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Unformatted text preview: EE5620 Stochastic Processes 1 EE5620  Homework #4 Solutions 1. The linear minimum meansquared error (LMMSE) estimator has the optimal worstcase performance. To see this, suppose the given first and second order moments of x and y are m x , m y , K x , K y and K xy . Then, e (ˆ x LMMSE ( y )) = max f x , y E £  x ˆ x LMMSE ( y  2 / = max f x , y E • fl fl fl fl fl fl x ‡ m x + K xy K 1 y ‡ y m y ··fl fl fl fl fl fl 2 ‚ = max f x , y K x K xy K 1 y K yx = K x K xy K 1 y K yx , where the maximization does not take any effect since the first 2 moments are given. Now let φ be any other estimator. Let g ( x , y ) be the jointly Gaussian density with first and second order moments m x , m y , K x , K y and K xy . Then, we have e ( φ ( y )) = max f x , y E £  x φ ( y )  2 / ≥ E • fl fl fl fl fl fl x φ ( y ) fl fl fl fl fl fl 2 ‚ fl fl fl fl fl g ( x , y ) ≥ E • fl fl fl fl fl fl x ˆ x MMSE ( y ) fl fl fl fl fl fl 2 ‚ fl fl fl fl fl g ( x , y ) = max f x , y E • fl fl fl fl fl fl x ˆ x LMMSE ( y ) fl fl fl fl fl fl 2 ‚ = e (ˆ x LMMSE ( y )) . 2. (a) From Lecture 7, we know that the LMMSE of U can be obtained as ˆ U LMMSE = m U + K UY K 1 Y ( Y m Y ) , where m U and m Y are the mean of U and Y , respectively, K UY is the cross covariance matrix of U and Y , and K Y is the covariance matrix of Y . The corresponding MSE of the LMMSE of U given Y is MSE = tr ( K U K UY K 1 Y K Y U ) , (1) EE5620 Stochastic Processes 2 where K U is the covariance matrix of U . Note that we can remove the “trace” operation since we are dealing with ”scalar,” instead of vector in this problem. To evaluate the MSE, we need the following: m U = E [ U ] = E [ X + αS ] = μ, K U = Var( U ) = E [( X μ + αS ) 2 ] = σ 2 X + α 2 σ 2 S , (X and S uncorrelated) m Y = E [ Y ] = E [ X ] + E [ S ] + E [ Z ] = μ, K Y = E [( Y μ ) 2 ] = σ 2 X + σ...
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This note was uploaded on 11/28/2010 for the course EE 301 taught by Professor Gfung during the Winter '10 term at National Chiao Tung University.
 Winter '10
 GFung

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