sol6[1]

# sol6[1] - ECE 255AN Fall 2011 Homework set 6 solutions 1(a...

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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 ) = λ and q be exponential distribution with mean λ . Then D ( p | q ) =- h ( p ) + Z p log( λ ) + Z p x λ log( e ) =- h ( p ) + log( λ ) + log ( e ) ≥ log( eλ ) ≥ h ( p ) Hence the maximum value of entropy is log( eλ ) and is achieved by exponential distribution. 3. A sequence belongs to typical set if and only if 2- ( h ( p )+ δ ) n ≤ p ( x n ) ≤ 2- ( h ( p )- δ ) n . (a) For Gaussian distribution with zero mean, p ( x n ) = 1 ( √ 2 πσ ) n e- ∑ n i =1 x 2 i 2 σ 2 and 2- h ( p ) n = 1 ( √ 2 πeσ ) n . Therefore a sequence, belongs to typical set if and only if, nσ 2 (1- ln(4) δ ) ≤ n X i =1 x 2 i ≤ nσ 2 (1 + ln(4) δ ) (b) For an exponential distribution with mean λ , p ( x n ) = 1 λ n e- ∑ n i =1 x i λ and 2- h ( p ) n = 1 ( eλ ) n . Hence a sequence belongs to typical set if and only if nλ (1- ln(2) δ ) ≤ n X i =1 x i ≤ nλ (1 + ln(2) δ ) 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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sol6[1] - ECE 255AN Fall 2011 Homework set 6 solutions 1(a...

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