# hw1_sol - EE595A Introduction to Information Theory Winter...

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EE595A Introduction to Information Theory University of Washington Winter 2004 Dept. of Electrical Engineering Handout 4: Problem Set 1: Solutions Prof: Jeff A. Bilmes <bilmes@ee.washington.edu> Lecture 3, Jan 27, 2004 4.1 Problems from Text Do problems 2.16, 2.3, 2.5, 2.6, 2.12, 2.10, 2.20 in Cover and Thomas. Problem 2.16 Example of joint entropy. Let p ( x, y ) be given by @ @ @ X Y 0 1 0 1 3 1 3 1 0 1 3 Find 1. H ( X ) , H ( Y ) . 2. H ( X | Y ) , H ( Y | X ) . 3. H ( X, Y ) . 4. H ( Y ) - H ( Y | X ) . 5. I ( X ; Y ) . 6. Draw a Venn diagram for the quantities in (a) through (e). Solution 2.16 Example of joint entropy 1. H ( X ) = 2 3 log 3 2 + 1 3 log 3 = 0 . 918 bits = H ( Y ) . 2. H ( X | Y ) = 1 3 H ( X | Y = 0) + 2 3 H ( X | Y = 1) = 0 . 667 bits = H ( Y | X ) . 3. H ( X, Y ) = 3 × 1 3 log 3 = 1 . 585 bits. 4. H ( Y ) - H ( Y | X ) = 0 . 251 bits. 5. I ( X ; Y ) = H ( Y ) - H ( Y | X ) = 0 . 251 bits. 6. See Figure 1. 4-1

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4-2 Figure 4.1: Venn diagram to illustrate the relationships of entropy and relative entropy H(X|Y) I(X;Y) H(Y|X) H(Y) H(X) Problem 2.3 Minimum entropy. What is the minimum value of H ( p 1 , ..., p n ) = H ( p ) as p ranges over the set of n -dimensional probability vectors? Find all p ’s which achieve this minimum. Solution 2.3 We wish to find all probability vectors p = ( p 1 , p 2 , . . . , p n ) which minimize H ( p ) = - X i p i log p i . Now - p i log p i 0 , with equality iff p i = 0 or 1 . Hence the only possible probability vectors which minimize H ( p ) are those with p i = 1 for some i and p j = 0 , j 6 = i . There are n such vectors, i.e., (1 , 0 , . . . , 0) , (0 , 1 , 0 , . . . , 0) , ..., (0 , . . . , 0 , 1) , and the minimum value of H ( p ) is 0. Problem 2.5 Entropy of functions of a random variable. Let X be a discrete random variable. Show that the entropy of a function of X is less than or equal to the entropy of X by justifying the following steps: H ( X, g ( X )) ( a ) = H ( X ) + H ( g ( X ) | X ) (4.1) ( b ) = H ( X ); (4.2) H ( X, g ( X )) ( c ) = H ( g ( X )) + H ( X | g ( X )) (4.3) ( d ) H ( g ( X )) . (4.4) Thus H ( g ( X )) H ( X ) . Solution 2.5 Entropy of functions of a random variable. 1. H ( X, g ( X )) = H ( X ) + H ( g ( X ) | X ) by the chain rule for entropies. 2. H ( g ( X ) | X ) = 0 since for any particular value of X, g(X) is fixed, and hence H ( g ( X ) | X ) = x p ( x ) H ( g ( X ) | X = x ) = x 0 = 0 . 3. H ( X, g ( X )) = H ( g ( X )) + H ( X | g ( X )) again by the chain rule. 4. H ( X | g ( X )) 0 , with equality iff X is a function of g ( X ) , i.e., g ( . ) is one-to-one. Hence H ( X, g ( X )) H ( g ( X )) . Combining parts (b) and (d), we obtain H ( X ) H ( g ( X )) . Problem 2.6
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## This note was uploaded on 10/12/2009 for the course EE 596A taught by Professor Jeffa.bilmes during the Winter '04 term at Washington University in St. Louis.

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hw1_sol - EE595A Introduction to Information Theory Winter...

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