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

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

View Full Document Right Arrow Icon
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
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
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 | .
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 ]}

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online