Results 1 to 11 of 11

Thread: Expected Values

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    34

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    I think your missing $\displaystyle E\hat f$ 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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2009
    Posts
    34
    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2009
    Posts
    34
    I am trying to show that the MSE = Bias squared + variance
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    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 $\displaystyle \theta$.

    After you correctly state MSE, all you do is add and subtract
    $\displaystyle E(\theta)$ inside of the MSE and you expand.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2009
    Posts
    34
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    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
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Mar 2009
    Posts
    34
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    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
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Mar 2009
    Posts
    34
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Joint Density expected values
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: Sep 4th 2011, 11:32 PM
  2. Expected values and Variances
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Jan 27th 2010, 05:18 AM
  3. expected values
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: Oct 8th 2009, 12:20 AM
  4. Expected values and variance
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 21st 2009, 04:21 PM
  5. Expected values
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: Mar 12th 2006, 02:36 AM

Search Tags


/mathhelpforum @mathhelpforum