Question.

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

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

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

Answer.

Asymptotic distribution of MLE: if \hat{\theta} is a maximum likelihood estimator, then \hat{\theta} converges in distribution to N(\theta, I_Y^{-1}(\theta)) and \sqrt{I_Y(\theta)}(\hat{\theta}}-\theta){\sim}N(0,1) is asymptotically pivotal for \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: I_y(\theta)=\frac{k}{{\alpha}^2}. Since I don't know \alpha, I am going to use \frac{k}{\hat{\alpha}^2} to approximate information.

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

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

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

z_{0.025}=-z_{0.975} right?

Solving for \alpha, I get

[\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}}]