Hw4_sol - Homework 4: Solutions Chapter 2 5.5 (a) Since...

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Homework 4: Solutions Chapter 2 5.5 (a) Since M n +1 = 2 U n +1 M n , and U n +1 is independent of M 0 ,...,M n we have using (1.3) that E ( M n +1 | M n = m n ,...,M 0 = m 0 ) = m n E (2 U n +1 | M n = m n ,...,M 0 = m 0 ) = m n . (b) - log U 1 , - log U 2 ,... are independent with P ( - log U i > x ) = P ( U i < e - x ) = e - x , so - log U i has an exponential distribution with mean 1. Using the strong law of large numbers gives (1 /n ) log X n → - 1. (c) For any sequence of positive numbers ( X n ), if (1 /n ) log X n → - 1 then 2 n X n 0. 5.6 (a) When X 0 = x , X 1 = Binomial( x/N,N ), so E x X 1 = x , E x X 2 1 = x (1 - x/N ) + x 2 , and E x ( X 1 ( N - X 1 )) = Nx - x + x 2 /N - x 2 = x ( N - x )(1 - 1 /N ). Part (a) follows from the Markov Property (b) Since Y n is a martingale, E x X n ( N - X n ) = x ( N - x )(1 - 1 /N ) n . The desired result now follows from the obsrvation that when 0 < X n < N we have N - 1 X n ( N - X n ) N 2 / 4. 5.13 The pedestrian lemma, (3.3) in Chapter 1, implies that
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Hw4_sol - Homework 4: Solutions Chapter 2 5.5 (a) Since...

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