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Math Help - Risk/Loss functions

  1. #1
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    Risk/Loss functions

    Let X1,...,Xn denote a random sample from a N(\mu , \sigma) distribution. Let Y = \Sigma \frac{(X_i - \overline{X})^2}{n}and let L(\theta , \delta (y)) = [\theta - \delta(y)]^2 and \delta (y) = by where b does not depend on y.

    Show that R(\theta^2, \delta) = \frac{\theta^2}{n^2}[(n^2 - 1)b^2 -2n(n-1)b + n^2]

    So:

    R(\theta^2, \delta) = E(L(\theta , \delta (y))) = E([\theta - \delta(y)]^2) = E((\theta -by)^2) = E(\theta^2 -2 \theta by +b^2y^2)

    Kinda stuck here. Not sure what the expectation of Y = \Sigma \frac{(X_i - \overline{X})^2}{n} is.

    Any help would be appreciated.
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  2. #2
    MHF Contributor matheagle's Avatar
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    first of all ....

    R(\theta^2, \delta) = E(L(\theta^2 , \delta (y))) = E([\theta^2 - \delta(y)]^2) = E((\theta^2 -by)^2)

    As for Y = \frac{\Sigma (X_i - \overline{X})^2}{n} we have....

    (n-1)S^2\sim\sigma^2 \chi^2_{n-1} and (n-1)S^2=\Sigma (X_i - \overline{X})^2

    So Y = \frac{\Sigma(X_i - \overline{X})^2}{n}

    = \left({\sigma^2\over n}\right) \frac{\Sigma(X_i - \overline{X})^2}{\sigma^2} \sim  {\sigma^2\over n} \chi^2_{n-1}

    So, if you want E(Y)={\sigma^2\over n} (n-1)
    Last edited by matheagle; January 10th 2010 at 10:03 PM.
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  3. #3
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    Thanks for the help, but I already solved the problem by doing the following:

    \frac{n-1}{n} \frac{1}{n-1} \Sigma (X_i - \overline{X})^2 = \frac{n-1}{n}s^2 and I know what the expecation and variance of the sample variance (sample from normal distribution) is.

    But thanks for your help none the less
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  4. #4
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by statmajor View Post
    Thanks for the help, but I already solved the problem by doing the following:

    \frac{n-1}{n} \frac{1}{n-1} \Sigma (X_i - \overline{X})^2 = \frac{n-1}{n}s^2 and I know what the expecation and variance of the sample variance (sample from normal distribution) is.

    But thanks for your help none the less

    that's the same thing
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