sol2[1]

sol2[1] - ECE 255AN Fall 2011 Homework set 2 - solutions 1....

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ECE 255AN Fall 2011 Homework set 2 - solutions 1. For the upper bound, observe that for all distributions over the positive integers, H ( X ) - E ( X ) = X x p ( x ) log 1 p ( x ) - X x p ( x ) log 2 x = X x p ( x ) log 2 - x p ( x ) log X x 2 - x log 1 = 0 . For the lower bound, if X is distributed monotonically over the positive integers then p ( x ) 1 /x , for example, p (2) 1 / 2. Hence, H ( X ) = E (log 1 p ( X ) ) E (log 1 1 /X ) E (log( X )) . 2. (a) ( X + Y,X - Y ) is in 1-1 correspondence with ( X,Y ), hence H ( X + Y,X - Y ) = H ( X,Y ) = H ( X ) + H ( Y ) = 2 log n . (b) 2 log n = H ( X,Y ) = H (min( X,Y ) , max( X,Y )) + H (( X,Y ) | (min( X,Y ) , max( X,Y ))) = H (min( X,Y ) , max( X,Y )) + P ( X,Y ∈ { n - 1 ,n } X 6 = Y ) · h ( . 5) = H (min( X,Y ) , max( X,Y )) + 2 n 2 . Solutions to Chapter 2 problems 8. Drawing with and without replacement. Intuitively, it is clear that if the balls are drawn with replacement, the number of possible choices for the i -th ball is larger, and therefore the conditional entropy is larger. But computing the conditional distributions is slightly involved. It is easier to compute the unconditional entropy. With replacement. In this case the conditional distribution of each draw is the same for
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sol2[1] - ECE 255AN Fall 2011 Homework set 2 - solutions 1....

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