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fisol - ECE1502F Information Theory Final Examination...

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ECE1502F — Information Theory Final Examination — Solutions December 11, 2007 1. (a) Let X and Y be independent Bernoulli(1/2) random variables and let Z = X Y , where denotes modulo-two addition. Since X and Y are independent, we have I ( X ; Y ) = 0. Since X is equally likely to be zero or one for each possible value of Z , Z provides no information about X , i.e., I ( X ; Z ) = 0. However, I ( X ; Y, Z ) = 1, since knowledge of both Y and Z determines X uniquely. Thus the answer to the question is no . (b) The same example as in the previous question provides a counterexample, i.e, I ( X ; Y ) = 0, yet I ( X ; Y | Z ) = 1. (c) True . To see this, let X be a postive real-valued random variable over with E [ X ] = m and probability density function p ( x ). and let Y be an exponential random variable with probability density function f ( x ) = (1 /m ) e - x/m . Then 0 D ( p || f ) = 0 p ( x ) ln[ p ( x ) /f ( x )] dx = 0 p ( x ) ln p ( x ) dx + 0 p ( x ) ln(1 /f ( x )) dx = - h ( X ) + 0 p ( x ) ln( me x/m ) dx = - h ( X ) + 1 m 0 xp ( x ) dx + ln( m ) 0 f ( x ) dx = - h ( X ) + 1 + ln( m ) = - h ( X ) + ln( me ) . Thus we find that h ( X ) ln( me ) with equality achieved if and only if p ( x ) = f ( x ) a.e., i.e., if and only if X is an exponential random variable. (d) To design a Huffman code over a quaternary alphabet it is necessary that the number of probability masses be one larger than an integer multiple of 3. Thus, we must add two dummy symbols. The resulting tree is shown in the figure, and corresponds to the code given in the table. 3 1 0 3 2 1 0 2 / 8 4 / 8 0 1 2 0 0 1 / 8 1 / 8 1 / 8 1 / 8 1 / 8 1 / 8 1 / 8 1 / 8 0 1 2 3 4 5 6 7 d d dummy symbols x 0 1 2 3 4 5 6 7 C ( x ) 0 1 20 21 22 23 30 31 (e) The channel is symmetric, and so has capacity C = log 2 (3) - H (1 / 2 , 1 / 4 , 1 / 4) = log 2 (3) - 3 / 2 0 . 085 bits per channel use. 1
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2. (a) The channel is neither symmetric nor weakly symmetric. (b) I ( X ; Y ) = H ( Y ) - H ( Y | X ) = H ( Y ) - H (1 / 2 , 1 / 4 , 1 / 4) = H ( Y ) - 3 / 2. To maximize mutual information, we must maximize H ( Y ) (by making Y as uniform as possible).
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