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HW12s[1] - ECE 534 Elements of Information Theory Fall 2010...

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ECE 534: Elements of Information Theory, Fall 2010 Homework 12 Solutions (ALL DUE TO Kenneth S. Palacio Baus) December 1, 2010 1. Problem 15.7. Convexity of capacity region of broadcast channel . Let C R 2 be the ca- pacity region of all achievable rate pairs R = ( R 1 , R 2 ) for the broadcast channel. Show that C is a convex set by using a time-sharing argument. Specifically, show that if R (1) and R (2) are achievable, λ R (1) + (1 - λ ) R (2) is achievable for 0 λ 1. Solution: We have two rate pairs that are achievable: R (1) = ( R (1) 1 , R (1) 2 ) and R (2) = ( R (2) 1 , (2) 2 ) for which, we have two sequences of codes: ((2 nR (1) 1 , 2 nR (1) 2 ) , n ) and ((2 nR (2) 1 , 2 nR (2) 2 ) , n ). Like it is done in the proof of Theorem 15.3.2 for the case of the multiple-access channel in the textbook, here we can apply a similar argument and construct a third codebook of length n at a rate λ R (1) +(1 - λ ) R (2) using the first codebook for the first λn symbols, and the second codebook for the last (1 - λ ) n . We have that the number of X 1 codewords for the new code is given by: 2 nλR (1) 1 2 n (1 - λ ) R (2) 1 = 2 n ( λR (1) 1 +(1 - λ ) R (2) 1 ) , And the number of X 2 codewords for the new code is given by: 2 nλR (1) 2 2 n (1 - λ ) R (2) 2 = 2 n ( λR (1) 2 +(1 - λ ) R (2) 2 ) , So, we have obtained a rate: λ R (1) + (1 - λ ) R (2) . Recalling that the overall probability of error is less that the sum of the probabilities of error for each of the segments: P ( n ) e P ( λn ) e (1) + P ((1 - λ ) n ) e (2) We see that the probability of error goes to 0 as n → ∞ , hence the rate is achievable. 1
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2. Problem 15.11. Converse for the degraded broadcast channel . The following chain of inequal- ities proves the converse for the degraded discrete memoryless broadcast channel. Provide reasons for each of the labeled inequalities. Setup for converse for degraded broadcast channel capacity: ( W 1 , W 2 ) indep X n ( W 1 , W 2 ) Y n 1 Y n 2 - Encoding: f n : 2 nR 1 × 2 nR 2 → X n - Decoding: g n : Y n 1 2 nR 1 , h n : Y n 2 2 nR 2 . Let U i = ( W 2 , Y i - 1 . Then nR 2 Fano I ( W 2 ; Y n 2 ) (1) = ( a ) n X i =1 I ( W 2 ; Y 2 i | Y i - 1 2 ) (2) = ( b ) X i ( H ( Y 2 i | Y i - 1 2 ) - H ( Y 2 i | W 2 , Y i - 1 2 )) (3) ( c ) X i ( H ( Y 2 i ) - H ( Y 2 i | W 2 , Y i - 1 2 , Y i - 1 1 )) (4) = ( d ) X i ( H ( Y 2 i ) - H ( Y 2 i | W 2 , Y i - 1 1 )) (5) = ( e ) X i I ( U i ; Y 2 i ) (6) Solution: Reasons for each of the labeled inequalities: (a) is given by the Chain rule. (b) corresponds to the definition of conditional mutual information. (c) conditioning reduces entropy. (d) since Y 2 i is conditionally independent of Y i - 1 2 given Y i - 1 1 .
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