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SOLUTIONS MANUAL FORAPPLIED PROBABILITYAND STOCHASTICPROCESSESSECOND EDITIONbyFrank BeicheltUniversity of theWitwatersrandJohannesburg, South AfricaCRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487- © 2016 by Taylor & Francis Group, LLC CRC Press i...

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SOLUTIONS MANUAL FOR APPLIED PROBABILITY AND STOCHASTIC PROCESSES SECOND EDITION

by

Frank Beichelt University of the Witwatersrand Johannesburg, South Africa

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SOLUTIONS MANUAL FOR APPLIED PROBABILITY AND STOCHASTIC PROCESSES SECOND EDITION

by

Frank Beichelt University of the Witwatersrand Johannesburg, South Africa

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa busi ness

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2016 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20160511 International Standard Book Number-13: 978-1-4822-5768-7 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validit y of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectif y in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access ww w.copyright.com (http://w ww.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at

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TABLE OF CONTENTS CHAPTER 1: RANDOM EVENTS AND THEIR PROBABILITIES

1

CHAPTER 2: ONE-DIMNSIONAL RANDOM VARIABLES

15

CHAPTER 3: MULTIDIMENSIONAL RANDOM VARIABLES

37

CHAPTER 4: FUNCTIONS OF RANDOM VARIABLES

49

CHAPTER 5: INEQUALITIES AND LIMIT THEOREMS

57

CHAPTER 6: BASICS OF STOCHASTIC PROCESSES

65

CHAPTER 7: RANDOM POINT PROCESSES

75

CHAPTER 8: DISCRETE-TIME MARKOV CHAINS

97

CHAPTER 9: CONTINUOUS-TIME MARKOV CHAINS

113

CHAPTER 10: MARTINGALES

143

CHAPTER 11: BROWNIAN MOTION

149

CHAPTER 12: SPECTRAL ANALYSIS OF STATIONARY PROCESSES

161

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CHAPTER 1 Random Events and their Probabilities 1.1) A random experiment consists of simultaneously flipping three coins. (1) What is the corresponding sample space? (2) Give the following events in terms of elementary events: A = 'head appears at least two times,' B = 'head appears not more than once,' C = 'no head appears.' (3) Characterize verbally the complementary events of A, B, and C. Solution (1) 1 head, 0 tail (head not). Ω = {(i, j, k); i, j, k = 0, 1}. Ω has 8 elements. (2) A = {(1, 1, 0), (1, 0, 1), (0, 1, 1), (0, 1, 1)}, B = {(1, 0, 0), (0, 1, 0), (0, 0, 1), (0, 0, 0)}, C = {(0, 0, 0)}. (3) A = 'head appears not more than once' (= B ). B = 'head appears at least two times' (= A ). C = 'at least one head appears'. 1.2) A random experiment consists of flipping a die to the first appearance of a '6.' What is the corresponding sample space? Solution The (countably infinite) sample space consists of all vectors (z 1 , z 2 , ..., z k− 1 , z k ) with property that zk = 6 and all z 1 , z 2 , ..., z k−1 are integers between 1 and 5; k = 1, 2, ... . 1.3) Castings are produced weighing either 1, 5, 10, or 20 kg. Let A, B, and C be the events that a casting weighs 1 or 5kg, exactly 10kg, and at least 10kg, respectively. Characterize verbally the events A ∩ B, A ∪ B, A ∩ C, and (A ∪ B) ∩ C. Solution A∩B A∪B

Impossible event. A casting weighs 1, 5, or 10kg.

A∩C (A ∪ B) ∩ C

A casting weighs 1 or 5kg. A casting weighs at least 10kg.

1.4) Three randomly chosen persons are to be tested for the presence of gene g. Three random events are introduced: A = 'none of them has gene g,' B = 'at least one of them has gene g,' C = 'not more than one of them has gene g.' Determine the corresponding sample space and characterize the events A ∩ B, B by elementary events.

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C, and B ∩ C

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2

SOLUTIONS MANUAL

Solution Let 1 and 0 indicate whether a person has gene g or not, respectively. Then the sample space Ω consists of all the 2 3 = 8 vectors (z 1 , z 2 , z 3 ) with zi =

1 if ap erson has gene g , 0 otherwise.

Ω is the same sample space as in exercise 1.1. A = {(0, 0, 0)}, B = A = Ω \A, C = {(0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1)}. A ∩ B = ∅ (impossible event), B C = B (since C ⊂ B ), B ∩ C = B ∪ C = A ∪ C = C (de Morgan rule, A ⊂ C). 1.5) Under which conditions are the following relations between events A and B true: (1) A ∩ B = Ω , (2) A ∪ B = Ω , (3) A ∪ B = A∩ B ? Solution (1) A = B = Ω . (2) A = B or B = A . More generally if A ⊇ B or A ⊇ B. (3) A = B . 1.6) Visualize by a Venn diagram that the following relations between random events A, B, and C are true: (1) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) , (2) (A ∩ B) ∪ (A ∩ B) = A , (3) A ∪ B = B ∪ (A ∩ B) . B

A∩ B∩ C

A

(A\B) ∩ C

C

1.7) (1) Verify by a Venn diagram that for three random events A, B, and C the following relation is true: (A\B) ∩ C = (A ∩ C)\(B ∩ C) . (2) Verify by the same Venn diagram that the relation (A ∩ B)\C = (A\C) ∩ (B\C) is true as well. 1.8) The random events A and B belong to a σ− algebra E. What events, generated by A and B, must belong to E (see definition 1.2)? Solution Ω , ∅, A, A, B, B, A ∪ B, A ∩ B, A ∪ B = A ∩ B, A ∩ B = A ∪ B (de Morgan rules (1.1)). Other events arise if in these events A and/or B are replaced with A and B : A∩ B = B\A, A ∩ B = A\B, A ∪ B = A\B, A ∪ B = B\A. Any unions of two or more of these events do not give rise to an event, which is different from the listed ones.

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1 RANDOM EVENTS AND THEIR PROBABILITIES

3

1.9) Two dice D1 and D2 are simultaneously thrown. The respective outcomes of D 1 and D 2 are ω1 and ω2 . Thus, the sample space is Ω = {(ω1 , ω2 ); ω1 , ω2 = 1, 2, ..., 6}. Let events A and B be defined as follows: A = 'ω1 is even and ω 2 is odd,' B = 'ω1 and ω2 satisfy ω1 + ω 2 = 9. ' What is the σ−algebra E generated by the events A and B ? Solution A = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}, B = {(3, 6), (4, 5), (5, 4), (6, 3)}. With these events A and B, the σ− algebra consists of all the events listed under exercise 1.8. 1.10) Let A and B be two disjoint random events, A ⊂ Ω , B ⊂ Ω . Check whether the set of events {A, B, A ∩ B, and A ∩ B } is (1) an exhaustive and (2) a disjoint set of events (Venn diagram). Solution This set is neither exhaustive nor disjoint. 1.11) A coin is flipped 5 times in a row. What is the probability of the event A that 'head' appears at least 3 times one after the other? Solution The underlying random experiment is a Laplace experiment the state space Ω of which has 2 5 = 32 elementary events. 'head' five times in a row: 1 elementary event 'head' four times in a row: 2 elementary events 'head' three times in a row: 3 elementary events Thus, 6 elementary events are favorable for the occurrence of A. Hence, P(A) = 6/32. 1.12) A die is thrown. Let A = {1, 2, 3} and B = { 3, 4, 6} be two random events. Determine the probabilities P(A ∪ B), P(A ∩ B), and P(B\A). Solution P(A) = P(B) = 0.5. P(A ∩ B) = P({3}) = 1/6. P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 0.5 + 0.5 − 1/6 = 5/6. P(B\A) = P(B) − P(A ∩ B) = 0.5 − 1/6 = 1/3. 1.13) A die is thrown 3 times. Determine the probability of the event A that the resulting sequence of three integers is strictly increasing. Solution The state space Ω of this random experiment comprises 6 3 = 216 elementary events. There are the following favorable elementary events: (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4.5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6). Hence, P(A) = 19/216.

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SOLUTIONS MANUAL

1.14) Two dice are thrown simultaneously. Let (ω 1 , ω 2 ) be an outcome of this random experiment ' A = ω1 + ω2 ≤ 10 ' and B = 'ω1 ⋅ ω2 ≥ 19 .' Determine the probability P(A ∩ B). Solution A = {(5, 6), (6, 5), (6, 6)}, B = {(4, 5), (5, 4), (5, 5), (5, 6), (6, 5), (6, 6)}. P(A) = 3/36, P(B) = 6/36. B = (A ∩ B) ∪ (A ∩ B) = (A ∩ B) ∪A since A ⊂ B. Hence, P(B) = P(A ∩ B) + P(A) so that P(A ∩ B) = 1/12. 1.15) What is the probability p 3 to get 3 numbers right with one ticket in the '6 out of 49' number lottery? Solution Hypergeometric distribution with N = 49, M = n = 6, m = 3 : ⎛ 6 ⎞ ⎛ 43 ⎞ ⎝ 3⎠ ⎝ 3 ⎠ p3 = = 0.01765. ⎛ 49 ⎞ ⎝6 ⎠ 1.16) A sample of 300 students showed the following results with regard to physical fitness and body weight: weight [kg] 60 <

[60-80]

80 >

48

64

11

fitness satisfactory

22

42

29

bad

19

17

48

good

One student is randomly chosen. It happens to be Paul. (1) What is the probability that the fitness of Paul is satisfactory? (2) What is the probability that the weight of Paul is greater than 80 kg? (3) What is the probability that the fitness of Paul is bad and that his weight is less than 60 kg? Solution (1) p = (22 + 42 + 29) /300 = 0.3100. (2) p = (11 + 29 + 48) /300 = 0.2933. (3) p = 19/300 = 0.0633. 1.17) Paul writes four letters and addresses the four accompanying envelopes. After having had a bottle of whisky, he puts the letters randomly into the envelopes. Determine the probabilities p k that k letters are in the 'correct' envelopes, k = 0, 1, 2, 3. Solution There are 4! = 24 possibilities (elementary events) to put the letters into the envelopes k = 0 : There are 9 favorable elementary events: p 0 = 9/24 ≈ 0.3750. k = 1 : There are 8 favorable elementary events: p 1 = 8/24 = 0.3333. k = 2 : There are 6 favorable elementary events: p 2 = 6/24 = 0.2500. k = 3 : p 3 = 1 − p 0 − p 1 − p 2 = 1/24 = 0.0416.

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