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Unformatted text preview: EECS 229A Spring 2007 * * Homework 2 solutions 1. Problem 2.48 on pg. 54 of the text. Solution : Sequence length (a) Since N is a deterministic function of X N , we have I ( N ; X N ) = H ( N ) H ( N  X N ) = H ( N ). Since P ( N = k ) = 2 k for k ≥ 1, we have H ( N ) = ∞ X k =1 2 k log 2 k = ∞ X k =1 k 2 k = 2 . (b) Since X N is a deterministic function of N , we have H ( X N  N ) = 0. (c) We have H ( X N ) = I ( X N ; N ) + H ( X N  N ) = I ( X N ; N ) = 2 . • Note that in the remaining three parts of the problem we are dealing with a different pair of random variables ( N,X N ) than in the first three parts. (d) Since N is a deterministic function of X N we have I ( N ; X N ) = H ( N ) H ( N  X N ) = H ( N ) = 1 3 log 3 + 2 3 log 3 2 = log 3 2 3 . (e) Conditioned on N = 6, X N is uniformly distributed on a set of size 2 6 , while, condi tioned on N = 12, X N is uniformly distributed on a set of size 2 12 . Thus H ( X N  N ) = H ( X N  N = 6) P ( N = 6)+ H ( X N  N = 12) P ( N = 12) = 1 3 6+ 2 3 12 = 10 . (f) We have H ( X N ) = I ( N ; X N ) + H ( X N  N ) = log 3 2 3 + 10 = log 3 + 28 3 . 2. Problem 3.3 on pg. 65 of the text. Solution : Piece of cake Let X k = 1 if the kth cut of the cake is done in the proportions ( 2 3 , 1 3 ), and let X k = 0 if it is done in the proportions ( 2 5 , 3 5 ). Then X k ,k ≥ 1 are i.i.d. with P ( X k = 1) = 3 4 and P ( X k = 0) = 1 4 . The size of the piece of cake after n cuts is n Y k =1 ( 2 3 ) X k ( 3 5 ) 1 X k = ( 3 5 ) n n Y k =1 ( 10 9 ) X k . 1 The logarithm of the size of the cake after n cuts is therefore n (log 3 5 + ( 1 n n X k =1 X k )log 10 9 ) . By the weak law or large numbers, for every > 0 we have P (  1 n n X k =1 X k 3 4  > ) → 0 as n → ∞ ....
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 Spring '10
 LIU
 Information Theory, Stochastic process, Markov chain, 1 k, 1 j, 1 9 k

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