160a09_hw1sol

# 160a09_hw1sol - Pstat160 Solution Homework#1 Problem 1 1...

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Pstat160 Winter 2009 Solution Homework #1 Problem 1. 1) Not independent, the ranges are dependent. 2) p X ( k ) = k j =0 e - λ λ k ( k + 1)! = ( k + 1) e - λ λ k ( k + 1)! = e - λ λ k k ! So X is Poisson ( λ ). 3) p Y | X = k ( j ) = p X,Y ( k, j ) p X ( k ) = 1 k + 1 , j = 0 , . . . , k So the conditional pdf of Y given X = k is uniform on 0 , . . . , k . 4) Note that if Y = j , X has range j, j + 1 , . . . P ( Y = 0) = p Y (0) = k =0 p X,Y ( k, j ) = k =0 e - λ λ k ( k + 1)! = 1 λ k =0 e - λ λ k +1 ( k + 1)! = 1 λ k =1 e - λ λ k k ! = 1 λ (1 - e - λ ) P ( Y = 1) = p Y (1) = k =1 p X,Y ( k, j ) = k =1 e - λ λ k ( k + 1)! = 1 λ k =1 e - λ λ k +1 ( k + 1)! = 1 λ k =2 e - λ λ k k ! = 1 λ (1 - e - λ - λe - λ ) Problem 2. (pb 29) Once the first flip is done, each new flip can result by a change over, this with probability 1/2. So the number of change over is Binomial( n - 1 , 1 / 2) Problem 3. (pb 56) Let X i = 1 if there is at least 1 coupon of type i in the collection. Then X = n i =1 X i . as usual, we have E [ X ] = n i =1 X i = n i =1 P ( X i = 1) But the probability that there is at least one coupon is 1-probability that there are none so P ( X i = 1) = 1 - P ( X i = 0) = 1 - (1 - p i ) k and we get E [ X ] = n - n i =1 (1 - p

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