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Minimum MSE estimation derivation

  1. Aug 4, 2011 #1
    Hello,

    Would anyone be-able to recommend a good, easy to read article which outlines MMSE and its derivation. Specifically I am having trouble finding this term


    [tex]
    + \int x'xP(x|y)dx-\left \| \int xP(x|y)dx \right \|^2
    [/tex]

    from

    [tex]
    E({\left | \left | X-z \right | \right |}^2|Y=y)
    =\int (x-z)'(x-z)P(x|y)dx\\
    =[z'-\int x'P(x|y)dx][z-\int xP(x|y)dx] + \int x'xP(x|y)dx-\left \| \int xP(x|y)dx \right \|^2
    [/tex]

    Thank you
     
  2. jcsd
  3. Aug 4, 2011 #2
    Shouldnt the term just be zero - I cant understand it's presence - are there any conditions in which it is not zero??
     
  4. Aug 5, 2011 #3
    For anyone who is interested - the last term

    [tex]
    + \int x'xP(x|y)dx-\left \| \int xP(x|y)dx \right \|^2
    [/tex]

    is necessary to account for the difference between E(x^2) and [E(x)]^2. When Z = E[x|Y=y] the term

    [tex]
    E({\left | \left | X-z \right | \right |}^2|Y=y)
    [/tex]

    Is a minimum and reduces to

    [tex]
    + \int x'xP(x|y)dx-\left \| \int xP(x|y)dx \right \|^2 =
    [/tex]

    Then

    [tex]
    E({\left | \left | X \right | \right |}^2|Y=y)-E(X|Y=y)^2\\
    =E({\left | \left | X \right | \right |}^2|Y=y)-{\left | \left | \hat{X} \right | \right |}^2
    [/tex]

    which is the average mean square error
     
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