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Unformatted text preview: EE562A Homework 3 Solutions Prepared by Hooman ShiraniMehr Problem 1: Suppose X is a meanzero random vector with covariance matrix K X = 1 4 3 4 20 22 3 22 36 Find a transformation A such that Y = AX has covariance matrix K Y = I . Solution H = h 11 h 21 h 22 h 31 h 32 h 33 ,H = h 11 h 21 h 31 h 22 h 32 h 33 k 11 = 1 = h 2 11 h 11 = 1 k 12 = 4 = h 11 h 21 h 21 = 4 k 13 = 3 = h 11 h 31 h 31 = 3 k 22 = 20 = h 2 21 + h 2 22 h 22 = 2 k 23 = 22 = h 21 h 31 + h 22 h 32 h 32 = 5 k 33 = 36 = h 2 31 + h 2 32 + h 2 33 h 32 = 2 H = 1 4 2 3 5 2 Problem 2: Let Z ( n ) be an i.i.d. Bernoulli sequence where P ( Z ( n ) = 1) = p, P ( Z ( n ) = 1) = q = 1 p. Let X ( n ) = n summationdisplay k =0 Z ( k ) where we take Z (0) = 0 = X (0) . Then X ( n ) is a discrete random walk. Find R X ( n,m ) for this random walk. Solution We can first assume n < m , then: R X ( n,m ) = E [ X ( n ) X ( m )] = E [ n summationdisplay k =0 Z ( k ) m summationdisplay k =0 Z ( k )] = n summationdisplay k =0 E [ Z ( k ) 2 ] + n summationdisplay k =0 m summationdisplay k =0 ,k negationslash = k E [ Z ( k ) Z ( k )] = n. 1 + nm. (2 p 1) 2 = n (1 + m (2 p 1) 2 ) if m < n , we can simply switch m and n and the answer become m (1 + n (2 p 1) 2 ) . Problem 3: Let X n , n 1 , denote a sequence of independent and identically distributed zeromean unitvariance Gaussian random variables. Define for n 2 Y n = 1 2 ( X n + X n 1 ) . a. Does this sequence converge in the mean square sense and, if so, to what limit? b. Does this sequence converge in distribution and, if so, to what distribution? Solution Since X n are i.i.d random variable, we immediately have: E [ Y n ] = 1 2 E [ X n + X n 1 ] = 0 E [ Y 2 n ] = 2 Y n = 1 4 E [ X 2 n + X 2 n 1 ] = 1 2 a) Since we dont know what random variable Y n converges to(if exists). We can use the Cauchys criterion: If Y n converges to some random variable Y , the lim n,m E [  Y n Y m  2 ] = 0 . And since Y n s are real in this problem, we have: lim n,m E [  Y n Y m  2 ] = E [ Y 2 n ] + E [ Y 2 m ] 2 E [ Y n Y m ] If we let m > n + 1 (we can always assume this), then E [ Y n Y m ] = E [ Y n ] E [ Y m ] = 0 . Then: lim n,m E [  Y n Y m  2 ] = E [ Y 2 n ] + E [ Y 2 m ] = 1 negationslash = 0 So Y n doesnt converge in mean square sense....
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 Summer '07
 ToddBrun

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