# Estimation of parameters using maximum likelihood method

1. Sep 10, 2008

### twoflower

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

$$f(x; \beta) = \beta^2x \mbox{e}^{-\beta x}$$

where $\beta > 0$ for $x > 0$ and $f(x) = 0$ otherwise.

Expected value of X is $EX = \frac{2}{\beta}$ and variance is $\mbox{var } X = \frac{2}{\beta^2}.$

Next, suppose we have random sample $X_1, X_2, ..., X_n$ from distribution defined by the given density function.

Infer estimate of the parameter $\beta$ using the method of maximum likelihood.

3. The attempt at a solution

$$L(\beta; X) = \prod_{i=1}^{n} f(X_i; \beta)$$

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

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

$$\frac{\partial \mbox{lik} (\beta; X)}{\partial \beta} = \frac{2n}{\beta} - \sum X_i$$

$$\frac{2n}{\widehat{\beta}} - \sum X_i = 0$$

$$\widehat{\beta} = \frac{2n}{\sum X_i} = \frac{2}{\overline{X_{n}}}$$

So $\widehat{\beta}$ is estimate of $\beta$.

Infer asymptotical distribution of $\widehat{\beta}$. Hint: Use Fisher information $\mathcal{F}_n = -\mbox{E} \frac{ \partial^2 \mbox{lik}(\beta) }{\partial \beta^2}$.

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

interval for parameter $\beta$ (possible unknown parameters replace with their estimates).

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

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

And now what? Can I continue like this?

$$\mathcal{F}_n = -\mbox{E} \left( -\frac{2n}{\beta^2} \right) = \frac{2n}{\beta^2}$$

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

$$\mathcal{F}_n = -\mbox{E} \left( -\frac{2n}{\beta^2} \right) = \mbox{E} \left{ \frac{2n}{\widehat{\beta}}^2} \right} = \mbox{E} \left{ \frac{2n\left( \overline{X_n}^2 \right)}{4} \right} = \mbox{E} \left{ \frac{1}{2} n \left( \overline{X_n}^2 \right) \right}$$

I don't know how to continue..