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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.

dylan
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1 Answers1

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$$E[\hat\theta]=3n\cdot E\left[\frac1{\sum_{i=1}^nX_i}\right]$$$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.

Shubham Johri
  • 17,739