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Thread: Unbiased estimator of the exponential distirbution

  1. #1
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    Unbiased estimator of the exponential distirbution

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

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

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

    therefore $\displaystyle 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 $\displaystyle E(W)=p$ then it does not follow that $\displaystyle E(W^{-1})=1/p$

    So, calculate $\displaystyle E(Y^{-1})$ and see what you get.
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  3. #3
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    Am I allowed to use $\displaystyle \frac{\theta^2 Y}{n}$
    Last edited by statmajor; Jan 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 $\displaystyle E[Y^{-1}]$

    In order to do that, recall that $\displaystyle E[h(Y)]=\int h(y) g(y) ~dy$, 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 $\displaystyle \theta$
    So its pdf is the one of a gamma distribution $\displaystyle (n,1/\theta)$ (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:

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

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

    So you're left with $\displaystyle E\left[(n-1)Y^{-1}\right]=\theta$
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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 $\displaystyle E[Y^{-1}]$

    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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