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Math Help - Expected Values

  1. #1
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    Expected Values

    Can anybody tell me why
    E(f hat (x) - f(x))^2
    = E(f hat (x) - f(x))^2 + var f hat (x)
    Thanks.
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  2. #2
    MHF Contributor matheagle's Avatar
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    I think your missing E\hat f somewhere.
    What you have here, is variance of something equal to zero, hence it's a constant.
    Are you trying to show that the MSE = Bias squared + variance?
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  3. #3
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    Thats exactly what im trying to do.
    I think the next step is split it into
    E(hat f -E(hat f) + E(hat f) -f)^2
    a = hat f -E(hat f)
    b = E(hat f) -f
    and then expand (a+b)^2
    but after that Im getting nowhere.
    Thanks
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by markrvr View Post
    Can anybody tell me why
    E(f hat (x) - f(x))^2
    = E(f hat (x) - f(x))^2 + var f hat (x)
    Thanks.

    you didn't read what I wrote.
    This ...
    E(f hat (x) - f(x))^2 = E(f hat (x) - f(x))^2 + var f hat (x)
    implies that var f hat (x)=0.
    which mean f hat (x) is constant

    Please write what you really want to prove.
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  5. #5
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    I am trying to show that the MSE = Bias squared + variance
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by markrvr View Post
    I am trying to show that the MSE = Bias squared + variance
    I guessed that two days ago, but what you wrote does not make sense.
    You're missing the paramter \theta.

    After you correctly state MSE, all you do is add and subtract
    E(\theta) inside of the MSE and you expand.
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  7. #7
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    Yea I did it like this
    E(hat f -E(hat f) + E(hat f) -f)^2
    and expanded but I'm having trouble showing that the middle term of the expanded square equals 0.
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  8. #8
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by markrvr View Post
    Yea I did it like this
    E(hat f -E(hat f) + E(hat f) -f)^2
    and expanded but I'm having trouble showing that the middle term of the expanded square equals 0.
    BECAUSE that is NOT MSE
    LOOK at your book/notes.

    E(f hat (x) - THETA)^2
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  9. #9
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    MSE was given to me as E(f hat (x) - f(x))^2 so your theta is f(x) here. This is a measure of how well f hat (x), which is an estimate of f(x), fits f(x) the function we are trying to estimate.
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  10. #10
    MHF Contributor matheagle's Avatar
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    BUT the problem here is understanding what a statistic is and what a parameter is.
    A parameter is NOT a function of the data, X. A statistic is a function of X.
    http://www.mathhelpforum.com/math-he...ror-proof.html
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  11. #11
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    No thats not the problem. It can still be shown for f(x) as f hat (x) is based on the data set X. The proof is used in kernel estimation of a probability density function.
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