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# 2s - EE 478 Multiple User Information Theory Handout#11...

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EE 478 Handout #11 Multiple User Information Theory October 14, 2008 Homework Set #2 Solutions 1. Solution: (a) We need to show that for any B 1 , B 2 and α [0 , 1], αC ( B 1 ) + (1 - α ) C ( B 2 ) C ( αB 1 + (1 - α ) B 2 ) . Let X 1 p 1 ( x ) achieves C ( B 1 ) and X 2 p 2 ( x ) achieves C ( B 2 ). Define Q Bern( α ) and X = X 1 , Q = 1; X 2 , Q = 0. (1) By this definition, p ( x ) = αp 1 ( x ) + (1 - α ) p 2 ( x ), and E B ( X ) = X x ∈X p ( x ) b ( x ) = α X x ∈X p 1 ( x ) b ( x ) + (1 - α ) X x ∈X p 2 ( x ) b ( x ) = αB 1 + (1 - α ) B 2 . On the other hand, by chain rule, I ( X, Q ; Y ) = I ( Q ; Y ) + I ( X ; Y | Q ) = I ( X ; Y ) + I ( Q ; Y | X ) . But Q X Y form a Markov chain, and therefore I ( Q ; Y | X ) = 0 and I ( X ; Y ) I ( X ; Y | Q ). Using this observation, and the definition of C ( αB 1 +(1 - α ) B 2 ) as maximum mutual information between X and Y among all the distributions of X that satisfy E B ( X ) αB 1 + (1 - α ) B 2 , yields C ( αB 1 + (1 - α ) B 2 ) I ( X ; Y ) I ( X ; Y | Q ) = αI ( X 1 ; Y ) + (1 - α ) I ( X 2 ; Y ) . (b) We need to show that n i =1 b ( x i ( w )) nB . Note that the codewords are chosen from ² -typical codewords. Therefore, n X i =1 b ( x i ( w )) = n X x ∈X π ( x | x n ( w )) b ( x ) n X x ∈X ( p ( x ) + ²p ( x )) b ( x ) n (E b ( X ) + ² E b ( X )) n ( B - δ ( ² ))(1 + ² ) = nB (1 - ² 1 - ² )(1 + ² ) = nB (1 - 2 ² 2 1 - ² ) < nB. 1

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The probability of error analysis is exactly like that of channel with no cost constraint. 2. Solution: Consider I ( X ; X + Z * ) = h ( X + Z * ) - h ( X + Z * | X ) = h ( X + Z * ) - h ( Z * ) h ( X * + Z * ) - h ( Z * ) = I ( X * ; X * + Z * ) , where the inequality follows from the fact that given the variance, the entropy is maximized by the normal distribution. To prove the other inequality, we use the entropy power inequality 2 2 h ( X + Z ) 2 2 h ( X ) + 2 2 h ( Z ) . I ( X * ; X * + Z ) = h ( X * + Z ) - h ( X * + Z | X * ) = h ( X * + Z ) - h ( Z ) = 1 2 log 2 2 h ( X * + Z ) - h ( Z ) 1 2 log 2 2 h ( X * ) + 2 2 h ( Z ) · - h ( Z ) = 1 2 log 2 πeP + 2 2 h ( Z ) · - 1 2 log 2 2 h ( Z ) = 1 2 log 1 + 2 πeP 2 2 h ( z ) 1 2 log 1 + 2 πeP 2 2 h ( Z * ) = 1 2 log 1 + P N = I ( X * ; X * + Z * ) . Alternatively, we can use the result of Question 3 directly to get I ( X * ; X * + Z ) = h ( X * ) - h ( X * | X * + Z ) 1 2 log(2 πeP ) - 1 2 log (2 πe ) PN P + N = 1 2 log 1 + P N = I ( X * ; X * + Z * ) . Combining the two inequalities, we have I ( X ; X + Z * ) I ( X * ; X * + Z * ) I ( X * ; X * + Z ) . Hence, using these inequalities, it follows directly that min Z max X I ( X ; X + Z ) max X I ( X ; X + Z * ) = I ( X * ; X * + Z * ) = min Z I ( X * ; X * + Z ) max X min Z I ( X * ; X * + Z ) . (2) 2
We will now prove the inequality in the other direction is a general result for all functions of two variables. For any function f ( a, b ) of two variables, for all b , for any a 0 , f ( a 0 , b ) min a f ( a, b ) .

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