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Unformatted text preview: UCSD ECE 153 Handout #41 Prof. YoungHan Kim Thursday, June 2, 2011 Solutions to Homework Set #7 (Prepared by TA Yu Xiang) 1. Symmetric random walk. Let X n be a random walk defined by 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) Approximate P { 10 X 100 10 } using the central limit theorem. (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 E( Z j ) = 0 and E( Z 2 j ) = 1, by the central limit theorem, P { 10 X 100 10 } = P braceleftBigg 1 parenleftBigg 1 100 100 summationdisplay i =1 Z i parenrightBigg 1 bracerightBigg 1 2 Q (1) = 2(1) 1 . 682 . (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 2. Discretetime Wiener process. Let Z n , n 0 be a discrete time white Gaussian noise (WGN) process, i.e., Z 1 ,Z 2 ,... are i.i.d. N (0 , 1). Define the process X n , n 1 as X = 0, and X n = X n 1 + Z n for n 1. (a) Is X n an independent increment process? Justify your answer. (b) Is X n a Markov process? Justify your answer. (c) Is X n a Gaussian process? Justify your answer. (d) Find the mean and autocorrelation functions of X n . (e) Specify the first and second order pdfs of X n . (f) Specify the joint pdf of X 1 ,X 2 , and X 3 . (g) Find E ( X 20  X 1 ,X 2 ,...,X 10 ). Solution: (a) Yes. The increments X n 1 , X n 2 X n 1 ,...,X n k 1 X n k are sums of different Z i random variables, and the Z i are IID. (b) Yes. Since the process has independent increments, it is Markov. (c) Yes. Any set of samples of X n , n 1 are obtained by a linear transformation of IID N (0 , 1) random variables and therefore all n th order distributions of X n are jointly Gaussian (it is not sufficient to show that the random variable X n is Gaussian for each n ). (d) We have E[ X n ] = E bracketleftBigg n summationdisplay i =0 Z i bracketrightBigg = n summationdisplay i =0 E[ Z i ] = n summationdisplay i =0 0 = 0 , R X ( n 1 ,n 2 ) = E[ X n 1 X n 2 ] = E n 1 summationdisplay i =1 Z i n 2 summationdisplay j =1 Z j = n 1 summationdisplay i =1 n 2 summationdisplay j =1 E[ Z i Z j ] = min( n 1 ,n 2 ) summationdisplay i =1 E( Z 2 i ) (IID) = min( n 1 ,n 2 ) . (e) As shown above, X n is Gaussian with mean zero and variance Var( X n ) = E[ X 2 n ] E 2 [ X n ] = R X ( n,n ) = n. Thus, X n N (0 ,n ). 2 (f) X n , n 1 is a zero mean Gaussian random process. Thus X 1 X 2 X 3 N E [ X 1 ] E [ X 2 ] E [ X 3 ] , R X (1 , 1) R X (1 , 2) R X (1 , 3) R X (2 , 1) R X (2 , 2) R X (2 , 3) R X (3 , 1) R X (3 , 2) R X (3 , 3) Substituting, we get X 1 X 2 X...
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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.
 Spring '11
 YoungHanKim

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