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Math Help - 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; August 2nd 2009 at 03:44 AM.
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    Hello,

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

    We have the global pdf :

    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)

    \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 \lambda is :

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

    The MLE \bar\lambda is such that :

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

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

    But obviously, \lambda\neq 0.

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

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

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

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