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Thread: Maximum Likelihood Estimator and Bias

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    Maximum Likelihood Estimator and Bias

    A R.V. X has pdf
    How would I find the mle of lambda based on realisations of x1..?
    Last edited by bluebiro; Aug 2nd 2009 at 03:44 AM.
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    Hello,

    Let $\displaystyle (x_1,\dots,x_n)$ a sample of iid rv with pdf f.

    We have the global pdf :

    $\displaystyle f(x_1,\dots,x_n)=\prod_{i=1}^n \frac 2\lambda \cdot x_i \cdot \exp\left(-\frac{x_i^2}{\lambda}\right)$

    $\displaystyle \ln f(x_1,\dots,x_n)=n (\ln(2)-\ln(\lambda))-\sum_{i=1}^n \frac{x_i^2}{\lambda}$

    Its derivative with respect to $\displaystyle \lambda$ is :

    $\displaystyle -\frac n\lambda+\frac{1}{\lambda^2}\sum_{i=1}^n x_i^2$

    The MLE $\displaystyle \bar\lambda$ is such that :

    $\displaystyle 0=-\frac{n}{\bar\lambda}+\frac{1}{\bar\lambda^2}\sum_ {i=1}^n x_i^2$

    --> $\displaystyle \bar\lambda=0 \text{ or } \bar\lambda=\frac 1n\sum_{i=1}^n x_i^2$

    But obviously, $\displaystyle \lambda\neq 0$.

    So $\displaystyle \boxed{\bar\lambda=\frac 1n\sum_{i=1}^n x_i^2}$

    -------------------------------------------
    To see if it's unbiased, prove that $\displaystyle \mathbb{E}(\bar\lambda)=\lambda$

    $\displaystyle \mathbb{E}(\bar\lambda)=\mathbb{E}\left(\frac 1n\sum_{i=1}^n x_i^2\right)=\mathbb{E}(x_1^2)$ because they're identically distributed.

    And $\displaystyle \mathbb{E}(x_1^2)=\int_0^\infty x^2 f(x) ~dx$

    Compute this integral and prove (or disprove) it equals $\displaystyle \lambda$
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