{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solns00 - University of Toronto November 8 2000 Department...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
University of Toronto Department of Electrical November 8, 2000 & Computer Engineering ECE1502S — Information Theory Midterm Test Solutions 1. ( Matching Distributions ) (a) Call a particular ordering of Q optimal if D ( P || Q ) is minimized. Suppose an optimal ordering exists in which i < j , but q i > q j . Let Q 0 be the distribution obtained by swapping the i th and j th probability masses. Then D ( P || Q ) - D ( P || Q 0 ) = p i log p i q i + p j log p j q j - p i log p i q j - p j log p j q i = p i log q j + p j log q i - p i log q i - p j log q j = ( p i - p j ) | {z } 0 (log q j - log q i ) | {z } < 0 0 , with equality if and only if p i = p j . We see that, in general, swapping the i th and j th probability masses reduces the relative entropy, so Q can be optimal in this situation only if p i = p j . But if p i = p j then swapping q i and q j does not affect the relative entropy. Thus sorting the probabilities yields an optimal ordering. (b) Consider now D ( Q || P ) and assume p 1 > 0. Again suppose that an optimal ordering exists in which i < j , but q i > q j . Let Q 0 be the distribution obtained by swapping the i th and j th probability masses. Then D ( Q || P ) - D ( Q 0 || P ) = q i log q i p i + q j log q j p j - q j log q j p i - q i log q i p j = q i log p j - q i log p i - q j log p j + q j log p i = ( q i - q j ) | {z } > 0 (log p j - log p i ) | {z } 0 = 0 with equality if and only if p i = p j . By the same argument as above, sorting the probabilities yields an optimal ordering. (c) If Q has one mass equal to zero, then by the result of (b), we can set q 1 = 0. We wish to select q 2 , q 3 , . . . , q m so that D ( Q || P ) is minimized. Setting up the Lagrangian L ( q 2 , . . . , q m , λ ) = m X i =1 q i ln( q i /p i ) + λ ( m X i =1 q i - 1) , differentiating with respect to q i ( i > 1) and setting the result to zero, we find that ln( q i /p i ) + 1 + λ = 0 , i.e., q i /p i is a constant, independent of i . The constant is chosen to make m i =2 q i = 1. We find that q i = ( 0 if i = 0; p i 1 - p 1 if 2 i m 1
Background image of page 1

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

View Full Document Right Arrow Icon
2. ( Huffman Coding with Costs ) The Huffman procedure minimizes p i l i , for a set of “weights” { p i } that sum to unity. To show that this procedure works for any arbitrary set of weights, simply divide by the sum of the weights. (a) Specifically, for a given set of non-negative weights W = { w 1 , w 2 , . . . , w m } , let Z ( W ) = m X i =1 w i .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}