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Unformatted text preview: ECE 255AN Fall 2011 Homework set 6  solutions 1. (a) Let Y = X + Z and f ( Y ) := X + Z mod (2) = X Z . Clearly, X Y f ( Y ) forms a Markov chain. Therefore, by data processing inequality, I ( X,Y ) I ( X,f ( Y )) for every input distribution p ( x ). Hence max p ( x ) I ( X,Y ) max p ( x ) I ( X,f ( Y )) and therefore capacity of X + Z is higher than that of X Z . (b) Let Z B ( p ). The relationship between capacities of Y 1 = XZ and Y 2 = X Z depends on the value of p . i. p 1 2 : Let X B ( 1 2 p ), then Y 1 B (1 / 2) C ( X : Y 1 ) I p ( x ) ( X : Y 1 ) = H ( Y 1 ) H ( Y 1  X ) = 1 1 2 p h ( p ) 1 h ( p ) = C ( X : Y 2 ) Hence the C ( X : Y 1 ) C ( X : Y 2 ). ii. p 1 2 : In this region, C ( X : Y 2 ) decreases with p and C ( X : Y 1 ) increases with p . There exist a point p < 1 / 2 at which they are equal. For p < p the C ( X : Y 2 ) > C ( X : Y 1 ) and for p > p , C ( X : Y 1 ) > C ( X : Y 2 ). 2. For any two distributions, p ( x ) and q ( x ), D ( p  q ) 0. Let p be any distribution with mean E ( X ) = and q be exponential distribution with mean . Then D ( p  q ) = h ( p ) + Z p log( ) + Z p x log( e ) = h ( p ) + log( ) + log ( e ) log( e ) h ( p ) Hence the maximum value of entropy is log( e ) and is achieved by exponential distribution. 3. A sequence belongs to typical set if and only if 2 ( h ( p )+ ) n p ( x n ) 2 ( h ( p ) ) n . (a) For Gaussian distribution with zero mean, p ( x n ) = 1 ( 2 ) n e n i =1 x 2 i 2 2 and 2 h ( p ) n = 1 ( 2 e ) n . Therefore a sequence, belongs to typical set if and only if, n 2 (1 ln(4) ) n X i =1 x 2 i n 2 (1 + ln(4) ) (b) For an exponential distribution with mean , p ( x n ) = 1 n e n i =1 x i and 2 h ( p ) n = 1 ( e ) n . Hence a sequence belongs to typical set if and only if n (1 ln(2) ) n X i =1 x i n (1 + ln(2) ) 1 Solutions to Chapter 2 problems 39. (a) H ( X,Y,Z ) = H ( X,Y ) + H ( Z  X,Y ) (1) H ( X,Y ) (2) = 2 . (3) So the minimum value for H...
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This document was uploaded on 01/19/2012.
 Fall '09

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