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!
 A: Given observed data $\mathbf{x} = (x_1,...,x_n)$ the log-likelihood is:
$$\ell_\mathbf{x}(\theta) = n ( \bar{x} \ln (\theta) - \theta ).$$
This has derivatives:
$$\begin{equation} \begin{aligned}
\frac{d \ell_\mathbf{x}}{d \theta}(\theta) 
&= n \bigg( \frac{\bar{x}}{\theta} - 1 \bigg), \\[6pt] 
\frac{d^2 \ell_\mathbf{x}}{d \theta^2}(\theta) 
&= - \frac{n \bar{x}}{\theta^2} <0. \\[6pt] 
\end{aligned} \end{equation}$$
This shows that the log-likelihood function is concave, so the MLE occurs at the unique critical point given by:
$$0 = \frac{d \ell_\mathbf{x}}{d \theta}(\hat{\theta}) 
\quad \quad \quad \implies \quad \quad \quad
\hat{\theta} = \bar{x}.$$
Since $\mathbb{E}(X) = \theta$ and $\mathbb{V}(X) = \theta$, it follows from the central limit theorem that when $n$ is large, we have the approximate distribution $\hat{\theta} \sim \text{N} ( \theta, \theta/n )$.  The estimator is unbiased, and so it has MSE given by:
$$\begin{equation} \begin{aligned}
\text{MSE}(\hat{\theta},\theta) 
= \mathbb{V} (\hat{\theta})
= \mathbb{V} (\bar{X} )
= \frac{\theta}{n}.
\end{aligned} \end{equation}$$
A: Your answers and understanding is correct.
For the 3rd answer - the MLE would tend to follow a normal distribution as per as Central limit theorem.
A: 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 $\theta/n$. 
