Math 180B Homework 9 Solutions

Transcription

1 Problem 1 (Pinsky & Karlin, Exercise 5.1.3). Let X and Y be independent Poisson distributed random variables with parameters α and β, respectively. Determine the conditional distribution of X, given that N X + Y n. Solution. Note that N X +Y has the Poisson distribution with parameter α+β, and that N X, so if k > n, then P (X k N n). If k n, then by the independence of X and Y we have so P (X k, N n) P (X k, Y n k) P (X k)p (Y n k) e α α k e β β n k (n k)!, P (X k, X + Y n) P (X k N n) P (N n) e α α k e β β n k (n k)! n! e (α+β) (α + β) n ( ) n α k β n k ( ) ( ) k ( n α k (α + β) n 1 α ) n k. k α + β α + β Problem 2 (Pinsky & Karlin, Exercise 5.1.5). Suppose that a random variable X is distributed according to a Poisson distribution with parameter λ. The parameter λ is itself a random variable, exponentially distributed with density f(x) θe θx for x. Find the probability mass function for X. Solution. Stated precisely, the assumption of the problem is that if x and k, 1, 2,..., then P (X k λ x) e x x k. 1

2 By the law of total probability, we have P (X k) P (X k λ x)f(x) dx e x x k θe θx dx θ x k e (θ+1)x dx. Making the change of variables u (θ + 1)x, simplifying, and using the fact that u k e u du Γ(k + 1), we get P (X k) θ (θ + 1) k+1 θ (θ + 1) k+1 θ θ + 1 u k e u du ( 1 θ θ + 1 Thus, X has the geometric distribution with parameter θ/(θ + 1). ) k Problem 3 (Pinsky & Karlin, Exercise 5.1.9). Let {X(t) : t } be a Poisson process having rate parameter λ 2. Determine the following expectations: (a) E[X(2)]. (b) E [ (X(1)) 2]. (c) E[X(1)X(2)]. Solution. (a) Since X(2) Poisson(4), we have E[X(2)] 4. (b) Since X(1) Poisson(2), we have E[X(1)] 2 and Var(X(1)) 2, so E [ (X(1)) 2] Var(X(1)) + (E[X(1)]) (c) Since a Poisson process has independent increments, the random variables X(1) and X(2) X(1) are independent. Moreover, X(2) X(1) Poisson(2). Therefore, we have E[X(1)X(2)] E [ (X(1)) 2 + X(1) ( X(2) X(1) )] E [ (X(1)) 2] + E[X(1)]E[X(2) X(1)]

3 Problem 4 (Pinsky & Karlin, Problem 5.1.2). Suppose that minor defects are distributed over the length of a cable as a Poisson process with rate α, and that, independently, major defects are distributed over the cable according to a Poisson process of rate β. Let X(t) be the number of defects, either major or minor, in the cable up to length t. Argue that X(t) must be a Poisson process of rate α + β. Solution. Let Y (t) be the number of minor defects over the length of the cable up to time t, and let Z(t) be the number of major defects over the length of the cable up to time t. Then {Y (t) : t } and {Z(t) : t } are independent Poisson processes with rates α and β, respectively, and X(t) Y (t) + Z(t) for all t. We now verify that {X(t) : t } is a Poisson process of rate α + β. First, we have X() Y () + Z(). Next, if s and t >, then Y (s + t) Y (s) Poisson(αt) and Z(s + t) Z(s) Poisson(βt), so, since Y (s + t) Y (s) and Z(s + t) Z(s) are independent, it follows that X(s + t) X(s) ( Y (s + t) Y (s) ) + ( Z(s + t) Z(s) ) Poisson((α + β)t). Finally, suppose we have time points t < t 1 < < t n. Then the random variables Y (t 1 ) Y (t ),..., Y (t n ) Y (t n 1 ), Z(t 1 ) Z(t ),..., Z(t n ) Z(t n 1 ) are independent, so since X(t i ) X(t i 1 ) ( Y (t i ) Y (t i 1 ) ) + ( Z(t i ) Z(t i 1 ) ) for all i 1,..., n, it follows that the random variables X(t 1 ) X(t ),..., X(t n ) X(t n 1 ) are independent. We conclude that {X(t) : t } is a Poisson process of rate α + β. 3

4 Problem 5 (Pinsky & Karlin, Problem 5.1.3). The generating function of a probability mass function p k P (X k), for k, 1,..., is defined by g X (s) E[s X ] p k s k for s < 1. k Show that the generating function for a Poisson random variable X with mean µ is given by g X (s) e µ(1 s). Solution. If X Poisson(µ), then we just compute its generating function: g X (s) P (X k)s k e µ µ k k k k s k e µ (µs) k e µ e µs e µ(1 s). Problem 6 (Pinsky & Karlin, Problem 5.1.9). Arrivals of passengers at a bus stop form a Poisson process X(t) with rate λ 2 per unit time. Assume that a bus departed at time t leaving no customers behind. Let T denote the arrival time of the next bus. Then, the number of passengers present when it arrives is X(T ). Suppose that the bus arrival time T is independent of the Poisson process and that T has the uniform probability density function 1 for t 1, f T (t) otherwise. (a) Determine E[X(T ) T t] and E [ (X(T )) 2 T t ]. (b) Determine the mean E[X(T )] and variance Var[X(T )]. Solution. (a) By the independence of T and {X(t) : t }, we have E[X(T ) T t] E[X(t) T t] E[X(t)] 2t 4

5 and E [ (X(T )) 2 ] [ ] [ T t E (X(t)) 2 T t E (X(t)) 2 ] 4t 2 + 2t. (b) By the law of total probability, we have E[X(T )] E[X(T ) T t]f T (t) dt 1 2t dt 1. Similarly, so E [ (X(T )) 2] 1 ( 4t 2 + 2t ) dt 7 3, Var(X(T )) E [ (X(T )) 2] (E[X(T )]) Problem 7 (Pinsky & Karlin, Exercise 5.2.4). Suppose that a book of 6 pages contains a total of 24 typographical errors. Develop a Poisson approximation for the probability that three particular successive pages are error-free. Solution. The typographical errors in the book can be modeled as the arrivals of a Poisson process with rate 24/6. If N is the number of typographical errors in three particular successive pages, then N approximately has the Poisson distribution with rate 3 24/6 1.2, so P (N ) e 1.2. Problem 8 (Pinsky & Karlin, Problem 5.2.4). Suppose that N points are uniformly distributed over the interval [, N). Determine the probability distribution for the number of points in the interval [, 1) as N. Solution. Let N be a positive integer, and let U 1,..., U N Uniform([, N)) be independent random variables, and for i 1,..., N, let X i be the indicator random variable of the event {U i [, 1)}. That is, 1 if U i [, 1), X i 1 {Ui [,1)} otherwise. 5

6 Since P (U i [, 1)) 1/N, we have X i Bernoulli(1/N) for each i, and X 1,..., X N are independent. If X denotes the number of the U i s in the interval [, 1), then X X X N Binomial(N, 1/N). By the Poisson approximation to the binomial distribution, it follows that as N, X approximately has the Poisson distribution with rate N 1/N 1. Problem 9 (Pinsky & Karlin, Problem 5.2.5). Suppose that N points are uniformly distributed over the surface of a circular disk of radius r. Determine the probability distribution for the number of points within a distance of one of the origin as N, r, N/(πr 2 ) λ. Solution. To simplify notation, let D R denote the surface of the disk of radius R centered at the origin in R 2. Let N be a positive integer, and let r >. Let U 1,..., U N be independent random variables distributed uniformly over D r. That is, each U i has density 1 if x 2 + y 2 r 2, πr f(x, y) 2 otherwise. For i 1,..., N, let X i be the indicator random variable of the event U i D 1. That is, 1 if U i D 1, X i 1 {Ui D 1 } otherwise. Since 1 P (U i D 1 ) D 1 πr dx dy 1 2 r, 2 we have X i Bernoulli(1/r 2 ) for each i, and X 1,..., X N are independent. If X denotes the number of the U i s within 1 of the origin, then X X X N Binomial(N, 1/r 2 ). By the Poisson approximation to the binomial distribution, it follows that as N and r, with N/(πr 2 ) λ, X approximately has the Poisson distribution with rate N 1/r 2 πλ. 6

7 Problem 1 (Pinsky & Karlin, Problem ). Let X and Y be jointly distributed random variables and B an arbitrary set. Fill in the details that justify the inequality P (X B) P (Y B) P (X Y ). Solution. We have {X B} {X B, Y B} {X B, Y / B} {Y B} {X Y }, and hence Therefore, P (X B) P ( {Y B} {X Y } ) P (Y B) + P (X Y ). P (X B) P (Y B) P (X Y ). Switching the roles of X and Y in the argument above, we also have P (Y B) P (X B) P (X Y ), and hence P (X B) P (Y B) P (X Y ). 7