Sample solution manual fundamentals of probability with stochastic processes Ghahramani 4th edition pdf PDF

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Authors: Saeed Ghahramani
Published: CRC 2019
Edition: 4th
Pages: 375
Type: pdf
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FOLFNKHUHWRGRZQORDG

Instructor's Solutions Manual

FUNDAMENTALS O F P R O B A B IL IT Y WITH

STOCH ASTIC PROCESSES

FOURTH EDITION

SAEED GHAHRAMANI Western New England University Springfield, Massachusetts, USA

A CHAPMAN & HALL BOOK

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FOLFNKHUHWRGRZQORDG

C ontents ◮

1

1.2 1.4 1.7

Sample Space and Events 1 Basic Theorems 4 Random Selection of Points from Intervals Review Problems 13

◮

Companion for Chapter 1

1B

Applications of Probability to Genetics

◮

2

2.2 2.3 2.4 2.5

Counting Principle 21 Permutations 24 Combinations 27 Stirling’ Formula 42 Review Problems 42

◮

3

3.1 3.2 3.3 3.4 3.5

Conditional Probability The Multiplication Rule Law of Total Probability Bayes’ Formula 60 Independence 66 Review Problems 76

◮

Companion for Chapter 3

3B

More on Applications of Probability to Genetics

Axioms of Probability

1

10

19 19

Combinatorial Methods

21

Conditional Probability and Independence

47

47 52 55

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80 80

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FOLFNKHUHWRGRZQORDG Contents

Distribution Functions and Discrete Random Variables

iii

85

◮

4

4.2 4.3 4.4 4.5 4.6

Distribution Functions 85 Discrete Random Variables 89 Expectations of Discrete Random Variables 94 Variances and Moments of Discrete Random Variables Standardized Random Variables 107 Review Problems 107

◮

5

5.1 5.2 5.3

Bernoulli and Binomial Random Variables Poisson Random Variable 117 Other Discrete Random Variables 125 Review Problems 132

◮

6

6.1 6.2 6.3

Probability Density Functions 137 Density Function of a Function of a Random Variable Expectations and Variances 144 Review Problems 151

◮

7

7.1 7.2 7.3 7.4 7.5 7.6

Uniform Random Variable 153 Normal Random Variable 158 Exponential Random Variables 166 Gamma Distribution 171 Beta Distribution 175 Survival Analysis and Hazard Function Review Problems 183

◮

8

8.1 8.2 8.3 8.4

Joint Distribution of Two Random Variables Independent Random Variables 200 Conditional Distributions 209 Transformations of Two Random Variables Review Problems 230

100

Special Discrete Distributions

110 110

Continuous Random Variables

137

Special Continuous Distributions

141

153

180

Bivariate Distributions

187 187

218

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FOLFNKHUHWRGRZQORDG Contents

◮

9

9.1 9.2 9.3

Joint Distribution of n > 2 Random Variables Order Statistics 248 Multinomial Distributions 253 Review Problems 255

◮

10

10.1 10.2 10.3 10.4 10.5

◮ 10B

◮ 11.1 11.2 11.3 11.4 11.5

◮ 12.2 12.3 12.4

◮ 12B

Multivariate Distributions

238 238

More Expectations and Variances

Expected Values of Sums of Random Variables Covariance 265 Correlation 274 Conditioning on Random Variables 276 Bivariate Normal Distribution 289 Review Problems 292

Companion for Chapter 10 Pattern Appearance

11

iv

260 260

299

299

Sums of Independent Random Variables and Limit Theorems

300

Moment-Generating Functions 300 Sums of Independent Random Variables 308 Markov and Chebyshev Inequalities 314 Laws of Large Numbers 318 Central Limit Theorem 321 Review Problems 326

12

330

Stochastic Processes

More on Poisson Processes 330 Markov Chains 335 Continuous-Time Markov Chains Review Problems 362

353

Companion for Chapter 12 Brownian Motion

367

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FOLFNKHUHWRGRZQORDG

Chapter 1

A xioms 1.2

Probability

of

SAMPLE SPACE AND EVENTS

1. {M, I, S, P} is a sample space for this experiment, and {I} is the event that the outcome is a vowel.

2. A sample space is S = {0, 1, 2, . . . , 57}. The desired event is E = {3, 4, 5, 6, 7, 8}. 3. E is the event of at least two heads. 4. E is the event that one die shows three times as many dots as the other. F is the event that the sum of the outcomes is exactly 6.

5. For 1 ≤ i, j ≤ 3, by (i, j ) we mean that Vann’s is i, and Paul’scard  card number  number is j. Clearly, A = (1, 2), (1, 3), (2, 3) and B = (2, 1), (3, 1), (3 , 2) . (a)

Since A ∩ B = ∅, the events A and B are mutually exclusive.

(b) None of (1, 1), (2, 2), (3, 3) belongs to A ∪ B. Hence A ∪ B not being the sample space shows that A and B are not complements of one another.

6. S = {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB}. 7. {x : 0 < x < 20}; {1, 2, 3, . . . , 19}. 8. Denote the dictionaries by d 1 , d2; the third book by a. The answers are {d1 d2 a, d1 ad2 , d2 d1 a, d2 ad1 , ad1 d2 , ad 2 d1 } and {d1 d2 a, ad1d2 }.

9. EF : One 1 and one even. E cF : One 1 and one odd. E cF c : Both even or both belong to {3, 5}.

10. S = {QQ, QN, QP, QD, DN, DP, N P, N N, P P }. (a) {QP }; (b) {DN, DP, N N }; (c) ∅.   1 ; 11. S = x : 7 ≤ x ≤ 9 6      1 x : 7 ≤ x ≤ 7 4 ∪ x : 7 34 ≤ x ≤ 8 14 ∪ x : 8 34 ≤ x ≤ 9 61 .

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FOLFNKHUHWRGRZQORDG Section 1.2

Sample Space and Events

2

12. (a) The sample space is S = {A 1 A2 , Ac1A2 , A1Ac2, Ac1Ac2 }. (b)

The event that the system is not operative at the random time is E = {A1cA2 , A1 Ac2 , Ac1 Ac2 }.

13. S is a sample space for the experiment of flipping a coin until two tails appear consecutively.

14. E ∪ F ∪ G = G: If E or F occurs, then G occurs. EF G = G: If G occurs, then E and F occur.

15. For 1 ≤ i ≤ 3, 1 ≤ j ≤ 3, by ai bj we mean passenger a gets off at hotel i and passenger b gets off at hotel j. The answers are {a i bj : 1 ≤ i ≤ 3, 1 ≤ j ≤ 3} and {a1 b1 , a2 b2 , a3b3 }, respectively.

16. Let a, ℓ, and f represent the outcomes in which the subject identifies almond, lemon, and flax as his or her favorite color, respectively. Let ∼a, ∼ℓ, and ∼f represent the outcomes in which the subject does not identify almond, lemon, and flax as his or her favorite color, respectively. The sample space of the experiment is  S = aℓf, (∼ a)ℓf, a(∼ ℓ)f, aℓ(∼ f ), (∼ a)(∼ ℓ)f, (∼ a)ℓ(∼ f ),a(∼ ℓ)(∼ f ),  (∼ a)(∼ ℓ)(∼ f) .

17. Let x, y, and z be the demand, in thousands, in a random month, for band saws, reciprocating saws, and hole saws, respectively. The sample space is   S = (x, y, z) : 30 ≤ x ≤ 36, 28 ≤ y ≤ 33, 300 ≤ z ≤ 600 . The desired event is   (x, y, z) ∈ S : x ≥ 33, z < 435 .

18. Let ad be the outcome that Alexia is dead in five years, and let aℓ be the outcome that she lives at that time. Define rd and rℓ similarly. A sample space for this experiment  . The event that at that time only one of is S = (ad, rd), (ad, rℓ), (aℓ, rd), (aℓ, rℓ)  them lives is E = (ad, rℓ), (aℓ, rd) .

19. (a) (E ∪ F )(F ∪ G) = (F ∪ E)(F ∪ G) = F ∪ EG. (b)

Using part (a), we have

(E ∪F )(E c ∪F )(E ∪F c ) = (F ∪EE c)(E ∪F c ) = F (E ∪ F c ) = F E ∪ F F c = F E.

20. (a) AB c C c ;

(b) A ∪ B ∪ C;

(e) AB c C c ∪ Ac B c C ∪ Ac BC c ;

(c) Ac B c C c ;

(d) ABC c ∪ AB c C ∪ Ac BC;

(f) (A − B) ∪ (B − A) = (A ∪ B) − AB.

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FOLFNKHUHWRGRZQORDG Section 1.2

Sample Space and Events

3

21. The event that the device is operative at that random time is E=

n 

Ai = A1 A2 · · · An .

i=1

22. If B = ∅, the relation is obvious. If the relation is true for every event A, then it is true for S, the sample space, as well. Thus S = (B ∩ S c ) ∪ (B c ∩ S) = ∅ ∪ B c = B c , showing that B = ∅.

23. Parts (a) and (d) are obviously true; part (c) is true by DeMorgan’s law; part (b) is false: throw a four-sided die; let F = {1, 2, 3}, G = {2, 3, 4}, E = {1, 4}.

24. Introducing a rectangular coordinate system with origin at the center of dartboard, we 

have that a sample space for the point at which a dart hits the board is S = (x, y) :  x2 + y 2 < 81 .   25. (a) The H, T T T H, ... . (b) The desired event is  sample space is S = H, T H, T T  E = H, T T H, T T T T H, T T T T T T H, ... .

26. (a)

∞

n=1

An ; (b)

37

n=1

An .

27. Clearly, E 1 ⊃ E2 ⊃ E3 ⊃ · · · ⊃ Ei ⊃ · · · . Hence

i=1 E i = E 1 = (−1/2, 1/2). Now the only point that belongs to all E i ’s is 0. For any other point, x, x ∈ (−1, 1), ∞ there is an i for which x ∈ / (−1/2 i , 1/2i ). So i=1 E i = {0}.

∞

28. Straightforward. 29. Straightforward. 30. Straightforward. 31. Let a1 , a2, and a3 be the first, the second, and the third volumes of the dictionary. Let a4 , a5 , a6 , and a7 be the remaining books. Let A = {a 1 , a2 , . . . , a7 }; the answers are   S = x1 x2 x3 x4 x5 x6 x7 : xi ∈ A, 1 ≤ i ≤ 7, and xi = xj if i = j and



 x1 x2 x3 x4 x5 x6 x7 ∈ S : xi xi+1 xi+2 = a1 a2 a3 for some i, 1 ≤ i ≤ 5 ,

respectively.

32. The sample space is S=

31 

c c c Ai2 Ai3 Ai1 Ai2 Ai3 ∪ Ai1 Ai2 Ai3 ∪ Ai1 Ai2c Ai3 ∪ Ai1 Ai2 Aci3 ∪ Ai1

i=1

 c c c c c c Ai3 Ai2 ∪ Ai1 Ai3 ∪ Ai1 Ai2 ∪ Aci1 Ai2 Ai3 .

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FOLFNKHUHWRGRZQORDG Section 1.4

33.

∞

m=1

∞

n=m

An .

34. Let B1 = A1 , B2 = A2 − A1 , B3 = A3 − (A1 ∪ A2 ), . . . , Bn = An − . . ..

1.4

4

Basic Theorems

n−1 i=1

Ai ,

BASIC THEOREMS

1. No; P (sum 11) = 2/36 while P (sum 12) = 1/36. 



2. Since P (AB) = 0, we have 1 ≥ P A ∪ B = P (A) + P (B). 3. 0.33 + 0.07 = 0.40. 4. No, they are not consistent. The first statement implies that the probability of success is 15/16, while the second statement implies that it is 1/16.

5. Let E be the event that an earthquake will damage the structure next year. Let H be the event that a hurricane will damage the structure next year. We are given that P (E) = 0.015, P (H ) = 0.025, and P (EH) = 0.0073. Since P (E ∪ H ) = P (E) + P (H ) − P (EH ) = 0.015 + 0.025 − 0.0073 = 0.0327, the probability that next year the structure will be damaged by an earthquake and/or a hurricane is 0.0327. The probability that it is not damaged by any of the two natural disasters is 0.9673.

6. Clearly, she made a mistake. Since EF ⊆ E, we must have P (EF ) ≤ P (E). However, in Tina’s calculations, P (EF ) =

3 1 > = P (E). 4 8

7. We are interested in the probability    of the  event A ∪ B − AB. Since AB ⊆ A ∪ B , we have P A ∪ B − AB = P A ∪ B − P (AB ) = 0.8 − 0.3 = 0.5.

8. Let A be the event that a randomly selected applicant has a high school GPA of at least 3.0. Let B be the event that this applicant’s SAT score is 1200 or higher. We have  P A ∪ B) = P (A) + P (B) − P (AB) = 0.38 + 0.30 − 0.15 = 0.53. Therefore, 53% of all applicants are admitted to the college.

9. Let J , B, and T be the events that Jacqueline, Bonnie, and Tina win, respectively. We are given that P (B) = (2/3)P (J ) and P (B) = (4/3)P (T ). Therefore, P (J ) = (3/2)P (B) and P (T ) = (3/4)P (B). Now P (J ) + P (B) + P (T ) = 1 implies that 3 3 P (B) + P (B) + P (B) = 1. 4 2 This gives P (B) = 4/13. Thus P (J ) = 6/13, and P (T ) = 3/13.

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