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Math Help - maximum likelihood estimator biased not unbiased

  1. #1
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    maximum likelihood estimator biased not unbiased

    a)derive the likelihood estimator of the geometric function f(x;\theta)=\theta(1-\theta)^{x-1},  x=1,2,3

    let x_{1},,,,x_{n} represent number of attacks,,

    b)determine the cramer-rao lower bound for an unbiased estimator of \Theta

    I have some answers for part a) which gave the MLE as \frac{n}{\sum x_i}=\frac{1}{\overline{x}}=\frac{1}{E(x)} which would work for part b) as it is unbiased...for it to be unbiased the proof is that either MSE(estimate)=var(estimate) or BIAS(estimate)=0 implies E(estimate)=0. hence E(\frac{1}{E(x)}-\Theta)<br />
=\frac{1}{\frac{1}{\theta}}-\Theta)=0

    however I cannot see how this estimator \frac{1}{E(x)} was obtained.
    My working is as follows and I obtained something completely different while the official working gave the anser above: any offerings as to how they did it?

    l(\Theta)=\log\Theta^n(1-\Theta)^{\sum x_{i}-n}
    =(\sum x_{i}-n)log(1-\Theta)+n\log(\Theta)
    l'(\Theta)=\frac{\sum x_{i}-n}{1-\Theta}+\frac{n}{\Theta}=0
    =\Theta(\sum x_{i}-n)+n(1-\Theta)=0
    =\Theta(\sum x_{i}-n)+n-n\Theta=0
    =\Theta(\sum x_{i}-2n)=-n
    \Theta=\frac{=n}{\sum x_{i}-2n}
    =\frac{n}{2n-\sum x_{i}}

    which is clearly biased
    Last edited by bluesblues; August 16th 2010 at 03:32 AM.
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