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# ex1_sol - Introduction to Information Theory(67548...

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Introduction to Information Theory (67548) December 21, 2008 Assignment 1: Solution Lecturer: Prof. Michael Werman Due: Important note: All logarithms are in base 2 unless speciﬁed otherwise. Problem 1 Joint Distribution and Entropy 1. H ( X,Y ) = - X x,y p ( x,y ) log( p ( x,y )) = 3 8 log(8) + 2 16 log(16) + 1 2 log(2) = 2 . 125 bits . H ( X | Y ) = - X x,y p ( x,y ) log( p ( x | y )) = 0 . 625 bits . H ( Y | X = c ) = - X y p ( c,y ) log( p ( y | c )) = 0 bits . It is easy to verify that the distribution of X is p ( a ) = 1 / 4 ,p ( b ) = 11 / 16 ,p ( c ) = 1 / 16, and therefore H ( X ) 1 . 12 bits. We have already calculated that H ( X | Y ) = 0 . 625 bits. As a result, I ( X ; Y ) = H ( X ) - H ( X | Y ) 1 . 12 - 0 . 625 = 0 . 495 bits . 2. For any discrete random variable (or a ﬁnite collection of discrete random variables), the distri- bution which maximizes entropy is the uniform distribution. The distribution which maximizes H ( X,Y ) is therefore p ( x,y ) = 1 / 9 for any x,y ∈ { a,b,c } , and the entropy if log(9) 3 . 17 bits. 3. It is easy to verify that if H ( X ) = 0, then X must be essentially constant (equal to a single value with probability 1). Let us assume w.l.o.g that X = a with probability 1. Under this constraint, the distribution which maximizes H ( X,Y ) is the uniform distribution for Y : namely, p ( a,y ) = 1 / 3 for any y ∈ { a,b,c } and 0 otherwise. In this case, H ( X,Y ) = log(3) 1 . 58 bits. Not surprisingly, notice that this is a lower entropy that in the previous question, where X was not constrained to be constant. 4. As explained in the answer to the previous question, H ( X ) = H ( Y ) = 0 implies that both X and Y receive the same value with probability 1, and thus their joint entropy H ( X,Y ) is 0 bits. Problem 2 Counterfeit Coins

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ex1_sol - Introduction to Information Theory(67548...

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