sol2[1]

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 01/19/2012.

Page1 / 4

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online