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# hw2sol - EE 376A Prof T Weissman Information Theory...

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EE 376A Information Theory Prof. T. Weissman Thursday, January 21, 2010 Homework Set #2 (Due: Thursday, January 28, 2010) 1. Prove that (a) Data processing decreases entropy: If Y = f ( X ) then H ( Y ) H ( X ). [ Hint: expand H ( f ( X ) , X ) in two different ways.] (b) Data processing on side information increases entropy: If Y = f ( X ) then H ( Z | X ) H ( Z | Y ). (c) Assume Y and Z are conditionally independent given X , denoted as Y X Z . In other words, P { Y = y | X = x, Z = z } = P { Y = y | X = x } for all x ∈ X , y Y , z ∈ Z . Prove that H ( Z | X ) H ( Z | Y ). Solution: (a) H ( X, f ( X )) = H ( X ) + H ( f ( X ) | X ) = H ( X ) H ( X, f ( X )) = H ( f ( X )) + H ( X | f ( X )) H ( f ( X ) | X ) = 0 since for any particular value of X , f ( X ) is fixed, and hence H ( f ( X ) | X ) = x p ( x ) H ( f ( X ) | X = x ) = x 0 = 0. Since H ( X | f ( X )) 0, we must have H ( f ( X )) H ( X ). (b) First, let us show that conditioning reduces conditional entropy: 0 I ( X 1 ; X 2 | X 3 ) = H ( X 1 | X 3 ) H ( X 1 | X 2 , X 3 ) Thus H ( X 1 | X 3 ) H ( X 1 | X 2 , X 3 ). For part (b) H ( Z | X ) = H ( Z | X, Y ) + I ( Z ; Y | X ) ( i ) = H ( Z | X, Y ) ( ii ) H ( Z | Y ) where (i) comes from the fact Y = f ( X ) and (ii) comes from conditioning reduces conditional entropy. (c) Y X Z implies that p ( y, z | x ) = p ( y | x ) p ( z | x ). I ( Z ; Y | X ) = summationdisplay x,y,z p ( x, y, z ) log p ( y, z | x ) p ( y | x ) p ( z | x ) = 0 . Now we can follow the steps in (b) to complete the proof. 1

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2. Entropy of a disjoint mixture. Let X 1 and X 2 be discrete random variables drawn according to probability mass functions p 1 ( · ) and p 2 ( · ) over the respective alphabets X 1 = { 1 , 2 , . . . , m } and X 2 = { m + 1 , . . . , n } . Let Θ be the result of a biased coin flip, i.e., P { Θ = 1 } = α and P { Θ = 0 } = 1 α . X 1 , X 2 and Θ are mutually independent. X = braceleftbigg X 1 , if Θ = 1 , X 2 , if Θ = 0 . (a) Find H ( X ) in terms of H ( X 1 ) and H ( X 2 ) and α. (b) Maximize over α to show that 2 H ( X ) 2 H ( X 1 ) + 2 H ( X 2 ) and interpret using the notion that 2 H ( X ) is the effective alphabet size. Solution: (a) We can do this problem by writing down the definition of entropy and expanding the various terms. Instead, we will use the algebra of entropies for a simpler proof. Since X 1 and X 2 have disjoint support sets, Θ is a function of X : Θ = f ( X ) = braceleftbigg 1 when X ∈ X 1 0 when X ∈ X 2 Then, we have H ( X ) = H ( X, f ( X )) = H (Θ) + H ( X | Θ) = H (Θ) + p (Θ = 1) H ( X | Θ = 1) + p (Θ = 0) H ( X | Θ = 0) = H ( α ) + αH ( X 1 ) + (1 α ) H ( X 2 ) where H ( α ) = α log α (1 α ) log(1 α ). (b). Solve: dH ( X ) α = H ( X 1 ) H ( X 2 ) + log 2 1 α α = 0 . We get α = 2 H ( X 1 ) 2 H ( X 1 ) +2 H ( X 2 ) defines α * . Note that H ( X ) is a concave function in α . Thus α * is the maximizer. H ( X ) | α = α * = 2 H ( X 1 ) + 2 H ( X 2 ) . 3. Example of joint entropy. Let p ( x, y ) be given by a64 a64 a64 X Y 0 1 0 1 3 1 3 1 0 1 3 2
Find (a) H ( X ) , H ( Y ) . (b) H ( X | Y ) , H ( Y | X ) . (c) H ( X, Y ) . (d) H ( Y ) H ( Y | X ) .

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