
−0.0751
0.1006
−0.0751
90
91
0.1495
0.1481
−1.1012
0.0962
Null
91
92
0.0263
0.9820
−1.4882
0.6383
Null
92
93
0.6774
0.4351
0.5573
0.2828
0.5573
93
94
0.3336
0.4401
−0.5227
0.2861
Null
94
95
0.3045
0.2773
−0.6143
0.1803
−0.6143
95
96
0.9500
0.5186
1.4136
0.3371
1.4136
96 (continued)
4.2
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
73
Table 4.5 (continued) i
u1[i]
u2[i]
97
0.7374
0.9651
98
0.6385
99
0.5705
100
0.5668
U1[i]
U2[i]
eta[i]
i
0.7459
0.6273
Null
97
0.9689
0.4352
0.6298
Null
98
0.4434
0.2214
0.2882
0.2214
99
0.8242
0.2100
0.5357
Null
100
Table 4.6 Showing etaa[i] and y[i] = (Cos(etaa[i]))/p[i]. Using prog. 4.6 i
etaa[i]
p[i]
y[i]
i
etaa[i]
p[i]
y[i]
1
1.1367
0.5487
0.7665
26
0.8621
0.4930
1.3202
2
1.1297
0.5472
0.7801
27
−0.2406
0.2695
3.6030
3
0.8157
0.4836
1.4171
28
0.2079
0.3604
2.7147
4
−0.3773
0.2419
3.8439
29
−0.5464
0.2076
4.1160
5
−0.9741
0.1209
4.6471
30
1.3489
0.5917
0.3720
6
−0.3580
0.2458
3.8110
31
1.1556
0.5525
0.7301
7
0.2769
0.3744
2.5690
32
0.3172
0.3826
2.4834
8
1.1809
0.5576
0.6817
33
1.0403
0.5291
0.9563
9
0.2015
0.3591
2.7281
34
−1.3757
0.0395
4.9035
10
1.5230
0.6269
0.0762
35
1.4450
0.6111
0.2052
11
1.3793
0.5978
0.3185
36
1.3621
0.5943
0.3485
12
0.5510
0.4300
1.9815
37
1.4592
0.6140
0.1814
13
1.4882
0.6199
0.1331
38
1.3348
0.5888
0.3971
14
0.6320
0.4464
1.8075
39
0.4008
0.3995
2.3045
15
0.5077
0.4212
2.0748
40
1.4146
0.6050
0.2571
16
1.2643
0.5745
0.5252
41
−0.7396
0.1684
4.3858
17
−0.4937
0.2183
4.0345
42
−0.3363
0.2502
3.7734
18
−1.0509
0.1054
4.7155
43
0.4162
0.4027
2.2715
19
−0.3472
0.2479
3.7925
44
0.7381
0.4679
1.5810
20
1.4548
0.6131
0.1887
45
−0.0688
0.3044
3.2777
21
−0.2722
0.2632
3.6601
46
0.3279
0.3848
2.4606
22
0.1775
0.3543
2.7782
47
0.3756
0.3944
2.3586
23
−0.1892
0.2800
3.5080
48
−0.0751
0.3031
3.2901
24
1.4214
0.6063
0.2455
49
0.5573
0.4312
1.9680
25
1.3281
0.5874
0.4092
50
−0.6143
0.1938
4.2161
51
1.4136
0.6048
0.2588
52
0.2214
0.3632
2.6862
74
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
2.0
1
1.5
1.0
0.5
0.0 1.0
0.5
0.0
0.5
1.0
1.5
Fig. 4.6 Showing that random variates ηa are linearly distributed in the interval −π/2 to +π/2. There are more variates for larger values of ηa
ListPlot[Table[{etaa[i],1},{i,0,51,1}], Frame-> True,FrameLabel-> {"\[Eta]a [i]","1"}] That random variate ηa is distributed as p(x) = 2x/π2 + 1/π in the interval −π/2 to +π/2 is evidenced by Fig. 4.6 which displays values of ηa against stagnant integer 1. We find that indeed there are more variates for larger values of ηa. We now up a normalized Gaussian variation for p(x) given by p(x) = take 2 √1 exp − 1 x−a where a is average and σ is variance of x. See Fig. 4.7 in which 2 σ σ 2π we have plotted F(x) = Cos(x) along with the normalized Gaussian probability density function p(x). As Fig. 4.7 shows, the Gaussian probability density function p(x) with a = 0 and σ = 0.5 is not too different from the function Cos(x). Hence it is ideal for use as p(x). 2 We now generate random variate g obeying p(g) = √1 exp − 21 g−a with a = σ σ 2π 0 and σ = 0.5 in the interval −π/2 < g < +π/2 using acceptance-rejection sampling, using Program number 4.7 and resulting Table 4.7. From Table 4.7, we manually gather accepted random variates ga which we use in Program number 4.8 and resulting Table 4.8 to evaluate the integral with the result 2.132 rather than the exact value 2.
4.2
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
75
1.0
0.8
0.6
0.4
0.2
0.0 1.5
1.0
0.5
0.0 x
0.5
1.0
1.5
2 Fig. 4.7 Showing F(x) = Cos(x) as undashed curve and Gaussian p(x) = √1 exp − 21 x−a σ σ 2π as dashed curve obtained using the command sig = 0.5; a = 0; p1 = Plot[Cos[x],{x,-Pi/2,Pi/2},PlotStyle-> {Black},Frame-> True, FrameLabel-> {"x","F(x) = Cos (x) or Gaussian (x)"}]; p2 = Plot[(1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((x-a)/sig)^2], {x,-Pi/2,Pi/2},PlotStyle-> {Dashed,Black}]; Show[p1,p2] in Mathematica 6.0
Program number 4.7 n = 654321 SeedRandom[n] sig = 0.5;a = 0; Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = -Pi/2 + Pi*u1[i],U2[i] = 0 + 0.8*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((U1[i]-a)/sig)^2], g[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]", "U1[i]","U2[i]","g[i]"}}]
76
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.7 Showing u1[i], u2[i], U1[i], U2[i] and g[i]. Using prog. 4.7 i
u1[i]
u2[i]
1
0.8618
0.4287
2
0.8596
3
0.7597
4
U1[i]
U2[i]
g[i]
i
1.1367
0.3430
Null
1
0.0342
1.1297
0.0274
1.1297
2
0.4609
0.8157
0.3687
Null
3
0.4189
0.4746
−0.2548
0.3797
−0.2548
4
5
0.3799
0.1687
−0.3773
0.1350
−0.3773
5
6
0.1899
0.0364
−0.9741
0.0292
−0.9741
6
7
0.3335
0.4257
−0.5232
0.3406
−0.5232
7
8
0.3860
0.2445
−0.3580
0.1956
−0.3580
8
9
0.5881
0.2322
0.2769
0.1858
0.2769
9
10
0.6743
0.9089
0.5476
0.7271
Null
10
11
0.8759
0.3658
1.1809
0.2927
Null
11
12
0.2945
0.6719
−0.6455
0.5375
Null
12
13
0.9104
0.9276
1.2893
0.7421
Null
13
14
0.0504
0.3510
−1.4125
0.2808
Null
14
15
0.1510
0.1761
−1.0963
0.1409
Null
15
16
0.3796
0.5239
−0.3782
0.4192
−0.3782
16
17
0.3586
0.4202
−0.4442
0.3362
−0.4442
17
18
0.5641
0.1372
0.2015
0.1097
0.2015
18
19
0.0774
0.8924
−1.3277
0.7140
Null
19
20
0.9848
0.4854
1.5230
0.3883
Null
20
21
0.3080
0.4945
−0.6033
0.3956
Null
21
22
0.9390
0.5923
1.3793
0.4738
Null
22
23
0.1921
0.3830
−0.9673
0.3064
Null
23
24
0.6754
0.2944
0.5510
0.2355
0.5510
24
25
0.9737
0.4501
1.4882
0.3600
Null
25
26
0.7833
0.7877
0.8901
0.6302
Null
26
27
0.7012
0.5937
0.6320
0.4750
Null
27
28
0.6616
0.5202
0.5077
0.4161
0.5077
28
29
0.9024
0.0050
1.2643
0.0040
1.2643
29
30
0.2183
0.2361
−0.8851
0.1889
Null
30
31
0.4477
0.7308
−0.1643
0.5846
−0.1643
31
32
0.3429
0.2260
−0.4937
0.1808
−0.4937
32 (continued)
4.2
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
77
Table 4.7 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
33
0.1079
0.4879
−1.2318
0.3903
Null
33
34
0.1655
0.1218
−1.0509
0.0975
Null
34
35
0.3544
0.3523
−0.4575
0.2819
−0.4575
35
36
0.3895
0.0911
−0.3472
0.0729
−0.3472
36
37
0.9631
0.5184
1.4548
0.4147
Null
37
38
0.6328
0.8258
0.4172
0.6606
Null
38
39
0.4101
0.6742
−0.2826
0.5394
−0.2826
39
40
0.2088
0.8631
−0.9148
0.6905
Null
40
41
0.2042
0.3027
−0.9292
0.2422
Null
41
42
0.0504
0.8463
−1.4126
0.6770
Null
42
43
0.4134
0.2451
−0.2722
0.1961
−0.2722
43
44
0.5565
0.0084
0.1775
0.0067
0.1775
44
45
0.3532
0.7547
−0.4612
0.6038
Null
45
46
0.1188
0.5403
−1.1976
0.4322
Null
46
47
0.5235
0.8228
0.0739
0.6582
0.0739
47
48
0.5526
0.8279
0.1653
0.6623
0.1653
48
49
0.6770
0.7266
0.5561
0.5812
Null
49
50
0.4398
0.2243
−0.1892
0.1794
−0.1892
50
51
0.9524
0.7628
1.4214
0.6102
Null
51
52
0.9227
0.7673
1.3281
0.6139
Null
52
53
0.7744
0.0388
0.8621
0.0311
0.8621
53
54
0.4234
0.3146
−0.2406
0.2517
−0.2406
54
55
0.2849
0.8526
−0.6756
0.6821
Null
55
56
0.2712
0.4589
−0.7187
0.3672
Null
56
57
0.5662
0.1270
0.2079
0.1016
0.2079
57
58
0.3261
0.1921
−0.5464
0.1537
−0.5464
58
59
0.9294
0.0971
1.3489
0.0777
Null
59
60
0.8678
0.0527
1.1556
0.0421
1.1556
60
61
0.6010
0.5101
0.3172
0.4081
0.3172
61
62
0.0491
0.8662
−1.4165
0.6930
Null
62
63
0.8311
0.3699
1.0403
0.2959
Null
63
64
0.6055
0.7086
0.3313
0.5669
0.3313
64 (continued)
78
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.7 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
65
0.0621
0.0293
−1.3757
0.0234
Null
65
66
0.9600
0.7918
1.4450
0.6334
Null
66
67
0.2790
0.3078
−0.6944
0.2462
−0.6944
67
68
0.9336
0.7173
1.3621
0.5738
Null
68
69
0.9645
0.1720
1.4592
0.1376
Null
69
70
0.1165
0.6613
−1.2048
0.5291
Null
70
71
0.9249
0.6311
1.3348
0.5049
Null
71
72
0.6276
0.1130
0.4008
0.0904
0.4008
72
73
0.1215
0.9735
−1.1892
0.7788
Null
73
74
0.0148
0.2807
−1.5243
0.2246
Null
74
75
0.1744
0.8147
−1.0229
0.6518
Null
75
76
0.9503
0.8674
1.4146
0.6939
Null
76
77
0.3100
0.7364
−0.5970
0.5891
Null
77
78
0.2646
0.1094
−0.7396
0.0875
−0.7396
78
79
0.1454
0.9416
−1.1139
0.7533
Null
79
80
0.3930
0.0684
−0.3363
0.0547
−0.3363
80
81
0.6325
0.3698
0.4162
0.2958
0.4162
81
82
0.3261
0.3225
−0.5462
0.2580
−0.5462
82
83
0.7350
0.4885
0.7381
0.3908
Null
83
84
0.4781
0.0687
−0.0688
0.0550
−0.0688
84
85
0.7332
0.7276
0.7327
0.5821
Null
85
86
0.2226
0.7403
−0.8713
0.5922
Null
86
87
0.6044
0.2942
0.3279
0.2354
0.3279
87
88
0.3337
0.8206
−0.5224
0.6565
Null
88
89
0.6196
0.1319
0.3756
0.1056
0.3756
89
90
0.4761
0.1548
−0.0751
0.1238
−0.0751
90
91
0.1495
0.1481
−1.1012
0.1184
Null
91
92
0.0263
0.9820
−1.4882
0.7856
Null
92
93
0.6774
0.4351
0.5573
0.3481
0.5573
93
94
0.3336
0.4401
−0.5227
0.3521
−0.5227
94
95
0.3045
0.2773
−0.6143
0.2219
−0.6143
95
96
0.9500
0.5186
1.4136
0.4149
Null
96 (continued)
4.2
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
79
Table 4.7 (continued) i
u1[i]
u2[i]
97
0.7374
0.9651
98
0.6385
99
0.5705
100
0.5668
U1[i]
U2[i]
g[i]
i
0.7459
0.7721
Null
97
0.9689
0.4352
0.7751
Null
98
0.4434
0.2214
0.3547
0.2214
99
0.8242
0.2100
0.6593
0.2100
100
Table 4.8 Showing ga[i], p[i] and y[i] = (Cos(ga[i]))/p[i]. Using prog. 4.8 i
ga[i]
p[i]
y[i]
i
ga[i]
p[i]
y[i]
1
1.1297
0.0621
6.8699
26
0.2079
0.7318
1.3370
2
−0.2548
0.7008
1.3810
27
−0.5464
0.4391
1.9457
3
−0.3773
0.6002
1.5489
28
1.1556
0.0552
7.3058
4
−0.9741
0.1196
4.6977
29
0.3172
0.6525
1.4562
5
−0.5232
0.4615
1.8769
30
0.3313
0.6406
1.4761
6
−0.3580
0.6175
1.5168
31
−0.6944
0.3041
2.5265
7
0.2769
0.6844
1.4054
32
0.4008
0.5786
1.5913
8
−0.3782
0.5994
1.5504
33
−0.7396
0.2672
2.7647
9
−0.4442
0.5378
1.6791
34
−0.3363
0.6364
1.4833
10
0.2015
0.7357
1.3318
35
0.4162
0.5642
1.6210
11
0.5510
0.4347
1.9599
36
−0.5462
0.4393
1.9451
12
0.5077
0.4765
1.8339
37
−0.0688
0.7904
1.2622
13
1.2643
0.0326
9.2479
38
0.3279
0.6435
1.4712
14
−0.1643
0.7560
1.3050
39
0.3756
0.6017
1.5461
15
−0.4937
0.4900
1.7970
40
−0.0751
0.7889
1.2640
16
−0.4575
0.5250
1.7089
41
0.5573
0.4287
1.9797
17
−0.3472
0.6269
1.4999
42
−0.5227
0.4620
1.8755
18
−0.2826
0.6801
1.4120
43
−0.6143
0.3751
2.1786
19
−0.2722
0.6880
1.4000
44
0.2214
0.7234
1.3487
20
0.1775
0.7491
1.3139
45
0.2100
0.7306
1.3388
21
0.0739
0.7892
1.2636
22
0.1653
0.7554
1.3057
23
−0.1892
0.7428
1.3223
24
0.8621
0.1805
3.6062
25
−0.2406
0.7106
1.3667
80
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Program number 4.8 n = 654321 SeedRandom[n] sig = 0.5;a = 0; Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = -Pi/2 + Pi*u1[i],U2[i] = 0 + 0.8*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((U1[i]-a)/sig)^2], g[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","g[i]"}}] ga[1] = g[2];ga[2] = g[4];ga[3] = g[5];ga[4] = g[6];ga[5] = g[7];ga[6] = g[8]; ga[7] = g[9];ga[8] = g[16];ga[9] = g[17];ga[10] = g[18];ga[11] = g[24]; ga[12] = g[28];ga[13] = g[29];ga[14] = g[31];ga[15] = g[32];ga[16] = g[35]; ga[17] = g[36];ga[18] = g[39];ga[19] = g[43];ga[20] = g[44];ga[21] = g[47]; ga[22] = g[48];ga[23] = g[50];ga[24] = g[53];ga[25] = g[54];ga[26] = g[57]; ga[27] = g[58];ga[28] = g[60];ga[29] = g[61];ga[30] = g[64];ga[31] = g[67]; ga[32] = g[72];ga[33] = g[78];ga[34] = g[80];ga[35] = g[81];ga[36] = g[82]; ga[37] = g[84];ga[38] = g[87];ga[39] = g[89];ga[40] = g[90];ga[41] = g[93]; ga[42] = g[94];ga[43] = g[95];ga[44] = g[99];ga[45] = g[100]; Table[{i = i + 1,ga[i], p[i] = (1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((ga[i]-a)/sig)^2], y[i] = (Cos[ga[i]])/p[i]},{i,0,44,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","ga[i]", "p[i]","y[i]"}}] Integral = Sum[y[i]/(45),{i,1,45}] ListPlot[Table[{ga[i],1},{i,0,44,1}],Frame-> True, FrameLabel-> {"ga [i]","1"}] 2 That random variate ga is distributed as p(x) = √1 exp − 21 x−a in the interval σ σ 2π −π/2 to +π/2 is evidenced by Fig. 4.8 which displays values of ga against stagnant integer 1. We find that indeed there are more variates near ga = 0 than elsewhere.
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
81
2.0
1
1.5
1.0
0.5
0.0 1.0
0.5
0.0
0.5
1.0
Fig. 4.8 Showing random variates ga are distributed in the interval −π/2 to +π/2 as p(x) = that 2 √1 exp − 1 x−a . There are more variates near ga = 0 2 σ σ 2π
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example III
As example III, we now take up the integral b I =
2 F(x)d x =
a
ex d x
(4.7)
0
where F(x) = ex . We re-write Eq. (4.7) as b I =
F(x) p(x)d x p(x)
(4.8)
a
which, as discussed in Sect. 3.6, implies that average value of F/p is the value of the integral, i.e. I =
N 1 F(xi ) N p(xi ) i=1
(4.9)
82
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
where x i ’s are random values of x in the interval a < x < b obeying probability density function p(x). We now take up a linear variation for p(x) given by p(x) = C x where C is a constant. 2 Normalization of p(x) requires 0 p(x)d x = 1 or, C = 1/2. As such normalized probability density function is p(x) = x/2. See Fig. 4.9 in which we have plotted F(x) = ex along with the normalized linear probability density function p(x) = x/2. The linear probability density function p(x) = x/2 follows the function ex to some extent. Hence it is not a bad choice. We now generate random variate η obeying p(x) = x/2 in the interval 0 < x < 2 using acceptance-rejection sampling, using Program number 4.9 and resulting Table 4.9. From Table 4.9, we manually gather accepted random variates ηa which we use in Program number 4.10 and resulting Table 4.10 to evaluate the integral with the result 6.460 rather than the exact value 6.39.
7 6 5 4 3 2 1 0 0.0
0.5
1.0
1.5
2.0
x Fig. 4.9 Showing F(x) = ex as undashed curve and p(x) = x/2 as dashed curve obtained using the command Plot[{Exp[x],x/2},{x,0,2},Frame-> True, FrameLabel-> {"x","F(x) = e^x or p(x)"}, PlotStyle-> {{Black},{Dashed, Black}}] in Mathematica 6.0
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
83
Table 4.9 Showing u1[i], u2[i], U1[i], U2[i] and eta[i]. Using prog. 4.9 i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
1
0.8618
0.4287
1.7237
0.4287
1.7237
1
2
0.8596
0.0342
1.7192
0.0342
1.7192
2
3
0.7597
0.4609
1.5193
0.4609
1.5193
3
4
0.4189
0.4746
0.8378
0.4746
Null
4
5
0.3799
0.1687
0.7598
0.1687
0.7598
5
6
0.1899
0.0364
0.3799
0.0364
0.3799
6
7
0.3335
0.4257
0.6669
0.4257
Null
7
8
0.3860
0.2445
0.7721
0.2445
0.7721
8
9
0.5881
0.2322
1.1763
0.2322
1.1763
9
10
0.6743
0.9089
1.3486
0.9089
Null
10
11
0.8759
0.3658
1.7518
0.3658
1.7518
11
12
0.2945
0.6719
0.5891
0.6719
Null
12
13
0.9104
0.9276
1.8208
0.9276
Null
13
14
0.0504
0.3510
0.1008
0.3510
Null
14
15
0.1510
0.1761
0.3021
0.1761
Null
15
16
0.3796
0.5239
0.7593
0.5239
Null
16
17
0.3586
0.4202
0.7172
0.4202
Null
17
18
0.5641
0.1372
1.1283
0.1372
1.1283
18
19
0.0774
0.8924
0.1548
0.8924
Null
19
20
0.9848
0.4854
1.9696
0.4854
1.9696
20
21
0.3080
0.4945
0.6159
0.4945
Null
21
22
0.9390
0.5923
1.8781
0.5923
1.8781
22
23
0.1921
0.3830
0.3842
0.3830
Null
23
24
0.6754
0.2944
1.3508
0.2944
1.3508
24
25
0.9737
0.4501
1.9474
0.4501
1.9474
25
26
0.7833
0.7877
1.5667
0.7877
Null
26
27
0.7012
0.5937
1.4024
0.5937
1.4024
27
28
0.6616
0.5202
1.3232
0.5202
1.3232
28
29
0.9024
0.0050
1.8049
0.0050
1.8049
29
30
0.2183
0.2361
0.4365
0.2361
Null
30
31
0.4477
0.7308
0.8954
0.7308
Null
31
32
0.3429
0.2260
0.6857
0.2260
0.6857
32 (continued)
84
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.9 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
33
0.1079
0.4879
0.2158
0.4879
Null
33
34
0.1655
0.1218
0.3310
0.1218
0.3310
34
35
0.3544
0.3523
0.7088
0.3523
0.7088
35
36
0.3895
0.0911
0.7789
0.0911
0.7789
36
37
0.9631
0.5184
1.9262
0.5184
1.9262
37
38
0.6328
0.8258
1.2656
0.8258
Null
38
39
0.4101
0.6742
0.8201
0.6742
Null
39
40
0.2088
0.8631
0.4176
0.8631
Null
40
41
0.2042
0.3027
0.4084
0.3027
Null
41
42
0.0504
0.8463
0.1007
0.8463
Null
42
43
0.4134
0.2451
0.8267
0.2451
0.8267
43
44
0.5565
0.0084
1.1130
0.0084
1.1130
44
45
0.3532
0.7547
0.7064
0.7547
Null
45
46
0.1188
0.5403
0.2376
0.5403
Null
46
47
0.5235
0.8228
1.0471
0.8228
Null
47
48
0.5526
0.8279
1.1052
0.8279
Null
48
49
0.6770
0.7266
1.3540
0.7266
Null
49
50
0.4398
0.2243
0.8796
0.2243
0.8796
50
51
0.9524
0.7628
1.9049
0.7628
1.9049
51
52
0.9227
0.7673
1.8455
0.7673
1.8455
52
53
0.7744
0.0388
1.5488
0.0388
1.5488
53
54
0.4234
0.3146
0.8468
0.3146
0.8468
54
55
0.2849
0.8526
0.5699
0.8526
Null
55
56
0.2712
0.4589
0.5424
0.4589
Null
56
57
0.5662
0.1270
1.1323
0.1270
1.1323
57
58
0.3261
0.1921
0.6521
0.1921
0.6521
58
59
0.9294
0.0971
1.8587
0.0971
1.8587
59
60
0.8678
0.0527
1.7357
0.0527
1.7357
60
61
0.6010
0.5101
1.2019
0.5101
1.2019
61
62
0.0491
0.8662
0.0982
0.8662
Null
62
63
0.8311
0.3699
1.6623
0.3699
1.6623
63
64
0.6055
0.7086
1.2109
0.7086
Null
64 (continued)
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
85
Table 4.9 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
65
0.0621
0.0293
0.1242
0.0293
0.1242
65
66
0.9600
0.7918
1.9199
0.7918
1.9199
66
67
0.2790
0.3078
0.5579
0.3078
Null
67
68
0.9336
0.7173
1.8672
0.7173
1.8672
68
69
0.9645
0.1720
1.9289
0.1720
1.9289
69
70
0.1165
0.6613
0.2330
0.6613
Null
70
71
0.9249
0.6311
1.8498
0.6311
1.8498
71
72
0.6276
0.1130
1.2552
0.1130
1.2552
72
73
0.1215
0.9735
0.2430
0.9735
Null
73
74
0.0148
0.2807
0.0296
0.2807
Null
74
75
0.1744
0.8147
0.3488
0.8147
Null
75
76
0.9503
0.8674
1.9006
0.8674
1.9006
76
77
0.3100
0.7364
0.6200
0.7364
Null
77
78
0.2646
0.1094
0.5292
0.1094
0.5292
78
79
0.1454
0.9416
0.2909
0.9416
Null
79
80
0.3930
0.0684
0.7859
0.0684
0.7859
80
81
0.6325
0.3698
1.2650
0.3698
1.2650
81
82
0.3261
0.3225
0.6523
0.3225
0.6523
82
83
0.7350
0.4885
1.4699
0.4885
1.4699
83
84
0.4781
0.0687
0.9562
0.0687
0.9562
84
85
0.7332
0.7276
1.4665
0.7276
1.4665
85
86
0.2226
0.7403
0.4453
0.7403
Null
86
87
0.6044
0.2942
1.2087
0.2942
1.2087
87
88
0.3337
0.8206
0.6675
0.8206
Null
88
89
0.6196
0.1319
1.2391
0.1319
1.2391
89
90
0.4761
0.1548
0.9522
0.1548
0.9522
90
91
0.1495
0.1481
0.2990
0.1481
0.2990
91
92
0.0263
0.9820
0.0526
0.9820
Null
92
93
0.6774
0.4351
1.3548
0.4351
1.3548
93
94
0.3336
0.4401
0.6673
0.4401
Null
94
95
0.3045
0.2773
0.6089
0.2773
0.6089
95
96
0.9500
0.5186
1.9000
0.5186
1.9000
96 (continued)
86
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.9 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
97
0.7374
0.9651
1.4749
0.9651
Null
97
98
0.6385
0.9689
1.2771
0.9689
Null
98
99
0.5705
0.4434
1.1410
0.4434
1.1410
99
100
0.5668
0.8242
1.1337
0.8242
Null
100
That random variate η a is distributed as p(x) = x/2 in the interval 0 to 2 is evidenced by Fig. 4.10 which displays values of ηa against stagnant integer 1. We find that indeed there are more variates for larger values of ηa. Program number 4.9 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + 2*u1[i],U2[i] = 0 + 1*u2[i], If[U2[i] < U1[i]/2,eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]", "U1[i]","U2[i]","eta[i]"}}] Program number 4.10 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + 2*u1[i],U2[i] = 0 + 1*u2[i], If[U2[i] < U1[i]/2,eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","eta[i]"}}] etaa[1] = eta[1];etaa[2] = eta[2];etaa[3] = eta[3];etaa[4] = eta[5]; etaa[5] = eta[6];etaa[6] = eta[8];etaa[7] = eta[9];etaa[8] = eta[11]; etaa[9] = eta[18];etaa[10] = eta[20];etaa[11] = eta[22];etaa[12] = eta[24]; etaa[13] = eta[25];etaa[14] = eta[27];etaa[15] = eta[28];etaa[16] = eta[29]; etaa[17] = eta[32];etaa[18] = eta[34];etaa[19] = eta[35];etaa[20] = eta[36]; etaa[21] = eta[37];etaa[22] = eta[43];etaa[23] = eta[44];etaa[24] = eta[50]; etaa[25] = eta[51];etaa[26] = eta[52];etaa[27] = eta[53];etaa[28] = eta[54]; etaa[29] = eta[57];etaa[30] = eta[58];etaa[31] = eta[59];etaa[32] = eta[60]; etaa[33] = eta[61];etaa[34] = eta[63];etaa[35] = eta[65];etaa[36] = eta[66];
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
Table 4.10 Showing etaa[i] and y[i] = (Exp[etaa[i]))/etaa[i]. Using prog. 4.10 i
etaa[i]
y[i]
i
etaa[i]
y[i]
1
1.7237
3.2518
26
1.8455
3.4306
2
1.7192
3.2457
27
1.5488
3.0384
3
1.5193
3.0073
28
0.8468
2.7541
4
0.7598
2.8137
29
1.1323
2.7403
5
0.3799
3.8489
30
0.6521
2.9436
6
0.7721
2.8032
31
1.8587
3.4516
7
1.1763
2.7564
32
1.7357
3.2683
8
1.7518
3.2908
33
1.2019
2.7677
9
1.1283
2.7390
34
1.6623
3.1711
10
1.9696
3.6392
35
0.1242
9.1141
11
1.8781
3.4827
36
1.9199
3.5525
12
1.3508
2.8580
37
1.8672
3.4651
13
1.9474
3.6000
38
1.9289
3.5679
14
1.4024
2.8985
39
1.8498
3.4373
15
1.3232
2.8381
40
1.2552
2.7952
16
1.8049
3.3682
41
1.9006
3.5199
17
0.6857
2.8951
42
0.5292
3.2079
18
0.3310
4.2065
43
0.7859
2.7922
19
0.7088
2.8662
44
1.2650
2.8009
20
0.7789
2.7976
45
0.6523
2.9434
21
1.9262
3.5631
46
1.4699
2.9586
22
0.8267
2.7649
47
0.9562
2.7210
23
1.1130
2.7345
48
1.4665
2.9553
24
0.8796
2.7398
49
1.2087
2.7709
25
1.9049
3.5271
50
1.2391
2.7863
51
0.9522
2.7215
52
0.2990
4.5103
53
1.3548
2.8609
54
0.6089
3.0192
55
1.9000
3.5188
56
1.1410
2.7431
87
88
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
2.0
1
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
Fig. 4.10 Showing that random variates ηa are linearly distributed in the interval 0 to 2. There are more variates for larger values of ηa
etaa[37] = eta[68];etaa[38] = eta[69];etaa[39] = eta[71];etaa[40] = eta[72]; etaa[41] = eta[76];etaa[42] = eta[78];etaa[43] = eta[80];etaa[44] = eta[81]; etaa[45] = eta[82];etaa[46] = eta[83];etaa[47] = eta[84];etaa[48] = eta[85]; etaa[49] = eta[87];etaa[50] = eta[89];etaa[51] = eta[90];etaa[52] = eta[91]; etaa[53] = eta[93];etaa[54] = eta[95];etaa[55] = eta[96];etaa[56] = eta[99]; Table[{i = i + 1,etaa[i],y[i] = (Exp[etaa[i]])/etaa[i]},{i,0,55,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","etaa[i]","y[i]"}}] Integral = Sum[2*y[i]/(56),{i,1,56}] ListPlot[Table[{etaa[i],1},{i,0,55,1}], Frame-> True,FrameLabel-> {"\[Eta]a [i]","1"}] That random variate ηa is distributed as p(x) = x/2 in the interval 0 to 2 is evidenced by Fig. 4.10 which displays values of ηa against stagnant integer 1. We find that indeed there are more variates for larger values of ηa.
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
89
We now take up anormalized Gaussian variation for p(x) given by 2 p(x) = √1 exp − 21 x−a where a is average and σ is variance of x. See Fig. 4.11 σ σ 2π in which we have plotted F(x) = ex along with the normalized Gaussian probability density function p(x). As Fig. 4.11 shows, the Gaussian probability density function p(x) with a = 1 and σ = 0.3 is different from the function ex in that it does not follow ex for x > 1. Hence it is not expected to yield a good estimate of the value of the definite integral. Other values of σ yield worse results. 2 We now generate random variate g obeying p(g) = √1 exp − 21 g−a with a = 1 σ σ 2π and σ = 0.3 in the interval 0 < g < 2 using acceptance-rejection sampling, using Program number 4.11 and resulting Table 4.11. From Table 4.11, we manually gather accepted
7 6 5 4 3 2 1 0
0.0
0.5
1.0
1.5
2.0
x 2 Fig. 4.11 Showing F(x) = ex as undashed curve and Gaussian p(x) = √1 exp − 21 x−a as σ σ 2π dashed curve obtained using the command sig = 0.3; a = 1; p1 = Plot[Exp[x],{x,0,2},PlotStyle-> {Black},Frame-> True, FrameLabel-> {"x","F(x) = e^x or Gaussian (x)"}, PlotRange-> {0,7}]; p2 = Plot[(1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((x-a)/sig)^2], {x,0,2},PlotStyle-> {Dashed,Black},PlotRange-> {0,7}]; Show[p1,p2] in Mathematica 6.0
90
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
random variates ga which we use in Program number 4.12 and resulting Table 4.12 to evaluate the integral with the result 10.00 rather than the exact value 6.389. Program number 4.11 n = 654321 SeedRandom[n] sig = 0.3;a = 1; Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + 2*u1[i],U2[i] = 0 + 1*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))* Exp[-0.5*((U1[i]-a)/sig)^2],g[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]", "U1[i]","U2[i]","g[i]"}}] Program number 4.12 n = 654321 SeedRandom[n] sig = 0.3;a = 1; Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + 2*u1[i],U2[i] = 0 + 1*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))* Exp[-0.5*((U1[i]-a)/sig)^2],g[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","g[i]"}}] ga[1] = g[2]; ga[2] = g[4]; ga[3] = g[5]; ga[4] = g[6]; ga[5] = g[7]; ga[6] = g[8]; ga[7] = g[9]; ga[8] = g[16]; ga[9] = g[17]; ga[10] = g[18]; ga[11] = g[21]; ga[12] = g[24]; ga[13] = g[28]; ga[14] = g[29]; ga[15] = g[31]; ga[16] = g[32]; ga[17] = g[35]; ga[18] = g[36]; ga[19] = g[38]; ga[20] = g[39]; ga[21] = g[43]; ga[22] = g[44]; ga[23] = g[45]; ga[24] = g[47]; ga[25] = g[48]; ga[26] = g[50]; ga[27] = g[53]; ga[28] = g[54]; ga[29] = g[57]; ga[30] = g[58]; ga[31] = g[60]; ga[32] = g[61]; ga[33] = g[64]; ga[34] = g[67]; ga[35] = g[72]; ga[36] = g[78]; ga[37] = g[80]; ga[38] = g[81]; ga[39] = g[82]; ga[40] = g[84]; ga[41] = g[87]; ga[42] = g[89];ga[43] = g[90];ga[44] = g[93];ga[45] = g[94];ga[46] = g[95]; ga[47] = g[99];ga[48] = g[100]; Table[{i = i + 1,ga[i], p[i] = (1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((ga[i]-a)/sig)^2],
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
91
Table 4.11 Showing u1[i], u2[i], U1[i], U2[i] and g[i]. Using prog. 4.11 i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
1
0.8618
0.4287
1.7237
0.4287
Null
1
2
0.8596
0.0342
1.7192
0.0342
1.7192
2
3
0.7597
0.4609
1.5193
0.4609
Null
3
4
0.4189
0.4746
0.8378
0.4746
0.8378
4
5
0.3799
0.1687
0.7598
0.1687
0.7598
5
6
0.1899
0.0364
0.3799
0.0364
0.3799
6
7
0.3335
0.4257
0.6669
0.4257
0.6669
7
8
0.3860
0.2445
0.7721
0.2445
0.7721
8
9
0.5881
0.2322
1.1763
0.2322
1.1763
9
10
0.6743
0.9089
1.3486
0.9089
Null
10
11
0.8759
0.3658
1.7518
0.3658
Null
11
12
0.2945
0.6719
0.5891
0.6719
Null
12
13
0.9104
0.9276
1.8208
0.9276
Null
13
14
0.0504
0.3510
0.1008
0.3510
Null
14
15
0.1510
0.1761
0.3021
0.1761
Null
15
16
0.3796
0.5239
0.7593
0.5239
0.7593
16
17
0.3586
0.4202
0.7172
0.4202
0.7172
17
18
0.5641
0.1372
1.1283
0.1372
1.1283
18
19
0.0774
0.8924
0.1548
0.8924
Null
19
20
0.9848
0.4854
1.9696
0.4854
Null
20
21
0.3080
0.4945
0.6159
0.4945
0.6159
21
22
0.9390
0.5923
1.8781
0.5923
Null
22
23
0.1921
0.3830
0.3842
0.3830
Null
23
24
0.6754
0.2944
1.3508
0.2944
1.3508
24
25
0.9737
0.4501
1.9474
0.4501
Null
25
26
0.7833
0.7877
1.5667
0.7877
Null
26
27
0.7012
0.5937
1.4024
0.5937
Null
27
28
0.6616
0.5202
1.3232
0.5202
1.3232
28
29
0.9024
0.0050
1.8049
0.0050
1.8049
29
30
0.2183
0.2361
0.4365
0.2361
Null
30
31
0.4477
0.7308
0.8954
0.7308
0.8954
31
32
0.3429
0.2260
0.6857
0.2260
0.6857
32 (continued)
92
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.11 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
33
0.1079
0.4879
0.2158
0.4879
Null
33
34
0.1655
0.1218
0.3310
0.1218
Null
34
35
0.3544
0.3523
0.7088
0.3523
0.7088
35
36
0.3895
0.0911
0.7789
0.0911
0.7789
36
37
0.9631
0.5184
1.9262
0.5184
Null
37
38
0.6328
0.8258
1.2656
0.8258
1.2656
38
39
0.4101
0.6742
0.8201
0.6742
0.8201
39
40
0.2088
0.8631
0.4176
0.8631
Null
40
41
0.2042
0.3027
0.4084
0.3027
Null
41
42
0.0504
0.8463
0.1007
0.8463
Null
42
43
0.4134
0.2451
0.8267
0.2451
0.8267
43
44
0.5565
0.0084
1.1130
0.0084
1.1130
44
45
0.3532
0.7547
0.7064
0.7547
0.7064
45
46
0.1188
0.5403
0.2376
0.5403
Null
46
47
0.5235
0.8228
1.0471
0.8228
1.0471
47
48
0.5526
0.8279
1.1052
0.8279
1.1052
48
49
0.6770
0.7266
1.3540
0.7266
Null
49
50
0.4398
0.2243
0.8796
0.2243
0.8796
50
51
0.9524
0.7628
1.9049
0.7628
Null
51
52
0.9227
0.7673
1.8455
0.7673
Null
52
53
0.7744
0.0388
1.5488
0.0388
1.5488
53
54
0.4234
0.3146
0.8468
0.3146
0.8468
54
55
0.2849
0.8526
0.5699
0.8526
Null
55
56
0.2712
0.4589
0.5424
0.4589
Null
56
57
0.5662
0.1270
1.1323
0.1270
1.1323
57
58
0.3261
0.1921
0.6521
0.1921
0.6521
58
59
0.9294
0.0971
1.8587
0.0971
Null
59
60
0.8678
0.0527
1.7357
0.0527
1.7357
60
61
0.6010
0.5101
1.2019
0.5101
1.2019
61
62
0.0491
0.8662
0.0982
0.8662
Null
62
63
0.8311
0.3699
1.6623
0.3699
Null
63
64
0.6055
0.7086
1.2109
0.7086
1.2109
64 (continued)
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
93
Table 4.11 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
65
0.0621
0.0293
0.1242
0.0293
Null
65
66
0.9600
0.7918
1.9199
0.7918
Null
66
67
0.2790
0.3078
0.5579
0.3078
0.5579
67
68
0.9336
0.7173
1.8672
0.7173
Null
68
69
0.9645
0.1720
1.9289
0.1720
Null
69
70
0.1165
0.6613
0.2330
0.6613
Null
70
71
0.9249
0.6311
1.8498
0.6311
Null
71
72
0.6276
0.1130
1.2552
0.1130
1.2552
72
73
0.1215
0.9735
0.2430
0.9735
Null
73
74
0.0148
0.2807
0.0296
0.2807
Null
74
75
0.1744
0.8147
0.3488
0.8147
Null
75
76
0.9503
0.8674
1.9006
0.8674
Null
76
77
0.3100
0.7364
0.6200
0.7364
Null
77
78
0.2646
0.1094
0.5292
0.1094
0.5292
78
79
0.1454
0.9416
0.2909
0.9416
Null
79
80
0.3930
0.0684
0.7859
0.0684
0.7859
80
81
0.6325
0.3698
1.2650
0.3698
1.2650
81
82
0.3261
0.3225
0.6523
0.3225
0.6523
82
83
0.7350
0.4885
1.4699
0.4885
Null
83
84
0.4781
0.0687
0.9562
0.0687
0.9562
84
85
0.7332
0.7276
1.4665
0.7276
Null
85
86
0.2226
0.7403
0.4453
0.7403
Null
86
87
0.6044
0.2942
1.2087
0.2942
1.2087
87
88
0.3337
0.8206
0.6675
0.8206
Null
88
89
0.6196
0.1319
1.2391
0.1319
1.2391
89
90
0.4761
0.1548
0.9522
0.1548
0.9522
90
91
0.1495
0.1481
0.2990
0.1481
Null
91
92
0.0263
0.9820
0.0526
0.9820
Null
92
93
0.6774
0.4351
1.3548
0.4351
1.3548
93
94
0.3336
0.4401
0.6673
0.4401
0.6673
94
95
0.3045
0.2773
0.6089
0.2773
0.6089
95
96
0.9500
0.5186
1.9000
0.5186
Null
96 (continued)
94
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.11 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
97
0.7374
0.9651
1.4749
0.9651
Null
97
98
0.6385
0.9689
1.2771
0.9689
Null
98
99
0.5705
0.4434
1.1410
0.4434
1.1410
99
100
0.5668
0.8242
1.1337
0.8242
1.1337
100
Table 4.12 Showing ga[i], p[i] and y[i] = (Exp(ga[i]))/p[i]. Using prog. 4.12 i
ga[i]
p[i]
1
1.7192
0.0751
2
0.8378
1.1490
3
0.7598
4 5
y[i]
i
ga[i]
p[i]
y[i]
74.2764
26
0.8796
1.2269
1.9643
2.0116
27
1.5488
0.2495
18.8599
0.9652
2.2150
28
0.8468
1.1673
1.9980
0.3799
0.1570
9.3111
29
1.1323
1.2065
2.5717
0.6669
0.7180
2.7133
30
0.6521
0.6789
2.8276
6
0.7721
0.9965
2.1719
31
1.7357
0.0658
86.2567
7
1.1763
1.1189
2.8977
32
1.2019
1.0603
3.1374
8
0.7593
0.9637
2.2171
33
1.2109
1.0386
3.2319
9
0.7172
0.8529
2.4022
34
0.5579
0.4490
3.8910
10
1.1283
1.2137
2.5463
35
1.2552
0.9261
3.7884
11
0.6159
0.5860
3.1595
36
0.5292
0.3881
4.3739
12
1.3508
0.6712
5.7514
37
0.7859
1.0309
2.1286
13
1.3232
0.7443
5.0453
38
1.2650
0.9003
3.9354
14
1.8049
0.0364
167.1220
39
0.6523
0.6792
2.8265
15
0.8954
1.2514
1.9565
40
0.9562
1.3157
1.9775
16
0.6857
0.7682
2.5843
41
1.2087
1.0439
3.2084
17
0.7088
0.8301
2.4472
42
1.2391
0.9679
3.5672
18
0.7789
1.0136
2.1499
43
0.9522
1.3130
1.9736
19
1.2656
0.8986
3.9452
44
1.3548
0.6608
5.8658
20
0.8201
1.1110
2.0439
45
0.6673
0.7189
2.7110
21
0.8267
1.1255
2.0309
46
0.6089
0.5685
3.2336
22
1.1130
1.2387
2.4571
47
1.1410
1.1908
2.6283
23
0.7064
0.8237
2.4604
48
1.1337
1.2042
2.5802
24
1.0471
1.3136
2.1691
25
1.1052
1.2505
2.4151
4.3
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
95
y[i] = (Exp[ga[i]])/p[i]},{i,0,47,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","ga[i]", "p[i]","y[i]"}}] Integral = Sum[y[i]/(48),{i,1,48}] ListPlot[Table[{ga[i],1},{i,0,47,1}],Frame-> True, FrameLabel-> {"ga [i]","1"}] 2 That random variate ga is distributed as p(x) = √1 exp − 21 x−a with a = 1 σ σ 2π and sig = 0.3 in the interval 0 to 2 is evidenced by Fig. 4.12 which displays values of ga against stagnant integer 1. We find that indeed there are more variates near ga = 1. 2.0
1
1.5
1.0
0.5
0.0 0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
Fig. 4.12 Showing that random variates ga are distributed in the interval 0 to 2 as p(x) = 2 √1 exp − 1 x−a with a = 1 and sig = 0.3. There are more variates near ga = 1 2 σ σ 2π
96
4.4
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example IV
As example IV, we now take up the integral b I =
5 F(x)d x =
a
loge xd x
(4.10)
1
where F(x) = loge x. We re-write Eq. (4.10) as b I =
F(x) p(x)d x p(x)
(4.11)
a
which, as discussed in Sect. 3.6, implies that average value of F/p is the value of the integral, i.e. I =
N 1 F(xi ) N p(xi )
(4.12)
i=1
where x i ’s are random values of x in the interval 1 < x < 5 obeying probability density function p(x). We now take up an exponential variation for p(x) given by p(x) = C ex where C 5 is a constant. Normalization of p(x) requires 1 p(x)d x = 1 or, C = 1/145. As such normalized probability density function is p(x) = ex /145. See Fig. 4.13 in which we have plotted F(x) = loge x along with the normalized probability density function p(x) = ex /145. The probability density function p(x) = ex /145 follows the function loge x well. Hence it is a good choice. We now generate random variate η obeying p(x) = ex /145 in the interval 1 < x < 5 using acceptance-rejection sampling, using Program number 4.13 and resulting Table 4.13. From Table 4.13, we manually gather accepted random variates ηa which we use in Program number 4.14 and resulting Table 4.14 to evaluate the integral with the result 3.953 rather than the exact value 4.045. The difference is attributable to the fact that F(x) and p(x) used are not proportional to each other. Program number 4.13 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 1 + 4*u1[i],U2[i] = 0 + 1*u2[i], If[U2[i] < (Exp[U1[i]])/145,eta[i] = U1[i],Null]},{i,0,99,1}];
4.4
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example …
97
1.5
1.0
0.5
0.0 1
2
3 x
4
5
Fig. 4.13 Showing F(x) = loge x as undashed curve and p(x) = ex /145 as dashed curve obtained using the command a = 1/145; p1 = Plot[Log[E,x],{x,1,5},Frame-> True, FrameLabel-> {"x","F(x) = Log(x) or p(x)"},PlotStyle-> {Black}]; p2 = Plot[a*Exp[x],{x,1,5},PlotStyle-> {Dashed,Black}, Frame-> True]; Show[p1,p2] in Mathematica 6.0
TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]", "U1[i]","U2[i]","eta[i]"}}] Program number 4.14 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 1 + 4*u1[i],U2[i] = 0 + 1*u2[i], If[U2[i] < (Exp[U1[i]])/145,eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2},
98
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.13 Showing u1[i], u2[i], U1[i], U2[i] and eta[i]. Using prog. 4.13 i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
1
0.8618
0.4287
4.4473
0.4287
4.4473
1
2
0.8596
0.0342
4.4384
0.0342
4.4384
2
3
0.7597
0.4609
4.0386
0.4609
Null
3
4
0.4189
0.4746
2.6756
0.4746
Null
4
5
0.3799
0.1687
2.5196
0.1687
Null
5
6
0.1899
0.0364
1.7598
0.0364
1.7598
6
7
0.3335
0.4257
2.3339
0.4257
Null
7
8
0.3860
0.2445
2.5442
0.2445
Null
8
9
0.5881
0.2322
3.3526
0.2322
Null
9
10
0.6743
0.9089
3.6973
0.9089
Null
10
11
0.8759
0.3658
4.5035
0.3658
4.5035
11
12
0.2945
0.6719
2.1781
0.6719
Null
12
13
0.9104
0.9276
4.6416
0.9276
Null
13
14
0.0504
0.3510
1.2016
0.3510
Null
14
15
0.1510
0.1761
1.6042
0.1761
Null
15
16
0.3796
0.5239
2.5185
0.5239
Null
16
17
0.3586
0.4202
2.4345
0.4202
Null
17
18
0.5641
0.1372
3.2565
0.1372
3.2565
18
19
0.0774
0.8924
1.3096
0.8924
Null
19
20
0.9848
0.4854
4.9391
0.4854
4.9391
20
21
0.3080
0.4945
2.2318
0.4945
Null
21
22
0.9390
0.5923
4.7561
0.5923
4.7561
22
23
0.1921
0.3830
1.7684
0.3830
Null
23
24
0.6754
0.2944
3.7016
0.2944
Null
24
25
0.9737
0.4501
4.8949
0.4501
4.8949
25
26
0.7833
0.7877
4.1333
0.7877
Null
26
27
0.7012
0.5937
3.8047
0.5937
Null
27
28
0.6616
0.5202
3.6464
0.5202
Null
28
29
0.9024
0.0050
4.6097
0.0050
4.6097
29
30
0.2183
0.2361
1.8731
0.2361
Null
30
31
0.4477
0.7308
2.7909
0.7308
Null
31
32
0.3429
0.2260
2.3714
0.2260
Null
32 (continued)
4.4
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example …
99
Table 4.13 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
33
0.1079
0.4879
1.4317
0.4879
Null
33
34
0.1655
0.1218
1.6620
0.1218
Null
34
35
0.3544
0.3523
2.4176
0.3523
Null
35
36
0.3895
0.0911
2.5579
0.0911
Null
36
37
0.9631
0.5184
4.8523
0.5184
4.8523
37
38
0.6328
0.8258
3.5312
0.8258
Null
38
39
0.4101
0.6742
2.6402
0.6742
Null
39
40
0.2088
0.8631
1.8352
0.8631
Null
40
41
0.2042
0.3027
1.8169
0.3027
Null
41
42
0.0504
0.8463
1.2014
0.8463
Null
42
43
0.4134
0.2451
2.6535
0.2451
Null
43
44
0.5565
0.0084
3.2261
0.0084
3.2261
44
45
0.3532
0.7547
2.4127
0.7547
Null
45
46
0.1188
0.5403
1.4752
0.5403
Null
46
47
0.5235
0.8228
3.0941
0.8228
Null
47
48
0.5526
0.8279
3.2105
0.8279
Null
48
49
0.6770
0.7266
3.7081
0.7266
Null
49
50
0.4398
0.2243
2.7592
0.2243
Null
50
51
0.9524
0.7628
4.8098
0.7628
4.8098
51
52
0.9227
0.7673
4.6909
0.7673
Null
52
53
0.7744
0.0388
4.0976
0.0388
4.0976
53
54
0.4234
0.3146
2.6936
0.3146
Null
54
55
0.2849
0.8526
2.1398
0.8526
Null
55
56
0.2712
0.4589
2.0849
0.4589
Null
56
57
0.5662
0.1270
3.2647
0.1270
3.2647
57
58
0.3261
0.1921
2.3043
0.1921
Null
58
59
0.9294
0.0971
4.7175
0.0971
4.7175
59
60
0.8678
0.0527
4.4713
0.0527
4.4713
60
61
0.6010
0.5101
3.4038
0.5101
Null
61
62
0.0491
0.8662
1.1964
0.8662
Null
62
63
0.8311
0.3699
4.3245
0.3699
4.3245
63
64
0.6055
0.7086
3.4219
0.7086
Null
64 (continued)
100
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.13 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
65
0.0621
0.0293
1.2485
0.0293
Null
65
66
0.9600
0.7918
4.8399
0.7918
4.8399
66
67
0.2790
0.3078
2.1158
0.3078
Null
67
68
0.9336
0.7173
4.7343
0.7173
4.7343
68
69
0.9645
0.1720
4.8579
0.1720
4.8579
69
70
0.1165
0.6613
1.4660
0.6613
Null
70
71
0.9249
0.6311
4.6995
0.6311
4.6995
71
72
0.6276
0.1130
3.5104
0.1130
3.5104
72
73
0.1215
0.9735
1.4859
0.9735
Null
73
74
0.0148
0.2807
1.0592
0.2807
Null
74
75
0.1744
0.8147
1.6976
0.8147
Null
75
76
0.9503
0.8674
4.8012
0.8674
Null
76
77
0.3100
0.7364
2.2399
0.7364
Null
77
78
0.2646
0.1094
2.0584
0.1094
Null
78
79
0.1454
0.9416
1.5817
0.9416
Null
79
80
0.3930
0.0684
2.5719
0.0684
2.5719
80
81
0.6325
0.3698
3.5300
0.3698
Null
81
82
0.3261
0.3225
2.3045
0.3225
Null
82
83
0.7350
0.4885
3.9398
0.4885
Null
83
84
0.4781
0.0687
2.9124
0.0687
2.9124
84
85
0.7332
0.7276
3.9329
0.7276
Null
85
86
0.2226
0.7403
1.8906
0.7403
Null
86
87
0.6044
0.2942
3.4175
0.2942
Null
87
88
0.3337
0.8206
2.3349
0.8206
Null
88
89
0.6196
0.1319
3.4783
0.1319
3.4783
89
90
0.4761
0.1548
2.9043
0.1548
Null
90
91
0.1495
0.1481
1.5980
0.1481
Null
91
92
0.0263
0.9820
1.1052
0.9820
Null
92
93
0.6774
0.4351
3.7096
0.4351
Null
93
94
0.3336
0.4401
2.3345
0.4401
Null
94
95
0.3045
0.2773
2.2178
0.2773
Null
95
96
0.9500
0.5186
4.7999
0.5186
4.7999
96 (continued)
4.4
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example … 101
Table 4.13 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
97
0.7374
0.9651
3.9497
0.9651
Null
97
98
0.6385
0.9689
3.5541
0.9689
Null
98
99
0.5705
0.4434
3.2819
0.4434
Null
99
100
0.5668
0.8242
3.2673
0.8242
Null
100
Table 4.14 Showing tabulated values of etaa[i] and y[i] = (Log[etaa[i]])/(Exp[etaa[i]]). Using prog. 4.14
i
etaa[i]
y[i]
1
4.4473
0.0175
2
4.4384
0.0176
3
1.7598
0.0973
4
4.5035
0.0167
5
3.2565
0.0455
6
4.9391
0.0114
7
4.7561
0.0134
8
4.8949
0.0119
9
4.6097
0.0152
10
4.8523
0.0123
11
3.2261
0.0465
12
4.8098
0.0128
13
4.0976
0.0234
14
3.2647
0.0452
15
4.7175
0.0139
16
4.4713
0.0171
17
4.3245
0.0194
18
4.8399
0.0125
19
4.7343
0.0137
20
4.8579
0.0123
21
4.6995
0.0141
22
3.5104
0.0375
23
2.5719
0.0722
24
2.9124
0.0581
25
3.4783
0.0385
26
4.7999
0.0129
102
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","eta[i]"}}] etaa[1] = eta[1]; etaa[2] = eta[2]; etaa[3] = eta[6]; etaa[4] = eta[11]; etaa[5] = eta[18]; etaa[6] = eta[20]; etaa[7] = eta[22]; etaa[8] = eta[25]; etaa[9] = eta[29]; etaa[10] = eta[37]; etaa[11] = eta[44]; etaa[12] = eta[51]; etaa[13] = eta[53]; etaa[14] = eta[57]; etaa[15] = eta[59]; etaa[16] = eta[60]; etaa[17] = eta[63]; etaa[18] = eta[66]; etaa[19] = eta[68]; etaa[20] = eta[69]; etaa[21] = eta[71]; etaa[22] = eta[72]; etaa[23] = eta[80]; etaa[24] = eta[84]; etaa[25] = eta[89]; etaa[26] = eta[96]; Table[{i = i + 1,etaa[i],y[i] = (Log[etaa[i]])/(Exp[etaa[i]])},{i,0,25,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","etaa[i]","y[i]"}}] Integral = Sum[145*y[i]/26,{i,1,26}] ListPlot[Table[{etaa[i],1},{i,0,25,1}],Frame-> True, FrameLabel-> {"\[Eta]a [i]","1"}] That random variate ηa is non-uniformly distributed like p(x) = ex /145 in the interval 1 to 5 is evidenced by Fig. 4.14 which displays values of ηa against stagnant integer 1. We find that indeed there are more variates for larger values of ηa.
4.5
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling: Example V
As example V, we now take up the integral b I =
3 F(x)d x =
a
where F(x) =
1 . 1+x 2
0
1 dx 1 + x2
(4.13)
We re-write Eq. (4.13) as b I =
F(x) p(x)d x p(x)
(4.14)
a
which, as discussed in Sect. 3.6, implies that average value of F/p is the value of the integral, i.e. I =
N 1 F(xi ) N p(xi ) i=1
(4.15)
4.5
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
103
2.0
1
1.5
1.0
0.5
0.0 2.0
2.5
3.0
3.5
4.0
4.5
5.0
Fig. 4.14 Showing that random variates ηa are non-uniformly distributed like p(x) = ex /145 in the interval 1 to 5. We find that indeed there are more variates for larger values of ηa
where x i ’s are random values of x in the interval 0 < x < 3 obeying probability density function p(x). We now take up an exponential variation for p(x) given by p(x) = C e−x where C is a 3 constant. Normalization of p(x) requires 0 p(x)d x = 1 or, C = 1.05. As such normalized probability density function is p(x) = 1.05e−x . See Fig. 4.15 in which we have plotted 1 −x . The F(x) = 1+x 2 along with the normalized probability density function p(x) = 1.05 e
1 probability density function p(x) = 1.05e−x follows the function 1+x 2 quite well. Hence it is a very good choice. We now generate random variate η obeying p(x) = 1.05 e−x in the interval 0 < x < 3 using acceptance-rejection sampling, using Program number 4.15 and resulting Table 4.15. From Table 4.15, we manually gather accepted random variates ηa which we use in Program number 4.16 and resulting Table 4.16 to evaluate the integral with the result 1.292 rather than the exact value 1.249. The absence of difference is attributable to the fact that F(x) and p(x) used are almost proportional to each other. That random variate ηa is non-uniformly distributed in the interval 0 to 3 is evidenced by Fig. 4.16 which displays values of ηa against stagnant integer 1.
104
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
x Fig. 4.15 Showing F(x) = 1 2 as undashed curve and p(x) = 1.05 e−x as dashed curve obtained 1+x using the command p1 = Plot[1/(1 + x^2),{x,0,3},PlotRange-> {0,1},Frame-> True, FrameLabel-> {"x","F(x) = 1/(1 + x^2) or p(x)"}, PlotStyle-> {Black}]; p2 = Plot[1.05*Exp[-x],{x,0,3},PlotStyle-> {Dashed,Black}]; Show[p1,p2] in Mathematica 6.0
Program number 4.15 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + 3*u1[i],U2[i] = 0 + 1*u2[i], If[U2[i] < 1.05*Exp[-U1[i]],eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]", "U1[i]","U2[i]","eta[i]"}}]
4.5
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
105
Table 4.15 Showing u1[i], u2[i], U1[i], U2[i] and eta[i]. Using prog. 4.15 i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
1
0.8618
0.4287
2.5855
0.4287
Null
1
2
0.8596
0.0342
2.5788
0.0342
2.5788
2
3
0.7597
0.4609
2.2790
0.4609
Null
3
4
0.4189
0.4746
1.2567
0.4746
Null
4
5
0.3799
0.1687
1.1397
0.1687
1.1397
5
6
0.1899
0.0364
0.5698
0.0364
0.5698
6
7
0.3335
0.4257
1.0004
0.4257
Null
7
8
0.3860
0.2445
1.1581
0.2445
1.1581
8
9
0.5881
0.2322
1.7644
0.2322
Null
9
10
0.6743
0.9089
2.0229
0.9089
Null
10
11
0.8759
0.3658
2.6277
0.3658
Null
11
12
0.2945
0.6719
0.8836
0.6719
Null
12
13
0.9104
0.9276
2.7312
0.9276
Null
13
14
0.0504
0.3510
0.1512
0.3510
0.1512
14
15
0.1510
0.1761
0.4531
0.1761
0.4531
15
16
0.3796
0.5239
1.1389
0.5239
Null
16
17
0.3586
0.4202
1.0759
0.4202
Null
17
18
0.5641
0.1372
1.6924
0.1372
1.6924
18
19
0.0774
0.8924
0.2322
0.8924
Null
19
20
0.9848
0.4854
2.9544
0.4854
Null
20
21
0.3080
0.4945
0.9239
0.4945
Null
21
22
0.9390
0.5923
2.8171
0.5923
Null
22
23
0.1921
0.3830
0.5763
0.3830
0.5763
23
24
0.6754
0.2944
2.0262
0.2944
Null
24
25
0.9737
0.4501
2.9212
0.4501
Null
25
26
0.7833
0.7877
2.3500
0.7877
Null
26
27
0.7012
0.5937
2.1036
0.5937
Null
27
28
0.6616
0.5202
1.9848
0.5202
Null
28
29
0.9024
0.0050
2.7073
0.0050
2.7073
29
30
0.2183
0.2361
0.6548
0.2361
0.6548
30
31
0.4477
0.7308
1.3431
0.7308
Null
31
32
0.3429
0.2260
1.0286
0.2260
1.0286
32 (continued)
106
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.15 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
33
0.1079
0.4879
0.3238
0.4879
0.3238
33
34
0.1655
0.1218
0.4965
0.1218
0.4965
34
35
0.3544
0.3523
1.0632
0.3523
1.0632
35
36
0.3895
0.0911
1.1684
0.0911
1.1684
36
37
0.9631
0.5184
2.8893
0.5184
Null
37
38
0.6328
0.8258
1.8984
0.8258
Null
38
39
0.4101
0.6742
1.2302
0.6742
Null
39
40
0.2088
0.8631
0.6264
0.8631
Null
40
41
0.2042
0.3027
0.6126
0.3027
0.6126
41
42
0.0504
0.8463
0.1511
0.8463
0.1511
42
43
0.4134
0.2451
1.2401
0.2451
1.2401
43
44
0.5565
0.0084
1.6695
0.0084
1.6695
44
45
0.3532
0.7547
1.0596
0.7547
Null
45
46
0.1188
0.5403
0.3564
0.5403
0.3564
46
47
0.5235
0.8228
1.5706
0.8228
Null
47
48
0.5526
0.8279
1.6579
0.8279
Null
48
49
0.6770
0.7266
2.0311
0.7266
Null
49
50
0.4398
0.2243
1.3194
0.2243
1.3194
50
51
0.9524
0.7628
2.8573
0.7628
Null
51
52
0.9227
0.7673
2.7682
0.7673
Null
52
53
0.7744
0.0388
2.3232
0.0388
2.3232
53
54
0.4234
0.3146
1.2702
0.3146
Null
54
55
0.2849
0.8526
0.8548
0.8526
Null
55
56
0.2712
0.4589
0.8137
0.4589
0.8137
56
57
0.5662
0.1270
1.6985
0.1270
1.6985
57
58
0.3261
0.1921
0.9782
0.1921
0.9782
58
59
0.9294
0.0971
2.7881
0.0971
Null
59
60
0.8678
0.0527
2.6035
0.0527
2.6035
60
61
0.6010
0.5101
1.8029
0.5101
Null
61
62
0.0491
0.8662
0.1473
0.8662
0.1473
62
63
0.8311
0.3699
2.4934
0.3699
Null
63
64
0.6055
0.7086
1.8164
0.7086
Null
64 (continued)
4.5
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
107
Table 4.15 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
65
0.0621
0.0293
0.1864
0.0293
0.1864
65
66
0.9600
0.7918
2.8799
0.7918
Null
66
67
0.2790
0.3078
0.8369
0.3078
0.8369
67
68
0.9336
0.7173
2.8007
0.7173
Null
68
69
0.9645
0.1720
2.8934
0.1720
Null
69
70
0.1165
0.6613
0.3495
0.6613
0.3495
70
71
0.9249
0.6311
2.7746
0.6311
Null
71
72
0.6276
0.1130
1.8828
0.1130
1.8828
72
73
0.1215
0.9735
0.3644
0.9735
Null
73
74
0.0148
0.2807
0.0444
0.2807
0.0444
74
75
0.1744
0.8147
0.5232
0.8147
Null
75
76
0.9503
0.8674
2.8509
0.8674
Null
76
77
0.3100
0.7364
0.9300
0.7364
Null
77
78
0.2646
0.1094
0.7938
0.1094
0.7938
78
79
0.1454
0.9416
0.4363
0.9416
Null
79
80
0.3930
0.0684
1.1789
0.0684
1.1789
80
81
0.6325
0.3698
1.8975
0.3698
Null
81
82
0.3261
0.3225
0.9784
0.3225
0.9784
82
83
0.7350
0.4885
2.2049
0.4885
Null
83
84
0.4781
0.0687
1.4343
0.0687
1.4343
84
85
0.7332
0.7276
2.1997
0.7276
Null
85
86
0.2226
0.7403
0.6679
0.7403
Null
86
87
0.6044
0.2942
1.8131
0.2942
Null
87
88
0.3337
0.8206
1.0012
0.8206
Null
88
89
0.6196
0.1319
1.8587
0.1319
1.8587
89
90
0.4761
0.1548
1.4282
0.1548
1.4282
90
91
0.1495
0.1481
0.4485
0.1481
0.4485
91
92
0.0263
0.9820
0.0789
0.9820
Null
92
93
0.6774
0.4351
2.0322
0.4351
Null
93
94
0.3336
0.4401
1.0009
0.4401
Null
94
95
0.3045
0.2773
0.9134
0.2773
0.9134
95
96
0.9500
0.5186
2.8499
0.5186
Null
96 (continued)
108
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table 4.15 (continued) i
u1[i]
u2[i]
U1[i]
U2[i]
eta[i]
i
97
0.7374
0.9651
2.2123
0.9651
Null
97
98
0.6385
0.9689
1.9156
0.9689
Null
98
99
0.5705
0.4434
1.7115
0.4434
Null
99
100
0.5668
0.8242
1.7005
0.8242
Null
100
Table 4.16 Showing etaa[i] and y[i] = (Exp[etaa[i]])/(1 + etaa[i]^2). Using prog. 4.16 i
etaa[i]
y[i]
i
etaa[i]
y[i]
1
2.5788
1.7230
26
2.6035
1.7370
2
1.1397
1.3597
27
0.1473
1.1341
3
0.5698
1.3346
28
0.1864
1.1644
4
1.1581
1.3599
29
0.8369
1.3580
5
0.1512
1.1372
30
0.3495
1.2640
6
0.4531
1.3052
31
1.8828
1.4460
7
1.6924
1.4059
32
0.0444
1.0433
8
0.5763
1.3358
33
0.7938
1.3568
9
2.7073
1.7995
34
1.1789
1.3603
10
0.6548
1.3472
35
0.9784
1.3591
11
1.0286
1.3592
36
1.4343
1.3727
12
0.3238
1.2512
37
1.8587
1.4401
13
0.4965
1.3181
38
1.4282
1.3722
14
1.0632
1.3592
39
0.4485
1.3037
15
1.1684
1.3601
40
0.9134
1.3590
16
0.6126
1.3417
17
0.1511
1.1371
18
1.2401
1.3618
19
1.6695
1.4020
20
0.3564
1.2672
21
1.3194
1.3650
22
2.3232
1.5957
23
0.8137
1.3575
24
1.6985
1.4069
25
0.9782
1.3591
4.5
Evaluation of Definite Integrals Using Acceptance-Rejection Sampling …
109
2.0
1
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
2.5
Fig. 4.16 Showing that random variates ηa are non-uniformly distributed like p(x) = 1.05 e−x in the interval 0 to 3. We find that indeed there are less number of variates for larger values of ηa
Program number 4.16 n = 654321 SeedRandom[n] Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = 0 + 3*u1[i],U2[i] = 0 + 1*u2[i], If[U2[i] < 1.05*Exp[-U1[i]],eta[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","eta[i]"}}] etaa[1] = eta[2];etaa[2] = eta[5];etaa[3] = eta[6];etaa[4] = eta[8]; etaa[5] = eta[14];etaa[6] = eta[15];etaa[7] = eta[18];etaa[8] = eta[23]; etaa[9] = eta[29];etaa[10] = eta[30];etaa[11] = eta[32];etaa[12] = eta[33]; etaa[13] = eta[34];etaa[14] = eta[35];etaa[15] = eta[36];etaa[16] = eta[41]; etaa[17] = eta[42];etaa[18] = eta[43];etaa[19] = eta[44];etaa[20] = eta[46]; etaa[21] = eta[50];etaa[22] = eta[53];etaa[23] = eta[56];etaa[24] = eta[57]; etaa[25] = eta[58];etaa[26] = eta[60];etaa[27] = eta[62];etaa[28] = eta[65]; etaa[29] = eta[67];etaa[30] = eta[70];etaa[31] = eta[72];etaa[32] = eta[74]; etaa[33] = eta[78];etaa[34] = eta[80];etaa[35] = eta[82];etaa[36] = eta[84]; etaa[37] = eta[89];etaa[38] = eta[90];etaa[39] = eta[91];etaa[40] = eta[95];
110
4 Evaluation of Definite Integrals Using Acceptance-Rejection …
Table[{i = i + 1,etaa[i],y[i] = (Exp[etaa[i]])/(1 + etaa[i]^2)},{i,0,39,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","etaa[i]","y[i]"}}] Integral = Sum[0.95*y[i]/40,{i,1,40}] ListPlot[Table[{etaa[i],1},{i,0,39,1}], Frame-> True,FrameLabel-> {"\[Eta]a [i]","1"}] That random variate ηa is non-uniformly distributed like p(x) = 1.05 e−x in the interval 0 to 3 is evidenced by Fig. 4.16 which displays values of ηa against stagnant integer 1. We find that indeed there are less number of variates for larger values of ηa.
5
Variational Monte Carlo Method Applied to Ground State of Simple Harmonic Oscillator Using Acceptance-Rejection Sampling Utilizing Mathematica
This chapter provides practical demonstration of obtaining ground state energy and eigenfunction of simple harmonic oscillator using variational Quantum Monte Carlo method. Necessary background on variational method has been covered so that readers need not consult Quantum Mechanics textbooks. Random variates have been obtained by acceptance-rejection method using Gaussian probability distribution function. Programs written in Mathematica have been used.
5.1
The Variational Method of Quantum Mechanics Applied to Ground State of Any Quantum Mechanical System
Let 0 , 1 , 2 , 3 ,. . . , N be exact, unknown, orthonormal eigenfunctions of known Hamiltonian H. Let E 0 < E 1 < E 2 < E 3 < · · · < E N be exact, unknown eigenvalues of energy respectively. That is, we have eigenvalue equation H n = E n n
(5.1)
which we cannot solve. Suppose we wish to estimate E0 and we wish to know approximate form of 0 . Let us choose by guess a suitable function 0 that is expected to resemble unknown 0 and write 0 =
N
Cn(0) n
(5.2)
n=0
and let us evaluate the quantity E 0 given by
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Chowdhury, Monte Carlo Methods Utilizing Mathematica®, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-23294-7_5
111
112
5 Variational Monte Carlo Method Applied to Ground State …
∗ H 0 dτ (0 , H 0 ) E 0 = = 0∗ 0 0 dτ (0 , 0 ) (0) (0) (0) (0) C m m , H C n n C m m , C n H n m=0 n=0 m=0 n=0 = = (0) (0) (0) (0) C m m , C n n C m m , C n n m=0
n=0
(0)
C m m ,
m=0
(0)
C n E n n m=0 n=0 = (0) (0) C m m , C n n m=0
=
(0)∗
(0)
m=0 n=0
n=0
using equation (5.1)
n=0
Cm Cn E n (m , n ) (0)∗ (0) Cm Cn (m , n )
=
m=0 n=0
(5.3)
(0)∗
(0) 2 Cn E n
(0)
Cm Cn E n δmn
m=0 n=0
(0)∗ (0) Cm Cn δmn
=
m=0 n=0
n=0
(0) 2 Cn
n=0
Thus (0) 2 Cn E n
E 0 − E 0 =
n=0
− E0 (0) 2 Cn
(0) 2 Cn (E n − E 0 )
=
n=0
n=0
(0) 2 Cn
(5.4)
n=0
(0) 2 In Eq. (5.4), E n ≥ E 0 . ∴ E n − E 0 ≥ 0; moreover, Cn ≥ 0. Thus Eq. (5.4) shows that E0 − E0 ≥ 0 or, E 0 ≥ E 0 . We find that for arbitrary 0 , calculated value of E 0 ≥ E 0 i.e. E 0 is larger than or equal to ground state energy E 0 . E 0 gives upper bound of ground state energy E 0 . If chosen 0 is different from actual unknown 0 , calculated value of E 0 calculated using Eq. (5.3) will be larger than E 0 . If chosen 0 happens to be same as unknown 0 , value of E 0 calculated using Eq. (5.3) becomes equal to E 0 . Thus we can choose arbitrary 0 and calculate E 0 using Eq. (5.3) and get upper bound of E 0 . Usually we choose a suitable 0 known as trial function containing one or more (0) adjustable parameters αi (0) ’s; e.g. 0 = e−α1 x where α1 (0) is an adjustable parameter. Then we use this expression of 0 in Eq. (5.3) and calculated expression of E 0 will ∂ E 0 contain the parameters αi (0) ’s. We then stipulate (0) = 0 for each αi (0) separately. We ∂αi
thus get values of αi (0) ’s for which E 0 is minimum. We can put these values of αi (0) ’s in the expression of E 0 . We then get lowest value of E 0 for the chosen 0 . This upper bound to E 0 will be close to actual unknown value of E 0 if trial function 0 has a form closely resembling that of actual unknown 0 . The entire process is repeated for several different functions as 0 to see which function leads to much lower value of E 0 . If
5.2 The Variational Method of Quantum Mechanics Applied to Ground …
113
we are happy that a calculated value of E 0 is low enough and that it is likely to be close to actual unknown value of E 0 , we can take E 0 ≈ E 0 and 0 ≈ 0 within some approximation.
5.2
The Variational Method of Quantum Mechanics Applied to Ground State of Simple Harmonic Oscillator
d Suppose, we know Hamiltonian operator H = − 2m + 21 mωc2 x 2 of a simple harmonic dx2 oscillator and we do not know eigenfunctions n and eigenvalues En of H, because suppose we cannot solve eigenvalue equation H n = E n n of the operator H. Suppose, we wish to estimate ground state energy E0 and wish to know approximate form of corresponding eigenfunction 0 . 2 Let us choose 0 = e−αx by guess as an approximate form of 0 and let us evaluate the quantity 2
E 0 = Now (0 , 0 ) = (e−αx , e−αx ) = 2
1 = 2
2
+∝ 2 x −1 e−2αx 2xd x −∝
1 = 2
+∝ 2 = (x 2 )−1/ 2 e−2αx d x 2
−∝
(0 , H 0 ) (0 , 0 )
(5.5)
e−αx e−αx d x = 2
+∝
2
+∝ 2 x −1 e−2αx d x 2
−∝
e−2αx d x 2
+∝ 2 = x −1 e−2αx d x 2
−∝
0
+∝ 2 = (x 2 )1/ 2−1 e−2αx d x 2
0
= (2α)
+∝
2
= (2α)−1/ 2 (1 2)
0 −1/ 2 √
π
=
+∝ x n−1 e−ax d x = a −n n using
π 2α
0
where stands for well known Gamma function Thus
π (0 , 0 ) = 2α
(5.6)
We also note that
+∝ 1 π 2 e−2αx d x = 2 2α 0
(5.7)
114
5 Variational Monte Carlo Method Applied to Ground State …
We now turn to (0 , H 0 ). H = T + V where T is operator of kinetic energy and V is operator of potential energy. 2 d 2 0 1 2 2 (5.8) (0 , H0 ) = (0 , (T + V )0 ) = 0 , − + x mω 0 c 2m d x 2 2 2 2 Now (0 , V 0 ) = 0 , 21 mωc2 x 2 0 = e−αx , 21 mωc2 x 2 e−αx
1 1 2 2 = mωc2 e−αx , x 2 e−αx = mωc2 2 2
+∝ +∝ 2 2 −2αx 2 2 x e d x = mωc x 2 e−2αx d x
−∝
=
1 mωc2 2
+∝
xe−2αx 2xd x 2
0
+∝ 1 2 2 = mωc (x 2 )1/ 2 e−2αx d(x 2 ) 2
0
0
+∝ 1 1 2 2 = mωc (x 2 )3/ 2−1 e−2αx d(x 2 ) = mωc2 (2α)−3/ 2 (3 2) 2 2 0
=
√ 1 mωc2 1 mωc2 (2α)−3/ 2 (1 2) (1 2) = mωc2 (2α)−3/ 2 (1 2) π = 2 2 8α
+∝ using 0 x n−1 e−ax d x = a −n n and (n + 1) = n n Thus
mωc2 π (0 , V 0 ) = 8α 2α
π 2α
(5.9)
We also note that
+∝ π 1 2 −2αx 2 x e dx = 8α 2α 0
2 d 2 0 Now (0 , T 0 ) = 0 , − 2m dx2
(5.10)
5.2 The Variational Method of Quantum Mechanics Applied to Ground …
115
2 −αx 2 d 2 −αx 2 2 d 2 −αx 2 2 e = − = e−αx , − e , e 2m d x 2 2m dx2 +∝ +∝ 2 2 2 2 2 −αx 2 d −αx 2 −αx 2 d e e d x = − e e−αx d x =− 2 2 2m dx m dx =−
2
−∝ +∝
0
2 d 2 −2αxe−αx d x e−αx dx
m 0
+∝
2 2 2 2 =− e−αx −2αe−αx − 2αx(−2αx)e−αx d x m 0
+∝
2 2 2 2 =− e−αx −2αe−αx + 4α 2 x 2 e−αx d x m 0 ⎡ ⎤ +∝ +∝ 2 ⎣ 2 2 −2α =− e−2αx d x + 4α 2 x 2 e−2αx d x ⎦ m 0 0
1 π 2 2 1 π α π 2 α π 2 −2α =− =− + 4α −α + = m 2 2α 8α 2α m 2 2α 2m 2α using Eqs. (5.7) and (5.10) Thus (0 , T 0 ) =
2 α 2m
π 2α
(5.11)
From Eqs. (5.11) and (5.9) , we gather that (0 , H 0 ) = (0 , (T + V )0 ) =
2 α mωc2 + 2m 8α
π 2α
Using Eqs. (5.12) and (5.6) in Eq. (5.5), we get
2 mωc2 α π + 2m 8α 2α (0 , H 0 ) 2 α mωc2 E 0 = = = + π (0 , 0 ) 2m 8α
(5.12)
(5.13)
2α
We now stipulate that E 0 for
∂ E 0 ∂α
= 0. Hence
α=
2 2m
−
mωc 2
mωc2 8α 2
= 0. Thus we get lowest value of
(5.14)
116
5 Variational Monte Carlo Method Applied to Ground State …
for the chosen 0 ( =e−αx ). Using Eq. (5.14) in Eq. (5.13), we get the lowest value of 2
E 0 =
2 α 2m
+
mωc2 8α
=
2 mωc 2m 2
+
mωc2 mω 8 2c
.
We take ground state energy as
E 0 ≈ E 0 =
1 ωc 2
(5.15)
This incidentally agrees exactly with known value of ground state energy E0 of simple 2 harmonic oscillator. The ground state eigenfunction 0 = e−αx becomes 1 mωc
0 ≈ 0 = e− 2
x2
(5.16)
which incidentally agrees exactly with known ground state eigenfunction 0 of energy of simple harmonic oscillator. Thus, to some extent, we have got familiar with use of variational method.
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational Quantum Monte Carlo Method and Acceptance-Rejection Sampling
We consider an electron residing in a simple harmonic oscillator potential V (x) =
1 1 mωc2 x 2 = kx 2 2 2
(5.17)
We first gather exact results from Quantum Mechanics textbooks. Eigenvalue equation of energy is H = E or, 2 d 2 1 2 2 − (5.18) + mωc x = E 2m d x 2 2 Eigenvalues of energy are En = (n + 1/2)ωc
(5.19)
where n = 0, 1, 2, 3, …. Ground state energy in unit of ωc is E 0 /(ωc ) = 0.5 and ground state eigenfunction is 1 mωc
0 = e − 2
x2
1
= e− 2 L x
2
(5.20)
where L=
mωc
(5.21)
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational …
117
We now gather results from variational method of Quantum Mechanics detailed in 2 2 Sect. 5.2: for trial wavefunction 0 = e−αx = e−β L x , ground state energy estimated or 0 ,H 0 ) evaluated using Eq. (5.5): E 0 = ( (0 ,0 ) is E0 /(ωc ) = 0.5 and value of β turns out to be 0.5 so that ground state eigenfunction is 1
0 = e−αx = e−β L x = e− 2 L x 2
2
2
(5.22)
We now move onto variational Monte Carlo method of evaluating ground state eigen2 d 2 + value and eigenfunction of energy. We have Hamiltonian operator H = − 2m dx2 1 2 2 2 mωc x or, 2 e−β L x lead
d H = − 2m + 21 kx 2 which together with trial wavefunction 0 = e−αx = dx2 2
2
2
to
0 H 0 = e−αx H e−αx = − 2
2
2 1 2 2 (−2α + 4α 2 x 2 )e−2αx + k x 2 e−2αx 2m 2
and ground state energy in unit of ωc as E0 (0 , H 0 ) = ωc ωc (0 , 0 ) 1 2 1 2 −2αx 2 2 2 −2αx 2 − dx + kx e = (−2α + 4α x )e π 2m 2 ωc 2α = F(x)d x = f (x) p(x)d x 2 1 2 1 −2αx 2 1 2 2 − dx = (−2α + 4α x ) + k x e π ωc 2m 2
(5.23)
2α
Here F(x) =
1
ωc
π 2α
−
f (x) =
2 1 2 2 (−2α + 4α 2 x 2 )e−2αx + k x 2 e−2αx 2m 2 2 1 1 − (−2α + 4α 2 x 2 ) + k x 2 ωc 2m 2
(5.24)
(5.25)
and 1 2 p(x) = e−2αx π 2α
(5.26)
118
5 Variational Monte Carlo Method Applied to Ground State …
60000 50000 40000 30000 20000 10000 0 0.00002
0.00001
0
0.00001
0.00002
x Fig. 5.1 Showing F(x) and p(x) using prog. 5.1
We find that average or expectation value of f is the ground state energy in unit of ωc of simple harmonic oscillator. But p(x) should look like F(x). We now plot the functions F(x) and p(x) together for k = 10−18 , β = 0.8, and using the corresponding values of √ mω ωc = (k / m), L = c , α = β L. We use Program number 5.1 in this regard. See Fig. 5.1. We find that in the interval − 20 × 10−6 < x < + 20 × 10−6 shown, F(x) and p(x) look similar. Hence Eq. (5.23) will give a good estimate of E 0 /(ωc ). And the 2 p(x) = √1π e−2αx will be ideal for use as probability density function. 2α
We now generate random variate g obeying p(x) = √1π e−2αx in the interval − 20 × 2
2α
10−6 < x < + 20 × 10−6 using acceptance-rejection sampling, using Program number 5.2 and resulting Table 5.1. From Table 5.1, we manually gather accepted random variates ga which we use in Program number 5.3 and resulting Table 5.2 to evaluate the integral with the result 0.563 for β = 0.8. Program number 5.1 beta = 0.8 k = 10^-18 m = 9.1*10^-31 hcut = (6.626*10^-34)/(2*3.1416) omegac = Sqrt[k/m] L = m*omegac/hcut
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational …
119
Table 5.1 Showing u1[i], u2[i], U1[i], U2[i] and g[i]. Using prog. 5.2 U1[i] × 106
U2[i]
g[i] × 106
i
30,010.2
Null
1
14.3841
2
Null
3
33,219.7
−3.2436
4
−4.8036
11,808.8
−4.8036
5
−12.4023
2551.2
−12.4023
6
29,798.2
−6.6611
7
17,113.7
−4.5581
8
3.5258
16,255.2
3.5258
9
0.9089
6.9726
63,625.1
Null
10
0.3658
15.0354
25,607.0
Null
11
0.2945
0.6719
−8.2187
47,031.6
Null
12
13
0.9104
0.9276
16.4157
64,931.4
Null
13
14
0.0504
0.3510
−17.9839
24,566.6
Null
14
15
0.1510
0.1761
−13.9585
12,330.0
Null
15
16
0.3796
0.5239
−4.8149
36,676.2
−4.8149
16
17
0.3586
0.4202
−5.6552
29,414.1
−5.6552
17
18
0.5641
0.1372
2.5654
9601.9
2.5654
18
19
0.0774
0.8924
−16.9043
62,470.6
Null
19
20
0.9848
0.4854
19.3914
33,975.0
Null
20
21
0.3080
0.4945
−7.6816
34,614.2
Null
21
22
0.9390
0.5923
17.5612
41,460.1
Null
22
23
0.1921
0.3830
−12.3158
26,810.1
Null
23
24
0.6754
0.2944
7.0160
20,607.0
7.0160
24
25
0.9737
0.4501
18.9486
31,503.6
Null
25
26
0.7833
0.7877
11.3334
55,142.0
Null
26
27
0.7012
0.5937
8.0473
41,562.5
Null
27
28
0.6616
0.5202
6.4639
36,410.6
6.4639
28
29
0.9024
0.0050
16.0972
349.3
16.0972
29
30
0.2183
0.2361
−11.2695
16,525.3
Null
30
31
0.4477
0.7308
−2.0914
51,152.8
−2.0914
31
32
0.3429
0.2260
−6.2859
15,820.6
−6.2859
i
u1[i]
u2[i]
1
0.8618
0.4287
14.4733
2
0.8596
0.0342
14.3841
2396.9
3
0.7597
0.4609
10.3863
32,259.9
4
0.4189
0.4746
−3.2436
5
0.3799
0.1687
6
0.1899
0.0364
7
0.3335
0.4257
−6.6611
8
0.3860
0.2445
−4.5581
9
0.5881
0.2322
10
0.6743
11
0.8759
12
32 (continued)
120
5 Variational Monte Carlo Method Applied to Ground State …
Table 5.1 (continued) i
u1[i]
u2[i]
U1[i] × 106
U2[i]
g[i] × 106
i
33
0.1079
0.4879
−15.6832
34,153.1
Null
33
34
0.1655
0.1218
−13.3800
8528.3
Null
34
35
0.3544
0.3523
−5.8246
24,661.9
−5.8246
35
36
0.3895
0.0911
−4.4213
6379.1
−4.4213
36
37
0.9631
0.5184
18.5234
36,287.1
Null
37
38
0.6328
0.8258
5.3122
57,804.8
Null
38
39
0.4101
0.6742
−3.5977
47,195.1
−3.5977
39
40
0.2088
0.8631
−11.6477
60,416.6
Null
40
41
0.2042
0.3027
−11.8314
21,188.8
Null
41
42
0.0504
0.8463
−17.9858
59,238.9
Null
42
43
0.4134
0.2451
−3.4653
17,159.9
−3.4653
43
44
0.5565
0.0084
2.2606
590.5
2.2606
44
45
0.3532
0.7547
−5.8727
52,829.8
Null
45
46
0.1188
0.5403
−15.2484
37,819.4
Null
46
47
0.5235
0.8228
0.9410
57,594.9
0.9410
47
48
0.5526
0.8279
2.1048
57,950.5
2.1048
48
49
0.6770
0.7266
7.0808
50,859.2
Null
49
50
0.4398
0.2243
−2.4085
15,698.6
−2.4085
50
51
0.9524
0.7628
18.0977
53,392.6
Null
51
52
0.9227
0.7673
16.9092
53,713.2
Null
52
53
0.7744
0.0388
10.9761
2719.4
10.9761
53
54
0.4234
0.3146
−3.0639
22,023.8
−3.0639
54
55
0.2849
0.8526
−8.6025
59,684.4
Null
55
56
0.2712
0.4589
−9.1511
32,126.5
Null
56
57
0.5662
0.1270
2.6466
8890.7
2.6466
57
58
0.3261
0.1921
−6.9575
13,447.8
−6.9575
58
59
0.9294
0.0971
17.1748
6794.7
Null
59
60
0.8678
0.0527
14.7133
3688.1
Null
60
61
0.6010
0.5101
4.0384
35,704.8
4.0384
61
62
0.0491
0.8662
−18.0358
60,633.2
Null
62
63
0.8311
0.3699
13.2452
25,891.2
Null
63
64
0.6055
0.7086
4.2186
49,602.9
4.2186
64 (continued)
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational …
121
Table 5.1 (continued) i
u1[i]
u2[i]
U1[i] × 106
g[i] × 106
i
65
0.0621
0.0293
−17.5153
2047.7
Null
65
66
0.9600
0.7918
67
0.2790
0.3078
18.3989
55,422.5
Null
66
−8.8417
21,543.3
−8.8417
67
68
0.9336
0.7173
17.3432
50,208.8
Null
68
69
0.9645
70
0.1165
0.1720
18.5788
12,036.8
Null
69
0.6613
−15.3398
46,294.1
Null
70
71 72
0.9249
0.6311
16.9949
44,175.3
Null
71
0.6276
0.1130
5.1038
7909.1
5.1038
72
73
0.1215
0.9735
−15.1410
68,146.4
Null
73
74
0.0148
0.2807
−19.4085
19,648.2
Null
74
75
0.1744
0.8147
−13.0242
57,031.1
Null
75
76
0.9503
0.8674
18.0118
60,715.6
Null
76
77
0.3100
0.7364
−7.6006
51,549.8
Null
77
78
0.2646
0.1094
−9.4165
7655.8
−9.4165
78
79
0.1454
0.9416
−14.1828
65,914.5
Null
79
80
0.3930
0.0684
−4.2813
4789.3
−4.2813
80
81
0.6325
0.3698
5.2995
25,883.9
5.2995
81
82
0.3261
0.3225
−6.9549
22,576.3
−6.9549
82
83
0.7350
0.4885
9.3983
34,196.5
Null
83
84
0.4781
0.0687
−0.8760
4809.9
−0.8760
84
85
0.7332
0.7276
9.3293
50,934.5
Null
85
86
0.2226
0.7403
−11.0940
51,817.5
Null
86
87
0.6044
0.2942
4.1747
20,595.0
4.1747
87
88
0.3337
0.8206
−6.6509
57,443.5
Null
88
89
0.6196
0.1319
4.7825
9235.9
4.7825
89
90
0.4761
0.1548
−0.9568
10,836.1
−0.9568
90
91
0.1495
0.1481
−14.0204
10,363.8
Null
91
92
0.0263
0.9820
−18.9479
68,741.9
Null
92
93
0.6774
0.4351
7.0961
30,459.3
7.0961
93
94
0.3336
0.4401
−6.6548
30,809.2
−6.6548
94
95
0.3045
0.2773
−7.8218
19,412.8
−7.8218
95
96
0.9500
0.5186
17.9990
36,302.8
Null
U2[i]
96 (continued)
122
5 Variational Monte Carlo Method Applied to Ground State …
Table 5.1 (continued) U2[i]
g[i] × 106
i
9.4971
67,554.9
Null
97
0.9689
5.5414
67,820.4
Null
98
0.4434
2.8194
31,038.3
2.8194
99
0.5668
0.8242
2.6732
57,691.2
2.6732
100
101
0.7161
0.5652
8.6434
39,563.4
Null
101
102
0.9431
0.7145
17.7231
50,014.2
Null
102
103
0.7682
0.5920
10.7287
41,436.6
Null
103
104
0.3412
0.8780
−6.3514
61,462.9
Null
104
105
0.5395
0.4653
1.5815
32,569.2
1.5815
105
106
0.4578
0.6364
−1.6861
44,549.1
−1.6861
106
107
0.4551
0.3154
−1.7949
22,074.6
−1.7949
107
108
0.2764
0.2152
−8.9455
15,063.4
−8.9455
108
109
0.8724
0.2879
14.8980
20,150.8
Null
109
110
0.5121
0.0132
0.4851
921.0
0.4851
110
111
0.2828
0.9446
−8.6884
66,123.1
Null
111
112
0.5885
0.0416
3.5382
2914.5
3.5382
112
113
0.1807
0.7244
−12.7704
50,709.7
Null
113
114
0.4751
0.8880
−0.9955
62,157.1
−0.9955
114
115
0.2541
0.6059
−9.8370
42,410.7
Null
115
116
0.0221
0.9858
−19.1146
69,007.1
Null
116
117
0.3934
0.1550
−4.2629
10,848.9
−4.2629
117
118
0.2128
0.9825
−11.4873
68,777.4
Null
118
119
0.2589
0.9184
−9.6452
64,285.6
Null
119
120
0.6655
0.6142
6.6212
42,995.5
Null
120
121
0.9326
0.1509
17.3032
10,565.1
Null
121
122
0.6705
0.5234
6.8218
36,637.7
Null
122
123
0.2799
0.7116
−8.8049
49,812.7
Null
123
124
0.3740
0.9556
−5.0383
66,894.7
Null
124
125
0.7299
0.8112
9.1966
56,781.8
Null
125
126
0.9289
0.0862
17.1551
6030.7
Null
126
127
0.4937
0.4432
−0.2525
31,024.0
−0.2525
127
128
0.4509
0.5649
−1.9629
39,542.1
−1.9629
i
u1[i]
u2[i]
97
0.7374
0.9651
98
0.6385
99
0.5705
100
U1[i] × 106
128 (continued)
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational …
123
Table 5.1 (continued) i
u1[i]
u2[i]
U1[i] × 106
U2[i]
g[i] × 106
i
129
0.0499
0.8580
−18.0045
60,060.7
Null
129
130
0.2225
0.0270
−11.1003
1889.6
−11.1003
130
131
0.8603
0.3734
14.4113
26,138.5
Null
131
132
0.6852
0.3196
7.4078
22,369.4
7.4078
132
133
0.0708
0.2959
−17.1688
20,713.4
Null
133
134
0.9198
0.6120
16.7904
42,837.3
Null
134
135
0.8673
0.5239
14.6908
36,672.0
Null
135
136
0.6730
0.2424
6.9202
16,967.0
6.9202
136
137
0.5179
0.8175
0.7168
57,226.1
0.7168
137
138
0.2558
0.6163
−9.7693
43,142.2
Null
138
139
0.0526
0.5841
−17.8964
40,889.3
Null
139
140
0.3158
0.1379
−7.3693
9649.5
−7.3693
140
141
0.4767
0.6441
−0.9309
45,087.6
−0.9309
141
142
0.6311
0.0228
5.2451
1595.1
5.2451
142
143
0.2654
0.8424
−9.3824
58,967.9
Null
143
144
0.8255
0.9899
13.0180
69,294.3
Null
144
145
0.5169
0.3570
0.6744
24,987.9
0.6744
145
146
0.9078
0.1674
16.3117
11,718.6
Null
146
147
0.7129
0.3596
8.5154
25,168.8
Null
147
148
0.8932
0.0561
15.7284
3928.0
Null
148
149
0.1703
0.5996
−13.1883
41,970.1
Null
149
150
0.9517
0.4264
18.0692
29,844.9
Null
150
151
0.6781
0.8373
7.1234
58,612.4
Null
151
152
0.6383
0.4100
5.5311
28,702.8
5.5311
152
153
0.8559
0.2554
14.2343
17,875.0
Null
153
154
0.3567
0.4564
−5.7337
31,949.7
−5.7337
154
155
0.5290
0.3290
1.1615
23,028.4
1.1615
155
156
0.5712
0.8274
2.8460
57,917.9
2.8460
156
157
0.6910
0.4119
7.6396
28,830.2
7.6396
157
158
0.2121
0.4267
−11.5169
29,871.9
Null
158
159
0.0268
0.4474
−18.9269
31,318.3
Null
159
160
0.1947
0.6496
−12.2108
45,474.1
Null
160 (continued)
124
5 Variational Monte Carlo Method Applied to Ground State …
Table 5.1 (continued) U2[i]
g[i] × 106
i
12.0035
36,955.0
Null
161
0.2609
−11.9901
18,266.0
Null
162
0.0161
−11.3802
1129.3
−11.3802
163
0.7723
0.9430
10.8902
66,009.0
Null
164
165
0.5086
0.7920
0.3448
55,443.0
0.3448
165
166
0.6609
0.2570
6.4366
17,987.2
6.4366
166
167
0.0900
0.1263
−16.4009
8841.2
Null
167
168
0.2807
0.7241
−8.7706
50,687.9
Null
168
169
0.0896
0.8471
−16.4170
59,298.1
Null
169
170
0.1760
0.6250
−12.9595
43,748.2
Null
170
171
0.5264
0.5202
1.0540
36,412.9
1.0540
171
172
0.8276
0.8793
13.1053
61,548.2
Null
172
173
0.2708
0.7876
−9.1699
55,133.3
Null
173
174
0.4623
0.2713
−1.5065
18,988.0
−1.5065
174
175
0.5351
0.6979
1.4051
48,853.8
1.4051
175
176
0.5660
0.4703
2.6405
32,922.7
2.6405
176
177
0.7876
0.3392
11.5042
23,742.9
Null
177
178
0.1836
0.9136
−12.6562
63,953.5
Null
178
179
0.3897
0.9718
−4.4112
68,028.7
Null
179
180
0.0294
0.0927
−18.8259
6487.0
Null
180
181
0.4954
0.3589
−0.1852
25,121.6
−0.1852
181
182
0.6253
0.0256
5.0138
1789.6
5.0138
182
183
0.9012
0.4223
16.0461
29,560.3
Null
183
184
0.0465
0.6019
−18.1392
42,131.6
Null
184
185
0.2193
0.4897
−11.2263
34,277.5
Null
185
186
0.1713
0.2252
−13.1469
15,766.8
Null
186
187
0.1002
0.6847
−15.9904
47,927.8
Null
187
188
0.4652
0.1787
−1.3928
12,512.2
−1.3928
188
189
0.9316
0.2757
17.2646
19,296.1
Null
189
190
0.7848
0.9415
11.3936
65,907.6
Null
190
191
0.3975
0.3792
−4.1004
26,543.6
−4.1004
191
192
0.2350
0.7681
−10.6004
53,769.5
Null
i
u1[i]
u2[i]
161
0.8001
0.5279
162
0.2002
163
0.2155
164
U1[i] × 106
192 (continued)
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational …
125
Table 5.1 (continued) i
u1[i]
u2[i]
U1[i] × 106
U2[i]
g[i] × 106
i
193
0.1693
0.2771
−13.2272
19,394.0
Null
193
194
0.3306
0.3072
−6.7750
21,504.7
−6.7750
194
195
0.8652
0.7051
14.6060
49,360.3
Null
195
196
0.7953
0.8995
11.8124
62,967.3
Null
196
197
0.4876
0.3045
−0.4979
21,313.5
−0.4979
197
198
0.5590
0.7461
2.3588
52,225.5
2.3588
198
199
0.1489
0.0499
−14.0447
3492.7
−14.0447
199
200
0.5615
0.6135
2.4603
42,946.1
2.4603
200
alpha = beta*L F = (1/(hcut*omegac*Sqrt[Pi/(2*alpha)]))* ((-hcut^2/(2*m))*(-2*alpha + 4*alpha^2*x^2)* Exp[-2*alpha*x^2] + (1/2)*k*x^2*Exp[-2*alpha*x^2]) p = (1/(Sqrt[Pi/(2*alpha)]))*Exp[-2*alpha*x^2] Plot[{F,p},{x,-20*10^-6,20*10^-6}, PlotStyle-> {{Black},{Dashed,Black}},Frame-> True, FrameLabel-> {"x", "F(x) or p(x)"}]
F(x) =
1
ωc
π 2α
−
2 1 2 2 (−2α + 4α 2 x 2 )e−2αx + k x 2 e−2αx 2m 2
and p(x) = √1π e−2αx as function of x for chosen value of spring constant k = 10−18 2
2α
that corresponds to ωc = 1.048 MHz, L = 7.237 × 109 .
mωc
= 9.046 × 109 , β = 0.8, α = βL =
126
5 Variational Monte Carlo Method Applied to Ground State …
1 − 2 (−2α + 4α 2 ga[i]2 ) + 1 k ga[i]2 . Using prog. 5.3 Table 5.2 Showing ga[i] and y[i] = ω 2m 2 c i
ga[i] × 106
1 2
ga[i] × 106
y[i]
i
y[i]
14.3841
−0.6598
51
−0.9955
0.7930
−3.2436
0.7258
52
−4.2629
0.6718
3
−4.8036
0.6372
53
−0.2525
0.7996
4
−12.4023
−0.2853
54
−1.9629
0.7728
5
−6.6611
0.4869
55
−11.1003
−0.0694
6
−4.5581
0.6534
56
7.4078
0.4128
7
3.5258
0.7123
57
6.9202
0.4621
8
−4.8149
0.6364
58
0.7168
0.7964
9
−5.6552
0.5743
59
−7.3693
0.4168
10
2.5654
0.7536
60
−0.9309
0.7939
11
7.0160
0.4527
61
5.2451
0.6059
12
6.4639
0.5052
62
0.6744
0.7968
13
16.0972
−1.0283
63
5.5311
0.5841
14
−2.0914
0.7691
64
−5.7337
0.5680
15
−6.2859
0.5212
65
1.1615
0.7905
16
−5.8246
0.5606
66
2.8460
0.7429
17
−4.4213
0.6621
67
7.6396
0.3882
18
−3.5977
0.7087
68
−11.3802
−0.1138
19
−3.4653
0.7153
69
0.3448
0.7992
20
2.2606
0.7639
70
6.4366
0.5077
21
0.9410
0.7938
71
1.0540
0.7922
22
2.1048
0.7687
72
−1.5065
0.7840
23
−2.4085
0.7591
73
1.4051
0.7861
24
10.9761
−0.0500
74
2.6405
0.7508
25
−3.0639
0.7338
75
−0.1852
0.7998
26
2.6466
0.7506
76
5.0138
0.6226
27
−6.9575
0.4584
77
−1.3928
0.7863
28
4.0384
0.6849
78
−4.1004
0.6814
29
4.2186
0.6744
79
−6.7750
0.4761
30
−8.8417
0.2484
80
−0.4979
0.7983
31
5.1038
0.6162
81
2.3588
0.7607
32
−9.4165
0.1744
82
−14.0447
−0.5918 (continued)
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational …
127
Table 5.2 (continued) i
ga[i] × 106
y[i]
−4.2813
0.6707
34
5.2995
0.6018
35
−6.9549
0.4587
36
−0.8760
0.7946
37
4.1747
0.6770
38
4.7825
0.6386
39
−0.9568
0.7935
40
7.0961
0.4447
41
−6.6548
0.4875
42
−7.8218
0.3683
43
2.8194
0.7439
44
2.6732
0.7496
45
1.5815
0.7824
46
−1.6861
0.7799
47
−1.7949
0.7773
48
−8.9455
0.2354
49
0.4851
0.7983
50
3.5382
0.7117
33
i 83
ga[i] × 106 2.4603
Program number 5.2 n = 654321 SeedRandom[n] beta = 0.8 k = 10^-18 m = 9.1*10^-31 hcut = (6.626*10^-34)/(2*3.1416) omegac = Sqrt[k/m] L = m*omegac/hcut alpha = beta*L Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = -20*10^-6 + (40*10^-6)*u1[i];10^6*U1[i], U2[i] = 0 + (70000)*u2[i], If[U2[i] < (1/(Sqrt[Pi/(2*alpha)]))*Exp[-2*alpha*(U1[i]^2)], g[i] = 10^6*U1[i],Null]},{i,0,199,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","g[i]"}}]
y[i] 0.7573
128
5 Variational Monte Carlo Method Applied to Ground State …
Program number 5.3 n = 654321 SeedRandom[n] beta = 0.8 k = 10^-18 m = 9.1*10^-31 hcut = (6.626*10^-34)/(2*3.1416) omegac = Sqrt[k/m] L = m*omegac/hcut alpha = beta*L Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i] = -20*10^-6 + (40*10^-6)*u1[i];10^6*U1[i], U2[i] = 0 + (70000)*u2[i], If[U2[i] < (1/(Sqrt[Pi/(2*alpha)]))*Exp[-2*alpha*(U1[i]^2)], g[i] = (10^6)*U1[i],Null]},{i,0,199,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","g[i]"}}] ga[1] = g[2];ga[2] = g[4];ga[3] = g[5];ga[4] = g[6];ga[5] = g[7];ga[6] = g[8]; ga[7] = g[9];ga[8] = g[16];ga[9] = g[17];ga[10] = g[18];ga[11] = g[24]; ga[12] = g[28];ga[13] = g[29];ga[14] = g[31];ga[15] = g[32];ga[16] = g[35]; ga[17] = g[36];ga[18] = g[39];ga[19] = g[43];ga[20] = g[44];ga[21] = g[47]; ga[22] = g[48];ga[23] = g[50];ga[24] = g[53];ga[25] = g[54];ga[26] = g[57]; ga[27] = g[58];ga[28] = g[61];ga[29] = g[64];ga[30] = g[67];ga[31] = g[72]; ga[32] = g[78];ga[33] = g[80];ga[34] = g[81];ga[35] = g[82];ga[36] = g[84]; ga[37] = g[87];ga[38] = g[89];ga[39] = g[90];ga[40] = g[93];ga[41] = g[94]; ga[42] = g[95];ga[43] = g[99];ga[44] = g[100];ga[45] = g[105];ga[46] = g[106]; ga[47] = g[107];ga[48] = g[108];ga[49] = g[110];ga[50] = g[112];ga[51] = g[114]; ga[52] = g[117];ga[53] = g[127];ga[54] = g[128];ga[55] = g[130];ga[56] = g[132]; ga[57] = g[136];ga[58] = g[137];ga[59] = g[140];ga[60] = g[141];ga[61] = g[142]; ga[62] = g[145];ga[63] = g[152];ga[64] = g[154];ga[65] = g[155];ga[66] = g[156]; ga[67] = g[157];ga[68] = g[163];ga[69] = g[165];ga[70] = g[166];ga[71] = g[171]; ga[72] = g[174];ga[73] = g[175];ga[74] = g[176];ga[75] = g[181];ga[76] = g[182]; ga[77] = g[188];ga[78] = g[191];ga[79] = g[194];ga[80] = g[197];ga[81] = g[198]; ga[82] = g[199];ga[83] = g[200];
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational …
129
Table[{i = i + 1,ga[i], y[i] = (1/(hcut*omegac))*((-hcut^2/(2*m))* (-2*alpha + (4*alpha^2)*((10^-6)*ga[i])^2) + 0.5*k*((10^-6)*ga[i])^2)},{i,0,82,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","ga[i]", "y[i]"}}] Integral = Sum[y[i]/(83),{i,1,83}] ListPlot[Table[{ga[i],1},{i,0,82,1}],Frame-> True, FrameLabel-> {"ga [i] × 10^6","1"}] That random variate ga is distributed as p(x) = √1π e−2αx is evidenced by Fig. 5.2 2
2α
which displays values of ga against stagnant integer 1. We find that indeed there are more variates near ga = 0. We have repeated all the calculations for various values of β namely 0.3 to 2 and obtained values of E 0 /(ωc ) in each case. Table 5.3 shows the results which are plotted in Fig. 5.3. We find from Fig. 5.3 that minimum value of E 0 /(ωc ) = 0.5 occurs for β 1 mωc 2 1 2 2 = 0.5. Thus the ground state eigenfunction is 0 = e−β L x = e− 2 L x = e− 2 x while ground state eigenvalue of energy is E 0 = 0.5ωc .
2.0
1
1.5
1.0
0.5
0.0 10
5
0
5
10
15
2 Fig. 5.2 Showing that random variates ga are distributed as p(x) = 1π e−2αx . There are more
variates near ga = 0
2α
130
5 Variational Monte Carlo Method Applied to Ground State …
Table 5.3 Showing values of ground state energy E 0 /(ωc ) of an electron in simple harmonic oscillator potential V (x) = 21 k x 2 with k = 10−18 , for various values of variational parameter β contained in trial wave function 0 = e−β L x = e−β 2
mωc
β
E 0 /(ωc )
0.3
0.578
0.5
0.500
0.8
0.563
1.1
0.631
1.5
0.812
2
0.966
x2
1 0.9 0.8
E0 /( ωc )
0.7 0.6 0.5 0.4 0
0.5
1
β
1.5
2
Fig. 5.3 Showing values of ground state energy E 0 /(ωc ) of an electron in simple harmonic oscillator potential V (x) = 21 k x 2 with k = 10−18 , for various values of variational parameter β contained in trial wavefunction 0 = e−β L x = e−β
mωc
x 2 . Minimum value of E /(ω ) = 0.5 occurs for β c 0 1 mωc 2 1 2 2 = 0.5. Thus the ground state eigenfunction is 0 = e−β L x = e− 2 L x = e− 2 x while ground state eigenvalue of energy is E 0 = 0.5ωc 2
Program number 5.3 rewritten (in Mathematica) n=654321; SeedRandom[n]; beta=0.8; a=-20*10^-6; b=20*10^-6; c=70000; k=10^-18; m=9.1*10^-31;
5.3
Ground State Energy of Simple Harmonic Oscillator Using Variational …
hcut=(6.626*10^-34)/(2*3.1416); omegac=Sqrt[k/m]; L=m*omegac/hcut; alpha=beta*L; Table[{i=i+1,u1[i]=RandomReal[],u2[i]=RandomReal[], U1[i]=a +(b-a)*u1[i];10^6*U1[i],U2[i]=0+(c-0)*u2[i], If[U2[i]{2,2}, TableHeadings->{None,{"i","u1[i]","u2[i]","U1[i] x 10^6","U2[i]", "g[i] x 10^6","i"}}] ga={}; i=0; While[i{None,{"i","ga[i] x 10^6","y[i]"}}] Integral=Sum[y[i]/(n1),{i,1,n1}] ListPlot[Table[{ga[[i]],1},{i,0,n1-1,1}], Frame->True,FrameLabel->{"ga [i] x 10^6","1"}]
131
Concluding Remarks
(1) The book is intended for undergraduate students of Mathematics, Statistics and Physics who wish to know how Monte Carlo Methods work with calculations done using a user friendly program Mathematica. (2) All treatments have been done as less manually as practicable. (3) All calculations except for gathering accepted values of random variate in acceptancerejection sampling have been done using Mathematica. (4) Definite integrals of a number of functions F(x) have been evaluated using uniform, linear, Gaussian and exponential probability density functions p(x). (5) It is shown that results agree with known exact values better if p(x) is proportional to F(x). (6) Deviation from the proportionality has been shown to result in worse agreement. (7) All necessary background materials are covered. (8) A separate chapter has been dedicated to variational Monte Carlo method applied to ground state of simple harmonic oscillator. Necessary background on variational method has been covered so that readers need not consult Quantum Mechanics textbooks. (9) The book is a good read; it is intended to turn the readers adept in using the method with Mathematica.
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Chowdhury, Monte Carlo Methods Utilizing Mathematica®, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-23294-7
133
Appendix: Handouts for Computational Lab and Automated Collection of Accepted Random Variates
A.1
Experiment I: Monte Carlo Integration Using Inverse Transform Sampling Using (a) Mathematica and (b) MS Excel
Let us consider the definite integral 5 I =
5 F(x)d x =
3
[1 − (x − 4)2 ]d x 3
Let us rewrite the expression as 5 I =
F(x) p(x)d x = p(x)
3
5 f (x) p(x)d x 3
If p is considered or taken as a normalized probability density function, knowledge of elementary statistics tells us that average or expectation value of f is the value of the integral. Thus we can write I ≈
N 1 F(xi ) N p(xi ) i=1
. where x i ’s are random values of x in the interval 3 < x < 5 obeying probability density function p(x). N should be large and according to what is called importance sampling, p(x) at least should follow F(x), if not be proportional to F(x), to get a good value of the integral using the prescription
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Chowdhury, Monte Carlo Methods Utilizing Mathematica®, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-23294-7
135
136
Appendix: Handouts for Computational Lab and Automated …
1.0
0.8
F, p
0.6
0.4
0.2
0.0 3.0
3.5
4.0
4.5
5.0
x Fig. A.1 Showing F(x) = 1 − (x − 4)2 as undashed curve and p(x) = 1/2 as dashed curve or line obtained using the following command in Mathematica F = 1-(x-4)^2; Plot[{F,1/2},{x,3,5},Frame-> True,PlotStyle-> {{Black},{Black,Dashed}}, FrameLabel-> {"x","F, p"}].
I ≈
N 1 F(xi ) N p(xi ) i=1
. See Fig. A.1 in which we have plotted F(x) = 1 − (x − 4)2 along with normalized probability density function of a uniform random variable in the interval 3–5 given by 5 p(x) = 1/2. Let p(x) = C, a constant. Normalization requires that 3 Cd x = 1 which gives C = 1/2. Uniform p(x) = 1/2 is not too different from a slowly varying function of x like F(x) = 1 − (x − 4)2 in the interval 3 < x < 5. Hence p(x) = 1/2 is a good choice. We now generate random variate η obeying p(x) = 1/2 in the interval 3 < x < 5 using what is called inverse transform sampling according to which values of η are obtainable η from the equation: 3 p(x)d x = u with p = 1/2. Here u stands for random numbers uniformly distributed in the interval 0–1. As such η = 2u + 3. Thus
Appendix: Handouts for Computational Lab and Automated …
I ≈
137
N N N N N 1 F(xi ) 2 2 1 F(ηi ) 2 F(ηi ) = [1 − (ηi − 4)2 ] = Yi = = N p(xi ) N 1/2 N N N i=1
i=1
i=1
i=1
i=1
Program A.1 (written in Mathematica) No = 100; n = 654321; SeedRandom[n]; Table[{i = i + 1,u[i] = RandomReal[],eta[i] = 2*u[i] + 3, Y[i] = 1-(eta[i]-4)^2},{i,0,No-1,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u[i]","eta[i]","Y[i]"}}] I1=Integral = Sum[2*Y[i]/No,{i,1,No}] ListPlot[Table[{eta[i],1},{i,0,No-1,1}],Frame-> True, FrameLabel-> {"\[Eta] [i]","1"}] Integrate[1-(x-4)^2,{x,3,5}]; I2 = N[%] PercentageError = (I1-I2)*100/I2 Using program A.1 and resulting Table A.1, we now evaluate the integral with the result 1.41127 rather than 1.33333. The 5.84% difference is attributable to the facts that F(x) and p(x) used are not proportional to each other and that we have used only 100 random variates. That random variate η is uniformly distributed in the interval 3–5 is evidenced by Fig. A.2 which displays values of η against stagnant integer 1. We have chosen n = 654321 by trial and error to get a set of values of ui and hence of ηi uniformly distributed in the interval 0 to 1 and 3 to 5 respectively. We used the command SeedRandom[n] to get ui ’s reproducibly. As a quality check of the random numbers u[i]’s, we can plot u[i]’s against a stagnant integer 1 to ascertain if u[i]’s are indeed nearly uniformly distributed in the interval 0–1. If not, we can run the spread-sheet as many times as necessary. In this regard, we have obtained Fig. A.3 using columns titled B and C of Table A.2.
138
Appendix: Handouts for Computational Lab and Automated …
Table A.1 Showing tabulated values of ui , ηi = 2ui + 3, Yi = 1 − (ηi − 4)2 from which we get N F(xi ) 1 N F(ηi ) = 2 N [1 − (η − 4)2 ] = 2 N Y as an approximate I = N1 i=1 i i=1 1/2 i=1 i=1 i N N p(xi ) = N 5 5 value of the integral I = 3 F(x)d x = 3 [1 − (x − 4)2 ]d x = 1.41127 whereas exact value is 1.33333. Error is 5.84%. Using p(x) = 1/2. Using prog. A.1 i
u[i]
η[i]
Y [i]
1
0.86183
4.72366
0.47631
2
0.42872
3.85743
0.97968
3
0.85960
4.71920
0.48275
4
0.03424
3.06848
0.13227
5
0.75966
4.51932
0.73031
6
0.46086
3.92171
0.99387
7
0.41891
3.83782
0.97370
8
0.47457
3.94913
0.99741
9
0.37991
3.75982
0.94231
10
0.16870
3.33739
0.56095
11
0.18994
3.37988
0.61546
12
0.03645
3.07289
0.14047
13
0.33347
3.66694
0.88907
14
0.42569
3.85138
0.97791
15
0.38605
3.77209
0.94806
16
0.24448
3.48896
0.73884
17
0.58815
4.17629
0.96892
18
0.23222
3.46443
0.71317
19
0.67431
4.34863
0.87846
20
0.90893
4.81786
0.33111
…
…
…
…
81
0.20422
3.40843
0.65005
82
0.30270
3.60539
0.84429
83
0.05035
3.10071
0.19128
84
0.84627
4.69254
0.52039
85
0.41337
3.82673
0.96998
86
0.24514
3.49028
0.74019
87
0.55651
4.11303
0.98723 (continued)
Appendix: Handouts for Computational Lab and Automated …
139
Table A.1 (continued) i
u[i]
η[i]
Y [i]
88
0.00844
3.01687
0.03346
89
0.35318
3.70637
0.91378
90
0.75471
4.50942
0.74049
91
0.11879
3.23758
0.41872
92
0.54028
4.08055
0.99351
93
0.52352
4.04705
0.99779
94
0.82278
4.64557
0.58324
95
0.55262
4.10524
0.98892
96
0.82786
4.65573
0.57002
97
0.67702
4.35404
0.87465
98
0.72656
4.45312
0.79468
99
0.43979
3.87957
0.98550
100
0.22427
3.44853
0.69588
2.0
1
1.5
1.0
0.5
0.0 3.0
3.5
4.0
4.5
5.0
Fig. A.2 Demonstrating that random variates ηi ’s are uniformly distributed in the interval 3–5
140
Appendix: Handouts for Computational Lab and Automated …
Table A.2 Showing that the same calculations as above can be performed using MS Excel spreadsheet. (But the option of getting same values of u[i]’s every time reproducibly is not available.) To run the spread sheet, we need to copy the bold face quantities and paste in MS Excel spread sheet. Using prog. A.2 A
B
C
D
E
1
i
u[i]
1
eta[i]
Y[i]
2
1
= RAND()
1
= 2*B2 + 3
= 1-(D2-4)^2
3
2
1
4
3
1
…
…
…
…
…
… 1
No
integral = 2 × (sum of Y[i]’s) / No
1.2
1
1
0.8
0.6
0.4
0.2
0 0.0
0.2
0.4
0.6
0.8
1.0
u[i] Fig. A.3 Figure to ascertain that random variates ui ’s used are nearly uniformly distributed in the interval 0–1
Appendix: Handouts for Computational Lab and Automated …
A.2
141
Experiment II: Monte Carlo Integration Using Acceptance-Rejection Sampling, Gaussian Pdf and Mathematica
Let us consider the definite integral 5 I =
5 F(x)d x =
3
[1 − (x − 4)2 ]d x 3
Let us rewrite the expression as 5 I =
F(x) p(x)d x = p(x)
3
5 f (x) p(x)d x 3
If p is considered or taken as a normalized probability density function, knowledge of elementary statistics tells us that average or expectation value of f is the value of the integral. Thus we can write I ≈
N 1 F(xi ) N p(xi ) i=1
where x i ’s are random values of x in the interval 3 < x < 5 obeying probability density function p(x). N should be large and according to what is called importance sampling, p(x) at least should follow F(x), if not be proportional to F(x), to get a good value of the N F(xi ) integral using the prescription I ≈ N1 i=1 p(xi ) . See Fig. A.4 in which we have plotted F(x) = 1 − (x − 4)2 along with normalized x−a 2 1 1 Gaussian probability density function p(x) = √ exp − 2 σ of random variable σ 2π x in the interval 3–5 for variance σ = 1/3 and average a = 4. The Gaussian pdf is not too different from the function F(x) = 1 − (x − 4)2 in the interval 3 < x < 5. Hence the Gaussian pdf is a good choice. 2 We now generate values of random variate g obeying p(x) = √1 exp − 21 x−a σ σ 2π in the interval 3 < x < 5 using what is called acceptance-rejection sampling. According to the sampling method, we first generate 2 sets of uniform random numbers u1 and u2 in the interval 0–1. We then obtain 2 sets of uniform random numbers U1 and U2 in the interval 3 to 5 and 0 to 1.2 respectively, using 3 + (5 − 3)u1 and 0 + (1.2 − 0)u2 respectively according to what we know from inverse transform sampling method. Here 3–5 is the span of the integration and 0 to 1.2 is the span of the Gaussian pdf. Thereafter, for each pair (U1, U2), we calculate p(x = U1) i.e. we calculate p for the value of U1;
142
Appendix: Handouts for Computational Lab and Automated …
1.2 1.0
F, p
0.8 0.6 0.4 0.2 0.0
3.0
3.5
4.0
4.5
5.0
x 2 Fig. A.4 Showing F(x) = 1 − (x − 4)2 as undashed curve and p(x) = √1 exp − 21 x−a σ σ 2π as dashed curve obtained using the following command in Mathematica, and using variance sig = 1/3 and average avg = 4 p1 = Plot[1-(x-4)^2,{x,3,5},PlotStyle-> {Black},Frame-> True, FrameLabel-> {"x","F, p"},PlotRange-> {0,1.2}]; sig = 1/3;avg = 4; p2 = Plot[(1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((x-avg)/sig)^2],{x,3,5}, PlotStyle-> {Dashed,Black},Frame-> True]; Show[p1,p2]
and if the U2 < p(x = U1), we accept the value of U1 as a value of random variate g and call it ga. Otherwise we reject the value of U1 as g. Using program A.2 and resulting Table A.3, we gather ga[i]’s as. ga[1]=g[2];ga[2]=g[4];ga[3]=g[5];ga[4]=g[6];ga[5]=g[7]; ga[6]=g[8];ga[7]=g[9];ga[8]=g[16];ga[9]=g[17];ga[10]=g[18]; …, ga[82]=g[176];ga[83]=g[181];ga[84]=g[182];ga[85]=g[188]; ga[86]=g[191];ga[87]=g[194];ga[88]=g[197];ga[89]=g[198]; ga[90]=g[199];ga[91]=g[200];
Appendix: Handouts for Computational Lab and Automated …
143
Table A.3 Table to help obtain accepted values of random variate g as per the acceptance-rejection sampling method. Using prog. A.2 i
u1[i]
u2[i]
U1[i]
U2[i]
g[i]
i
1
0.8618
0.4287
4.7237
0.5145
Null
1
2
0.8596
0.0342
4.7192
0.0411
4.7192
2
3
0.7597
0.4609
4.5193
0.5530
Null
3
4
0.4189
0.4746
3.8378
0.5695
3.8378
4
5
0.3799
0.1687
3.7598
0.2024
3.7598
5
6
0.1899
0.0364
3.3799
0.0437
3.3799
6
7
0.3335
0.4257
3.6669
0.5108
3.6669
7
8
0.3860
0.2445
3.7721
0.2934
3.7721
8
9
0.5881
0.2322
4.1763
0.2787
4.1763
9
10
0.6743
0.9089
4.3486
1.0907
Null
10
…
…
…
…
…
…
…
191
0.3975
0.3792
3.7950
0.4550
3.7950
191
192
0.2350
0.7681
3.4700
0.9218
Null
192
193
0.1693
0.2771
3.3386
0.3325
Null
193
194
0.3306
0.3072
3.6613
0.3687
3.6613
194
195
0.8652
0.7051
4.7303
0.8462
Null
195
196
0.7953
0.8995
4.5906
1.0794
Null
196
197
0.4876
0.3045
3.9751
0.3654
3.9751
197
198
0.5590
0.7461
4.1179
0.8953
4.1179
198
199
0.1489
0.0499
3.2978
0.0599
3.2978
199
200
0.5615
0.6135
4.1230
0.7362
4.1230
200
Program A.2 (written in Mathematica) n = 654321; SeedRandom[n]; sig = 1/3;avg = 4; Table[{i = i + 1, u1[i] = RandomReal[], u2[i] = RandomReal[], U1[i]=3+(5-3)*u1[i], U2[i]=0+(1.2-0)*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((U1[i]-avg)/sig)^2],
144
Appendix: Handouts for Computational Lab and Automated …
g[i] = U1[i],Null],i}, {i,0,199,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","g[i]","i"}}] Program A.3 (written in Mathematica) F = 1-(x-4)^2; p1 = Plot[1-(x-4)^2,{x,3,5},PlotStyle-> {Black},Frame-> True, FrameLabel-> {"x","F, p"},PlotRange-> {0,1.2}]; sig = 1/3;avg = 4; p2 = Plot[(1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((x-avg)/sig)^2],{x,3,5}, PlotStyle-> {Dashed,Black},Frame-> True]; Show[p1,p2] n = 654321; SeedRandom[n]; Table[{i = i + 1, u1[i] = RandomReal[], u2[i] = RandomReal[], U1[i]=3+(5-3)*u1[i], U2[i]=0+(1.2-0)*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((U1[i]-avg)/sig)^2], g[i] = U1[i],Null],i},{i,0,199,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","g[i]", "i"}}] ga[1]=g[2];ga[2]=g[4];ga[3]=g[5];ga[4]=g[6];ga[5]=g[7];ga[6]=g[8]; ga[7] = g[9];ga[8] = g[16];ga[9] = g[17];ga[10] = g[18];ga[11] = g[21]; ga[12] = g[24];ga[13] = g[28];ga[14] = g[29];ga[15] = g[30];ga[16] = g[31]; ga[17] = g[32];ga[18] = g[34];ga[19] = g[35];ga[20] = g[36];ga[21] = g[39]; ga[22] = g[43];ga[23] = g[44];ga[24] = g[47];ga[25] = g[48];ga[26] = g[50]; ga[27] = g[53];ga[28] = g[54];ga[29] = g[57];ga[30] = g[58];ga[31] = g[60]; ga[32] = g[61];ga[33] = g[64];ga[34] = g[65];ga[35] = g[67];ga[36] = g[72]; ga[37] = g[78];ga[38] = g[80];ga[39] = g[81];ga[40] = g[82];ga[41] = g[84]; ga[42] = g[87];ga[43] = g[89];ga[44] = g[90];ga[45] = g[93];ga[46] = g[94]; ga[47] = g[95];ga[48] = g[99];ga[49] = g[100];ga[50] = g[105];ga[51] = g[106]; ga[52] = g[107];ga[53] = g[108];ga[54] = g[110];ga[55] = g[112];ga[56] = g[114]; ga[57] = g[117];ga[58] = g[122];ga[59] = g[127];ga[60] = g[128];ga[61] = g[130]; ga[62] = g[132];ga[63] = g[136];ga[64] = g[137];ga[65] = g[140];ga[66] = g[141]; ga[67] = g[142];ga[68] = g[145];ga[69] = g[147];ga[70] = g[148];ga[71] = g[152]; ga[72] = g[154];ga[73] = g[155];ga[74] = g[156];ga[75] = g[157];ga[76] = g[163]; ga[77] = g[165];ga[78] = g[166];ga[79] = g[171];ga[80] = g[174];ga[81] = g[175]; ga[82] = g[176];ga[83] = g[181];ga[84] = g[182];ga[85] = g[188];ga[86] = g[191];
Appendix: Handouts for Computational Lab and Automated …
145
ga[87] = g[194];ga[88] = g[197];ga[89] = g[198];ga[90] = g[199];ga[91] = g[200]; Table[{i = i + 1, ga[i], p[i] = (1/(sig*Sqrt[2*Pi]))*Exp[-0.5*((ga[i]-avg)/sig)^2], Y[i] = (1-(ga[i]-4)^2)/p[i]},{i,0,90,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","ga[i]","p[i]","Y[i]"}}] ListPlot[Table[{ga[i],1},{i,0,90,1}],Frame-> True, FrameLabel-> {"ga [i]","1"}] I1=Integral=Sum[Y[i]/(91),{i,1,91}] Integrate[1-(x-4)^2,{x,3,5}]; I2=N[%] PercentageError = (I1-I2)*100/I2 Program A.3 is the complete program to calculate the integral. In 200 attempts, we could get N = 91 accepted values ga of the Gaussian random variate g. Using program A.3 and resulting Table A.4, we now evaluate the integral. The 4.3% error is attributable to the facts that F(x) and p(x) used are not proportional to each other and that we have used only 91 values of the random variate. That random variate ga is distributed in the interval 3–5 as per Gaussian pdf is evidenced by Fig. A.5 which displays values of ga against stagnant integer 1. We have chosen n = 654321 by trial and error to get sets of values of u1 and u2 uniformly distributed in the interval 0–1. We used the command SeedRandom[n] to get u[i]’s reproducibly.
A.3
Automated Collection of Accepted Random Variates
Here we rewrite Program number 4.4 such that accepted random variates are collected by the program, rather than manually. Program number 4.4 rewritten (in Mathematica) n = 654321 SeedRandom[n] sig = 0.5;a = Pi/2; Table[{i = i + 1,u1[i] = RandomReal[],u2[i] = RandomReal[], U1[i]=0+Pi*u1[i],U2[i]=0+0.8*u2[i], If[U2[i] < (1/(sig*Sqrt[2*Pi]))* Exp[-0.5*((U1[i]-a)/sig)^2],g[i] = U1[i],Null]},{i,0,99,1}]; TableForm[%,TableSpacing-> {2,2}, TableHeadings-> {None,{"i","u1[i]","u2[i]","U1[i]","U2[i]","g[i]"}}] ga = {};
146
Appendix: Handouts for Computational Lab and Automated …
Table A.4 Showing tabulated values of ga[i], calculated values of p[i] =
2 √1 exp − 1 ga[i]−a and Y[i] = (1 − (ga[i] − 4)2 )/ p[i] from which we get σ 2 σ 2π
I = N1
N
F(xi ) 1 N F(ga[i]) = 1 N i=1 p(xi ) = N i=1 i=1 p[i] N
√1 σ 2π
2 1−(ga[i]−4) 1 N Y
2 = N i=1 i 1 ga[i]−a exp − 2 σ
as an approximate value of the integral I = 35 F(x)d x = 35 [1 − (x − 4)2 ]d x = 1.39061 whereas exact value is 1.33333. Error is 4.3%. Using prog. A.3 i
ga[i]
p[i]
Y [i]
1
4.7192
0.1167
4.1359
2
3.8378
1.0632
0.9158
3
3.7598
0.9232
1.0207
4
3.3799
0.2121
2.9020
5
3.6669
0.7265
1.2238
6
3.7721
0.9474
1.0007
7
4.1763
1.0406
0.9311
8
3.7593
0.9221
1.0217
9
3.7172
0.8352
1.1016
10
4.1283
1.1114
0.8849
…
…
…
…
82
4.1320
1.1065
0.8880
83
3.9907
1.1964
0.8358
84
4.2507
0.9020
1.0390
85
3.9304
1.1710
0.8498
86
3.7950
0.9906
0.9671
87
3.6613
0.7141
1.2396
88
3.9751
1.1935
0.8374
89
4.1179
1.1242
0.8771
90
3.2978
0.1301
3.8959
91
4.1230
1.1180
0.8809
i = 0; While[i {2,2}, TableHeadings-> {None,{"i","ga[i]", "p[i]","y[i]"}}] Integral = Sum[y[i]/(n1),{i,1,n1}] ListPlot[Table[{ga[[i]],1},{i,0,n1-1,1}],Frame-> True, FrameLabel-> {"ga [i]","1"}]
References
1. A Primer for the Monte Carlo Method IIya M. Sobol’ CRC Press (1994) 2. The Monte Carlo Method I. M. Sobol’ The University of Chicago Press (1974) 3. The Monte Carlo Method: The Method of Statistical Trials Edited by Yu. A. Shreider, Pergamon Press (1966) 4. Monte Carlo Methods: A Hands-On Computational Introduction Utilizing Excel Sujaul Chowdhury Morgan & Claypool (2021) Springer
© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 S. Chowdhury, Monte Carlo Methods Utilizing Mathematica®, Synthesis Lectures on Mathematics & Statistics, https://doi.org/10.1007/978-3-031-23294-7
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