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# hw1sol - EE 376A/Stat 376A Handout#7 Information Theory...

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Unformatted text preview: EE 376A/Stat 376A Handout #7 Information Theory Thursday, January 20, 2011 Prof. T. Cover Solutions to Homework Set #1 1. Entropy of Hamming Code. Consider information bits X 1 ,X 2 ,X 3 ,X 4 ∈ { , 1 } chosen at random, and check bits X 5 ,X 6 ,X 7 chosen to make the parity of the circles even. X 4 X 2 X 6 X 3 X 7 X 5 X 1 Thus, for example, 1 1 1 becomes 1 1 1 1 That is, 1011 becomes 1011010. 1 (a) What is the entropy of H ( X 1 ,X 2 ,...,X 7 )? Now we make an error (or not) in one of the bits (or none). Let Y = X ⊕ e , where e is equally likely to be (1 , ,..., 0) , (0 , 1 , ,..., 0) ,..., (0 , ,..., , 1), or (0 , ,..., 0), and e is independent of X . (b) What is the entropy of Y ? (c) What is H ( X | Y )? (d) What is I ( X ; Y )? Solution: Entropy of Hamming Code. (a) By the chain rule, H ( X 1 ,X 2 ,...,X 7 ) = H ( X 1 ,X 2 ,X 3 ,X 4 ) + H ( X 5 ,X 6 ,X 7 | X 1 ,X 2 ,X 3 ,X 4 ) . Since X 5 ,X 6 ,X 7 are all deterministic functions of X 1 ,X 2 ,X 3 ,X 4 , we have H ( X 5 ,X 6 ,X 7 | X 1 ,X 2 ,X 3 ,X 4 ) = 0 . And since X 1 ,X 2 ,X 3 ,X 4 are independent Bernoulli(1 / 2) random variables, H ( X 1 ,X 2 ,...,X 7 ) = H ( X 1 ) + H ( X 2 ) + H ( X 3 ) + H ( X 4 ) = 4 . (b) We will expand H ( X ⊕ e , X ) in two different ways, using the chain rule. On one hand, we can write H ( X ⊕ e , X ) = H ( X ⊕ e ) + H ( X | X ⊕ e ) = H ( X ⊕ e ) . In the last step, H ( X | X ⊕ e ) = 0 because X is a deterministic function of X ⊕ e , since the (7,4) Hamming code can correctly decode X when there is at most one error. (You can check this by experimenting with all possible error patterns satisfying this constraint.) On the other hand, we can also expand H ( X ⊕ e , X ) as follows: H ( X ⊕ e , X ) = H ( X ) + H ( X ⊕ e | X ) = H ( X ) + H ( X ⊕ e ⊕ X | X ) = H ( X ) + H ( e | X ) = H ( X ) + H ( e ) = 4 + H ( e ) = 4 + log 2 8 = 7 . 2 The second equality follows since XORing with X is a one-to-one function. The third equality follows from the well-known property of XOR that y ⊕ y = 0. The fourth equality follows since the error vector e is independent of X . The fifth equality follows since from part (a), we know that H ( X ) = 4. The sixth equality follows since e is uniformly distributed over eight possible values: either there is an error in one of seven positions, or no error at all. Equating our two different expansions for H ( X ⊕ e , X ), we have H ( X ⊕ e , X ) = H ( X ⊕ e ) = 7 . The entropy of Y = X ⊕ e is 7 bits. (c) As mentioned before, X is a deterministic function of X ⊕ e , since the (7,4) Ham- ming code can correctly decode X when there is at most one error. So H ( X | Y ) = 0. (d) I ( X ; Y ) = H ( X )- H ( X | Y ) = H ( X ) = 4 ....
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hw1sol - EE 376A/Stat 376A Handout#7 Information Theory...

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