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hw6sol - UCSD ECE 153 Prof Young-Han Kim Handout#24...

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UCSD ECE 153 Handout #24 Prof. Young-Han Kim Thursday, December 4, 2008 Solutions to Homework Set #6 (Prepared by TA Halyun Jeong) 1. Read Sections 7.1–7.4, 9.1–9.6, 10.1–10.2, 10.4 in the text. Try to work on all examples. 2. Symmetric random walk. Let X n be a random walk defined as X 0 = 0 X n = n summationdisplay i =1 Z i , where Z 1 , Z 2 , . . . are i.i.d. with P( Z 1 = 1) = P( Z 1 = 1) = 1 2 . (a) Find P { X 10 = 10 } . (b) Find P { max 1 i< 20 X i = 10 | X 20 = 0 } . (c) Find P { X n = k } . Solution: (a) Since the event { X 10 = 10 } is equivalent to { Z 1 = · · · = Z 10 = 1 } , we have P { X 10 = 10 } = 2 10 . (b) Since the event { max X i = 10 , X 20 = 0 } is equivalent to { Z 1 = · · · = Z 10 = 1 , Z 11 = · · · = Z 20 = 1 } , we have P { max 1 i< 20 X i = 10 | X 20 = 0 } = 2 20 P ( X 20 = 0) = 1 ( 20 10 ) . (c) As shown in class, P { X n = k } = P { ( n + k ) / 2 heads in n independent coin tosses } = parenleftbigg n n + k 2 parenrightbigg 2 n for n k n with n + k even. 1
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3. Moving average process. Let Y n = 1 2 Z n 1 + Z n for n 1 , where Z 0 , Z 1 , Z 2 , . . . are i.i.d. ∼ N (0 , 1). Find the mean and autocorrelation function of Y n . Solution: EY n = 1 2 EZ n 1 + EZ n = 0 . R Y ( m, n ) = E ( Y m Y n ) = E bracketleftbiggparenleftbigg 1 2 Z n 1 + Z n parenrightbiggparenleftbigg 1 2 Z m 1 + Z m parenrightbiggbracketrightbigg = 1 2 E [ Z 2 n 1 ] , n m = 1 1 4 E [ Z 2 n 1 ] + E [ Z 2 n ] , n = m 1 2 E [ Z 2 n ] , m n = 1 0 , otherwise = 5 4 , n = m 1 2 , | n m | = 1 0 , otherwise 4. Gauss-Markov process. Let X 0 = 0 and X n = 1 2 X n 1 + Z n for n 1, where Z 1 , Z 2 , . . . are i.i.d. ∼ N (0 , 1). Find the mean and autocorrelation function of X n . Solution: It is easy to show that E ( X n ) = E ( X n 1 ) = · · · = E ( X 0 ) = 0. (Why?) For the autocorrelation function, first note that E ( X 2 n ) = 1 4 E ( X 2 n 1 ) + E ( Z 2 n ) + E ( X n 1 Z n ) = 1 4 E ( X 2 n 1 ) + 1 . Therefore, EX 2 1 = 1 4 · 0 + 1 = 1 . Similarly, EX 2 2 = 1 4 E ( X 2 1 ) + 1 = 1 4 · 1 + 1 = 5 4 . In general, EX 2 n = n 1 i =0 1 4 i = 4 3 (1 ( 1 4 ) n +1 ) . Moreover, we can write X n = 1 2 k X n k + 1 2 k 1 Z n k +1 + 1 2 k 2 Z n k +2 + · · · + Z n , and thus R X ( n, n k ) = EX n X n k = 1 2 k EX 2 n k = 4 3 parenleftBig 1 parenleftBig 1 4 parenrightBig n k +1 parenrightBig · 1 2 k for all n and all k 0.
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