sol2 - ECE 5620 Spring 2011 Homework #2: Solutions 1. 2. 1...

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Unformatted text preview: ECE 5620 Spring 2011 Homework #2: Solutions 1. 2. 1 3. 4. 2 5. 3 6. 7. (a) I ( X 1 ,X 2 ,...,X n ; Y ) = n X i =1 I ( X i ; Y | X 1 ,...,X i- 1 ) = n X i =1 [ H ( X i | X 1 ,...,X i- 1 )- H ( X i | X 1 ,...,X i- 1 ,Y )] ( i ) = n X i =1 [ H ( X i )- H ( X i | X 1 ,...,X i- 1 ,Y )] ( ii ) n X i =1 [ H ( X i )- H ( X i | Y )] = n X i =1 I ( X i ; Y ) , where ( i ) follows from the independence of the X 1 ,...,X n and ( ii ) follows because conditioning reduces entropy. (b) For equality in ( ii ) we require H ( X i | X 1 ,...,X i- 1 ,Y ) = H ( X i | Y ) for all i, which is equivalent to X i Y ( X 1 ,...,X i- 1 ) for all i. 8. Mimicking the alternative proof of the concavity of unconditional en- tropy, let ( X 1 ,Y 1 ) have joint PMF p 1 ( x,y ) and let ( X 2 ,Y 2 ) have joint PMF p 2 ( x,y ) . Suppose that T has PMF Pr( T = 1) = = 1- Pr( T = 2) . 4 Suppose that ( X 1 ,Y 1 ) , ( X 2 ,Y 2 ) , and T are all independent. Observe that ( X T ,Y T ) has joint PMF p ( x,y ) =...
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This note was uploaded on 10/03/2011 for the course ECE 5620 at Cornell University (Engineering School).

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sol2 - ECE 5620 Spring 2011 Homework #2: Solutions 1. 2. 1...

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