oldfinal2sol - UCSD ECE153 Prof. Young-Han Kim Handout #37...

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UCSD ECE153 Handout #37 Prof. Young-Han Kim Thursday, June 3, 2010 Solutions to Final (Spring 2008) 1. Coin with random bias (20 points). You are given a coin but are not told what its bias (probability of heads) is. You are told instead that the bias is the outcome of a random variable P Unif[0 , 1]. Assume P does not change during the sequence of tosses. (a) What is the probability that the Frst three ±ips are heads? (b) What is the probability that the second ±ip is heads given that the Frst ±ip is heads? Solution: (a) We have P { Frst three ±ips are heads | P = p } = p 3 . Hence, by the law of total probability, P { Frst three ±ips are heads } = i 1 0 p 3 dp = 1 / 4 . (b) Since P { the Frst ±ip is heads } = i 1 0 pdp = 1 / 2 and P { Frst two ±ips are heads } = i 1 0 p 2 dp = 1 / 3 , we have P { second ±ip heads | Frst ±ip heads } = 2 / 3 . 2. Estimation (20 points). Let X 1 and X 2 be independent identically distributed random vari- ables. Let Y = X 1 + X 2 . (a) ²ind E [ X 1 - X 2 | Y ]. (b) ²ind the minimum mean squared error estimate of X 1 given an observed value of Y = X 1 + X 2 . (Hint: Consider E [ X 1 + X 2 | X 1 + X 2 ].) Solution: (a) By symmetry, E [ X 1 | Y ] = E [ X 2 | Y ]. Hence, E [ X 1 - X 2 | Y ] = 0. (b) On one hand, we have E [ X 1 + X 2 | X 1 + X 2 ] = X 1 + X 2 = Y . On the other hand, we have E [ X 1 + X 2 | X 1 + X 2 ] = E [ X 1 | Y ] + E [ X 2 | Y ] = 2 E [ X 1 | Y ] . Hence, the MMSE estimate of X 1 given Y is E [ X 1 | Y ] = Y 2 . 1
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3. Stationary process (20 points). Consider the Gaussian autoregressive random process X k +1 = 1 3 X k + Z k , k = 0 , 1 , 2 ,..., where Z 0 ,Z 1 ,Z 2 ,... are i.i.d. N (0 , 1). (a) Find the distribution on X 0 that makes this a stationary stochastic process. (b) What is the resulting autocorrelation R X ( n )? Solution: (a) Consider X 0 N (0 , 9 / 8). It is easy to see that X n N (0 , 9 / 8), which implies that X n is stationary (why?). (b) We have X n = 1 3 X n - 1 + Z n - 1 = p 1 3 P 2 X n - 2 + 1 3 Z n - 2 + Z n - 1 = p 1 3 P n X 0 + n s k =1 p 1 3 P n - k Z k - 1 . Hence, R X ( n ) = EX 0 X n = p 9 8 Pp 1 3 P n for n 0, and in general R X ( n ) = EX 0 X n = p 9 8 Pp 1 3 P | n | .
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oldfinal2sol - UCSD ECE153 Prof. Young-Han Kim Handout #37...

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