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Estimation of parameters using maximum likelihood method

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Let's have random value X defined by its density function:

    [tex]
    f(x; \beta) = \beta^2x \mbox{e}^{-\beta x}
    [/tex]

    where [itex]\beta > 0[/itex] for [itex]x > 0[/itex] and [itex]f(x) = 0[/itex] otherwise.

    Expected value of X is [itex]EX = \frac{2}{\beta}[/itex] and variance is [itex]\mbox{var } X = \frac{2}{\beta^2}.[/itex]

    Next, suppose we have random sample [itex]X_1, X_2, ..., X_n[/itex] from distribution defined by the given density function.

    Infer estimate of the parameter [itex]\beta[/itex] using the method of maximum likelihood.

    3. The attempt at a solution

    [tex]
    L(\beta; X) = \prod_{i=1}^{n} f(X_i; \beta)
    [/tex]

    [tex]
    \mbox{lik}(\beta; X) = \log L(\beta; X) = \sum \log f(X_i; \beta) = ... = 2n\log \beta + \sum \left( \log X_i - \beta X_i\right)
    [/tex]

    Now we'll find derivative of [itex]\mbox{lik}(\beta; X)[/itex] and find point where it's equal to zero (to find maximum of likelihood function).

    [tex]
    \frac{\partial \mbox{lik} (\beta; X)}{\partial \beta} = \frac{2n}{\beta} - \sum X_i
    [/tex]

    [tex]
    \frac{2n}{\widehat{\beta}} - \sum X_i = 0
    [/tex]

    [tex]
    \widehat{\beta} = \frac{2n}{\sum X_i} = \frac{2}{\overline{X_{n}}}
    [/tex]

    So [itex]\widehat{\beta}[/itex] is estimate of [itex]\beta[/itex].

    Next task is the following:

    Infer asymptotical distribution of [itex]\widehat{\beta}[/itex]. Hint: Use Fisher information [itex]\mathcal{F}_n = -\mbox{E} \frac{

    \partial^2 \mbox{lik}(\beta) }{\partial \beta^2}[/itex].

    Using infered asymptotical distribution [itex]\widehat{\beta} \sim \mathbf{N}(\beta, \mathcal{F}_{n}^{-1})[/itex], try to designate a confidence

    interval for parameter [itex]\beta[/itex] (possible unknown parameters replace with their estimates).


    With this I'm having little troubles. I compute Fisher information:

    [tex]
    \mathcal{F}_n = -\mbox{E} \frac{ \partial^2 \mbox{lik}(\beta) }{\partial \beta^2} = -\mbox{E} \left( -\frac{2n}{\beta^2} \right)
    [/tex]

    And now what? Can I continue like this?

    [tex]
    \mathcal{F}_n = -\mbox{E} \left( -\frac{2n}{\beta^2} \right) = \frac{2n}{\beta^2}
    [/tex]

    or am I supposed to expand it and continue like this?

    [tex]
    \mathcal{F}_n = -\mbox{E} \left( -\frac{2n}{\beta^2} \right) = \mbox{E} \left{ \frac{2n}{\widehat{\beta}}^2} \right} = \mbox{E} \left{

    \frac{2n\left( \overline{X_n}^2 \right)}{4} \right} = \mbox{E} \left{ \frac{1}{2} n \left( \overline{X_n}^2 \right) \right}
    [/tex]

    I don't know how to continue..
     
  2. jcsd
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