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Math Help - Bias of Estimator

  1. #1
    Senior Member Danneedshelp's Avatar
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    Bias of Estimator

    Suppose that Y_{1},Y_{2},...,Y_{n} constitute a random sample from a normal distribution with parameters \mu and \sigma^{2}.

    Q: Show that S=\sqrt{S^{2}} is a biased estimator of \sigma. Moreover, adjust S to form an unbiased estimator for \sigma.

    [Hint: Recall the distribution of \frac{(n-1)S^{2}}{\sigma^{2}} and the result E(Y^{a}) =\frac{\beta^{a}\Gamma(\alpha\\+a)}{\Gamma(\alpha)  }].

    A: Let U=\frac{(n-1)S^{2}}{\sigma^{2}}. Since U~ \chi^{2}(n-1) we have that E(U)=n-1 and V(U)=2(n-1). Furthermore, using the result in the hint and the fact that U~ Gamma(\frac{n-1}{2},2) we have that

    E(\sqrt{U}) =E(U^{\frac{1}{2}})

    =\frac{(2)^{\frac{1}{2}}\Gamma(\frac{(n-1)}{2}+\frac{1}{2})}{\Gamma(\frac{(n-1)}{2})}

    =\frac{\sqrt{2}\Gamma(\frac{n}{2})}{\Gamma(\frac{(  n-1)}{2})}.

    So, \sqrt{\frac{\sigma^{2}}{(n-1)}}\frac{\sqrt{2}\Gamma(\frac{n}{2})}{\Gamma(\fra  c{(n-1)}{2})}

    =\sigma\sqrt{\frac{2}{(n-1)}}\frac{\Gamma(\frac{n}{2})}{\Gamma(\frac{(n-1)}{2})}.

    I am not sure what do do fro the second part of the question.

    Am I doing this correctly?
    Last edited by Danneedshelp; February 10th 2010 at 07:44 PM.
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  2. #2
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    I don't know how much help this will be, but from my stats textbook I found in the section on estimators that:
    E(S)=\sigma\sqrt{\frac{2}{n-1}}\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\neq\sigma

    Which proves that S is a biased estimator. However, note that it is asymptotically unbiased.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by cpbrunner View Post
    I don't know how much help this will be, but from my stats textbook I found in the section on estimators that:
    E(S)=\sigma\sqrt{\frac{2}{n-1}}\frac{\Gamma(n/2)}{\Gamma((n-1)/2)}\neq\sigma

    Which proves that S is a biased estimator. However, note that it is asymptotically unbiased.
    Thanks, I found the algebra error in my derivation. Even so, I am still stuck on the second part. My thinking is I need to multiply by the reciprocal of the first part of the result, but I don't know what to do about getting rid of the gamma stuff.

    Thanks for your help.
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  4. #4
    MHF Contributor matheagle's Avatar
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    USE \Gamma(a)=(a-1)\Gamma(a-1) which is just integration by parts.

    Also use \Gamma(.5)=\sqrt{\pi}

    So \Gamma(2.5)=(1.5)(.5)\sqrt{\pi}
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  5. #5
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    Quote Originally Posted by Danneedshelp View Post
    Thanks, I found the algebra error in my derivation. Even so, I am still stuck on the second part. My thinking is I need to multiply by the reciprocal of the first part of the result, but I don't know what to do about getting rid of the gamma stuff.

    Thanks for your help.
    Yeah, you've got it. Multiply by the reciprocal for an unbiased estimator. The gamma stuff is just a constant as matheagle pointed out. If n is even, \Gamma(n/2) will just be (n/2-1)!. If n is odd, \Gamma(n/2) will be (n/2 - 1)(n/2 - 2)\ldots(0.5)\sqrt{\pi}.
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