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Math Help - Unbiased estimator of the exponential distirbution

  1. #1
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    Unbiased estimator of the exponential distirbution

    Let f(x; \theta) = \theta e^{- \theta x} and Y = \Sigma X_i. What function of Y is an unbiased estimator of theta?

    So I know the mean of the exponential distirbutions is \frac{1}{\theta}

    E(Y) = E(\Sigma X_i) = \frac{n}{\theta}

    therefore E(nY^{-1}) = \theta .

    Is this correct? Something about it is bothing me, or am I just being paranoid?
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  2. #2
    MHF Contributor matheagle's Avatar
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    well, you should be a bit paranoid

    If E(W)=p then it does not follow that E(W^{-1})=1/p

    So, calculate E(Y^{-1}) and see what you get.
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  3. #3
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    Am I allowed to use \frac{\theta^2 Y}{n}
    Last edited by statmajor; January 27th 2010 at 08:01 AM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    NOPE
    you cannot estimate an unknown parameter with itself
    Otherwise theta would always be the unbiased estimator of theta
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  5. #5
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    So what shall I do?
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  6. #6
    Moo
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    Hello,

    Do as you've been told above : find E[Y^{-1}]

    In order to do that, recall that E[h(Y)]=\int h(y) g(y) ~dy, where g is the pdf of Y.
    And also see that Y is the sum of n independent rv following an exponential distribution with parameter \theta
    So its pdf is the one of a gamma distribution (n,1/\theta) (see here : Exponential distribution - Wikipedia, the free encyclopedia)
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  7. #7
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    Now I get it. Thank you both for your help.
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  8. #8
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    Just to make sure I didnt make any more mistakes:

    \int^{\infty}_o\frac{\theta^n}{\Gamma (n)}y^{n -1 -1}e^{-y \theta} dy = \frac{\theta \Gamma(n-1)}{\Gamma (n)} (after a change of variables t = y \theta)

    So E(\frac{\Gamma (n)}{\Gamma (n-1)}Y) = \theta
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  9. #9
    Moo
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    Yeah, looks correct.
    But note that \Gamma(n)=(n-1)! and \Gamma(n-1)=(n-2)!

    So you're left with E\left[(n-1)Y^{-1}\right]=\theta
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  10. #10
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Do as you've been told above : find E[Y^{-1}]

    I'm still in shock over this.
    Katie ordering someone to follow my advice?
    What a difference from a year ago.
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