solns5 - ECE 1502 Information Theory Problem Set 5...

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Unformatted text preview: ECE 1502 Information Theory Problem Set 5 solutions 1 April 11, 2006 9.1 Differential Entropy. (a) Exponential distribution. h ( f ) =- Z e- x [ln - x ] dx (1) =- ln + 1 nats. (2) = log e bits. (3) (b) Laplace density. h ( f ) =- Z - 1 2 e- | x | [ln 1 2 + ln - | x | ] dx (4) =- ln 1 2- ln + 1 (5) = ln 2 e nats. (6) = log 2 e bits. (7) (c) Sum of two normal distributions. The sum of two normal random variables is also normal, so applying the result derived the class for the normal distribution, since X 1 + X 2 N ( 1 + 2 , 2 1 + 2 2 ), h ( f ) = 1 2 log 2 e ( 2 1 + 2 2 ) bits. (8) 9.3 Mutual information for correlated normals. X Y N 2 , 2 2 2 2 (9) Using the expression for the entropy of a multivariate normal derived in class h ( X,Y ) = 1 2 log(2 e ) 2 | K | = 1 2 log(2 e ) 2 4 (1- 2 ) . (10) Since X and Y are individually normal with variance 2 , h ( X ) = h ( Y ) = 1 2 log 2 e 2 . (11) Hence I ( X ; Y ) = h ( X ) + h ( Y )- h ( X,Y ) =- 1 2 log(1- 2 ) . (12) 1 Solutions to problems from the text are supplied courtesy of Joy A. Thomas. 1 (a) = 1. In this case, X = Y , and knowing X implies perfect knowledge about Y . Hence the mutual information is infinite, which agrees with the formula. (b) = 0. In this case, X and Y are independent, and hence I ( X ; Y ) = 0, which agrees with the formula. (c) =- 1. In this case, X =- Y , and again the mutual information is infinite as in the case when = 1. 9.6 Scaling. Let Y = A X . Then the density of Y is g ( y ) = 1 | A | f ( A- 1 y ) . (13) Hence h ( A X ) =- Z g ( y ) ln g ( y ) d y (14) =- Z 1 | A | f ( A- 1 y ) ln f ( A- 1 y )- log | A | d y (15) =- Z 1 | A | f ( x ) [ln f ( x )- log | A | ] | A | d x (16) = h ( X ) + log | A | . (17) 10.3 The two look Gaussian channel.The two look Gaussian channel....
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solns5 - ECE 1502 Information Theory Problem Set 5...

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