# MSE and MLE with Poisson distribution

Consider a situation which is modeled by the Poisson distribution, $P(X_i=k|\theta)=\displaystyle\frac{e^{-\theta}\theta^k}{k!}$. Find:

(a) The MLE of $\theta$.

(b) The MSE of the MLE.

(c) The approximate distribution of the MLE from the Central Limit Theorem.

Okay, so I think I solved (a):

$f(X_1,X_2,...,X_n|\theta)=\displaystyle\frac{e^{-n\theta}\theta^{\Sigma X_i}}{X_1!\cdot...\cdot X_n!}$

so, $\displaystyle\frac{d(lnf)}{\theta}=-n+\displaystyle\frac{\Sigma X_i}{\theta} = 0$

and therefore $\hat\mu=\displaystyle\frac{\Sigma_{i=1}^n X_i}{n}=\boxed{\bar X}$.

If this is correct, how should I go about solving part (b)? I mean, if the MLE that I calculated is unbiased, then isn't the MSE of it just its variance? So, would the answer just be $Var(\bar X)$? Thank you!

As per central limit theorem $(\bar x- E(\bar x))/ \operatorname{Std}(\bar x)$ follows standard normal distribution with mean 0 and standard deviation of 1.
I.e. $(\bar x- E(\bar x))/ \operatorname{Std}(\bar x)$ follows $N(0,1)$ or equivalently $\bar x$ follows Normal with mean $E(\bar x)$ and variance $\operatorname{var}(\bar x)$.
Note that $E(\bar x) = \theta$ and as $\operatorname{Var}(x) = \theta$ for a Poisson distribution, $\operatorname{Var}(\bar x) = \theta/n$ which basically mean $\bar x$ follows normal distribution with mean $\theta$ and variance $\theta/n$.
• Thank you so much! So, for part (c), would my answer be $\displaystyle\frac{1}{\sigma_x\sqrt{2\pi}} e^{-\frac{x^2}{2Var(X)}}$? – Bing Feb 16 '17 at 5:34