Find the MSE of the MLE 

Let $X_1,\ldots,X_n$ be i.i.d exponentially distributed r.v's with pdf: 
$$ f_\theta(x) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$$
    such that $x\ge 0$.


I have found that the MLE is given by $$ \hat{\theta}(X)= \frac{1}{n}\sum_{i=1}^n X_i.$$
Note that $\hat{\theta}$ is unbiased as
$$ E(\hat{\theta}) = E\left(\frac{1}{n}\sum_{i=1}^n X_i\right) = \theta $$ 
since the r.v's are i.i.d. 
But then I am having difficulties computing the MSE as 
$$ MSE = E_{\theta}\left( (\theta-\hat{\theta})^2 \right) $$ 
$$= E(\hat{\theta^2})+\theta^2-2\theta E(\hat{\theta})$$ 
$$ =E(\hat{\theta^2}) -\theta^2$$
So then $E(\hat{\theta^2}) = E\left(\frac{1}{n^2}\sum_{i=1}^n X_i^2\right) =E(X^2) = \int_0^\infty f_\theta dx = ... =  \frac{2\theta^2}{n^2} $
$ \Rightarrow MSE = \frac{2\theta^2}{n^2} - \theta^2$
Supposedly the final answer should be $\frac{\theta^2}{n}$ but im not getting this at all... 
 A: Note that $$E[\hat{\theta}^2] \neq E\left[\frac{1}{n^2}\sum_{i=1}^n X_i^2\right].$$
An easy way to compute $E[(\theta-\hat{\theta})^2]$ is to observe that $E[(\theta-\hat{\theta})^2] = E[(\hat{\theta}-\theta)^2]=var(\hat{\theta})$. But
$$var(\hat{\theta})=\frac{1}{n^2}\sum_{i=1}^{n}var(X_i)=\frac{n \theta^2}{n^2}=\frac{\theta^2}{n}.$$
A: It's not true that
$$ E(\hat{\theta^2}) = E\left(\frac{1}{n^2}\sum_{i=1}^n X_i^2\right).$$
Rather, one has
\begin{align}
E(\hat{\theta^2}) &= E\left[\left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2\right] \\
&= E\left[\frac{1}{n^2}\sum_{i,j} X_iX_j \right] \\
&= \frac{1}{n^2}E\left[\sum_{i=1}^n X_i^2 + \sum_{i\ne j} X_iX_j \right] \\
&= \frac{1}{n^2}E\left[\sum_{i=1}^n X_i^2 \right] + \frac{1}{n^2}E\left[\sum_{i\ne j} X_iX_j \right] \\
&= \frac{1}{n^2} \sum_{i=1}^nE\left[ X_i^2  \right] + \frac{1}{n^2} \sum_{i\ne j} E\left[X_iX_j \right] \\
&= \frac{1}{n^2} \sum_{i=1}^nE\left[ X^2  \right] + \frac{1}{n^2} \sum_{i\ne j} E[X_i]E[X_j] \\
&= \frac{1}{n^2} \sum_{i=1}^n 2\theta^2 + \frac{1}{n^2} \sum_{i\ne j} \theta^2 \\
&= \frac{1}{n^2} n2\theta^2 + \frac{1}{n^2} (n^2 - n)\theta^2 \\
&= \frac{\theta^2}{n^2} (n^2 - n + 2n) \\
&= \theta^2 + \frac{\theta^2}{n}.
\end{align}
