I have the poisson distribution $X_a\sim $Poisson$(a\mu)$ for $a=1....n$ and I have found the maximum likelihood estimator as:
$$\hat{\mu}=\frac{\sum_{a=1}^{n}{x_a}}{\sum_{a=1}^{n}a}$$
However I am not sure what the best way to find the expected value of this is?
I would assume that $E(\hat{\mu})=\frac{x}{a}$ and $Var(\hat{\mu})=\frac{a\mu}{n}$ but I am not too sure. I was wondering if anyone can check this and see if I have gone wrong?
Assuming independence...
The likelihood is the following
$$L(\mu) \propto e^{-\mu\frac{n(n+1)}{2}}\mu^{\Sigma X}$$
the loglikelihood is
$$l(\mu)=-\mu\frac{n(n+1)}{2}+\Sigma X log\mu$$
Derivating...
$$l^*(\mu)=-\frac{n(n+1)}{2}+\frac{\Sigma X}{\mu}$$
Setting the score=0 and solving it in $\mu$ you get
$$\hat{\mu}_{ML}=\frac{2\Sigma X}{n(n+1)}$$
observing that $\Sigma X \sim Po\Big(\frac{n(n+1)}{2}\mu\Big)$ you get
$$E(\hat{\mu})=\mu$$
...it's unbiased.
I let you calculate its variance (it is very very easy)...
EDIT:
$$V(\hat{\mu})=\frac{2\mu}{n(n+1)}$$