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

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Unformatted text preview: 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 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 3. Moving average process. Let Y n = 1 2 Z n − 1 + Z n for n ≥ 1 , where Z ,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 , otherwise =      5 4 , n = m 1 2 , | n − m | = 1 , otherwise 4. Gauss-Markov process. Let X = 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. (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 ....
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This note was uploaded on 10/26/2011 for the course MATH 180C taught by Professor Eggers during the Winter '09 term at Aarhus Universitet.

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

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