# construct an interval estimator (Gamma) using asymptotic distribution of MLE

• Feb 13th 2011, 09:13 PM
Volga
construct an interval estimator (Gamma) using asymptotic distribution of MLE
Question.

Let $\displaystyle Y_1,... Y_n$ be a random sample from a distribution with density function

$\displaystyle f_y(y)=\frac{{\theta}^k}{(k-1)!}y^{k-1}e^{{-\theta}y}, y>0$
$\displaystyle 0$ $\displaystyle otherwise$

where $\displaystyle k\geq1$ is a known integer, and $\displaystyle \alpha>0$ is an unknown parameter. Using the asymptotic distribution of the maximum likelihood estimator, construct an interval estimator for $\displaystyle \alpha$ with approximate confidence coefficient of 0.95.

Asymptotic distribution of MLE: if $\displaystyle \hat{\theta}$ is a maximum likelihood estimator, then $\displaystyle \hat{\theta}$ converges in distribution to $\displaystyle N(\theta, I_Y^{-1}(\theta))$ and $\displaystyle \sqrt{I_Y(\theta)}(\hat{\theta}}-\theta){\sim}N(0,1)$ is asymptotically pivotal for $\displaystyle \theta$.

I found information for Gamma distribution random sample already here
http://www.mathhelpforum.com/math-he...tml#post617259

so I'll just use it here: $\displaystyle I_y(\theta)=\frac{k}{{\alpha}^2}$. Since I don't know $\displaystyle \alpha$, I am going to use $\displaystyle \frac{k}{\hat{\alpha}^2}$ to approximate information.

$\displaystyle \sqrt{\frac{k}{\hat{\alpha}^2}}{\sim}N(0,1)$

Let $\displaystyle Z=\sqrt{\frac{k}{\hat{\alpha}^2}}$

$\displaystyle P(z_{0.025}{\leq}Z{\leq}z_{0.975})=0.95$

$\displaystyle z_{0.025}=-z_{0.975}$ right?

Solving for $\displaystyle \alpha$, I get

$\displaystyle [\hat{\alpha}-z_{0.975}\sqrt{\frac{k}{\hat{\alpha}^2}}{\leq}\alp ha{\leq}\hat{\alpha}+z_{0.975}\sqrt{\frac{k}{\hat{ \alpha}^2}}]$