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Thread: 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 $\displaystyle f(x;\theta)=\theta(1-\theta)^{x-1}, x=1,2,3$

    let $\displaystyle x_{1},,,,x_{n}$ represent number of attacks,,

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

    I have some answers for part a) which gave the MLE as $\displaystyle \frac{n}{\sum x_i}=\frac{1}{\overline{x}}=\frac{1}{E(x)}$ 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 $\displaystyle E(\frac{1}{E(x)}-\Theta)
    =\frac{1}{\frac{1}{\theta}}-\Theta)=0$

    however I cannot see how this estimator $\displaystyle \frac{1}{E(x)}$ 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?

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

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