oldfinal2sol

oldfinal2sol - UCSD ECE153 Handout#39 Prof Young-Han Kim...

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Unformatted text preview: UCSD ECE153 Handout #39 Prof. Young-Han Kim Thursday, June 2, 2011 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 first three flips are heads? (b) What is the probability that the second flip is heads given that the first flip is heads? Solution: (a) We have P { first three flips are heads | P = p } = p 3 . Hence, by the law of total probability, P { first three flips are heads } = integraldisplay 1 p 3 dp = 1 / 4 . (b) Since P { the first flip is heads } = integraldisplay 1 pdp = 1 / 2 and P { first two flips are heads } = integraldisplay 1 p 2 dp = 1 / 3 , we have P { second flip heads | first flip 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) Find E [ X 1- X 2 | Y ]. (b) Find 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 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 ,Z 1 ,Z 2 ,... are i.i.d. ∼ N (0 , 1). (a) Find the distribution on X that makes this a stationary stochastic process. (b) What is the resulting autocorrelation R X ( n )? Solution: (a) Consider X ∼ 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 = parenleftbigg 1 3 parenrightbigg 2 X n- 2 + 1 3 Z n- 2 + Z n- 1 = parenleftbigg 1 3 parenrightbigg n X + n summationdisplay k =1 parenleftbigg 1 3 parenrightbigg n- k Z k- 1 ....
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This note was uploaded on 09/16/2011 for the course ECE ECE153 taught by Professor Younghankim during the Spring '11 term at UCSD.

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oldfinal2sol - UCSD ECE153 Handout#39 Prof Young-Han Kim...

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