Suppose that $X_1, X_2, …, X_n$ is a random sample from a distribution with pdf
$f(x;\theta) = \frac{\theta^3}{2}x^2e^{-\theta x}, 0<x<\infty$
Find MLE $\hat{\theta}$ of $\theta$ and determine if it is unbiased
Show that the MLE is a complete sufficient statistics for $\theta$.
Find the UMVUE of $\theta$
Here's what I did:
The likelihood function
$L(\theta) = \prod_{i=1}^{n} \frac{\theta^3}{2}x^2e^{-\theta x}$
Take log and differentiate,
$l'(\theta) = \frac{3n}{\theta}-n\bar{x}$
Therefore, $\hat{\theta}^{MLE} = \frac{3}{\bar{X}}$
I know that to find unbiasedness I need to show $E[\hat{\theta}] = \theta$, but I can't seem to do the computation.
Best Answer
$$E[\hat\theta]=3n\cdot E\left[\frac1{\sum_{i=1}^nX_i}\right]$$$X_i$ are IID Erlang distributions $\gamma(\theta,3)$, and so is $Y=\sum_{i=1}^nX_i\implies Y\sim\gamma(\theta,3n)$. So $$E[1/Y]=\int_0^\infty\frac1yf_Y(y)dy=\int_0^\infty\frac1y\left(\frac{\theta^{3n}y^{3n-1}e^{-{\theta y}}}{\Gamma(3n)}\right)dy=\theta\cdot\frac{\Gamma(3n-1)}{\Gamma(3n)}=\frac{\theta}{3n-1}$$giving that the MLE is biased.