{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw4sol - EE5620 Stochastic Processes 1 EE5620 Homework#4...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
EE5620 Stochastic Processes 1 EE5620 - Homework #4 Solutions 1. The linear minimum mean-squared error (LMMSE) estimator has the optimal worst-case 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 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 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)
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 2 S + σ 2 N K UY = E [( U - μ )( Y - μ )] = E [( X - μ + αS )( X - μ + S + Z )] = σ 2 X + ασ 2 S Substituting all these into (1) yields MSE = ( σ
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern