Construct a confidence interval for $\theta$ Let $X_{1}, \cdots, X_{n}$ be a random c.i.i.d sample such as, given $\theta$, $X_{1} \sim \mathcal{N}(0,\theta)$. Construct a confidence interval for $\theta$ using asymptotic results.
This question is asking to use Fisher information to find an approximated confidence interval for $\theta$. I already found that $\hat{\theta}_{\text{MLE}}$ is
$$ \frac{\sum_{i=1}^n X_{i}^2}{n}$$
But, I can't find theta's Fisher information because I don't know the expected value of $\sum_{i=1}^n X_{i}^2$. Could someone help me to fully solve this question?
 A: $$
\def\thest{\hat{\theta}_\text{MLE}}\\
\def\eqdef{\stackrel{\text{def}}{=}}\\
\def\var{\text{Var}}
$$
The only thing Fisher information gives you that's useful for calculating confidence intervals for large $\ n\ $ is the variance of $\ \thest\ $. But you can calculate this  directly without wading through the formula for Fisher Information.
Since $\ E\big(X_i\big)=0\ $, then $\ E\left(  X_i^2\right)=\var\big(X_i\big)=\theta\ $, and $\ E\left(\sum_\limits{i=1}^nX_i^2\right)=$$\,\sum_\limits{i=1}^nE\big(X_i^2\big)=n\theta\ $, and since $\ X_i\ $ are independent,
\begin{align} \var\left(\sum_\limits{i=1}^nX_i^2\right)&=\sum_\limits{i=1}^n\var\big(X_i^2\big)\\
&=\sum_\limits{i=1}^n\left(E\big(X_i^4)-E\big(X_i^2\big)^2\right)\\
&=\sum_\limits{i=1}^n\big(3\theta^2-\theta^2\big)\\
&=2n\theta^2\ .
\end{align}
Thus,
\begin{align}
E\big(\thest\big)&=E\left(\frac{1}{n}\sum_\limits{i=1}^nX_i^2\right)\\
&=\theta\ ,\\
\var\big(\thest\big)&=\var\left(\frac{1}{n}\sum_\limits{i=1}^nX_i^2\right)\\
&=\frac{2\theta^2}{n}\ ,
\end{align}
and so for sufficiently large $\ n\ $ the distribution of
$$
\sqrt{\frac{n}{2}}\left(\frac{\thest-\theta}{\theta}\right)
$$
is approximately standard normal.  To find a confidence interval of size $\ 1-\alpha\ $, therefore, first find $\ z_{\alpha/2}\ $ such that
$$
\mathcal{N}(0,1)\left(z_{\alpha/2}\right)=1-\frac{\alpha}{2}\ .
$$
Then, for sufficiently large $\ n\ $,
$$
P\left(-z_{\alpha/2}\le\sqrt{\frac{n}{2}}\left(\frac{\thest-\theta}{\theta}\right)\le z_{\alpha/2}\right)\approx1-\alpha\ ,
$$
which can be rewritten as
$$
P\left(\frac{\thest}{1+z_{\alpha/2}\sqrt{\frac{2}{n}}}\le\theta\le\frac{\thest}{1-z_{\alpha/2}\sqrt{\frac{2}{n}}}\right)\approx1-\alpha\ ,
$$
provided $\ z_{\alpha/2}<\sqrt{\frac{n}{2}}\ $. Unless this latter inequality is satisfied by a wide margin, however, $\ n\ $ is not large enough for the normal approximation to be valid.  The lower and upper limits of the $\ 1-\alpha\ $ confidence interval are therefore $\ \frac{\thest}{1+z_{\alpha/2}\sqrt{\frac{2}{n}}}\ $ and $\ \frac{\thest}{1-z_{\alpha/2}\sqrt{\frac{2}{n}}}\ $, respectively.
Addendum
I should point out that the centre of the above confidence interval isn't $\ \thest\ $, as one would normally like it to be, but
$$
\frac{\thest}{1-\frac{2z_{\alpha/2}^2}{n}}\ .
$$
However, when $\ n\ $ is large enough for the normal approximation to be valid, the difference is negligible, especially given that the limits of the interval are only approximations anyway.
While it's possible to construct a confidence interval with centre $\ \thest\ $, there's no point in doing this with the normal approximation, because it's no easier than doing it with the exact distribution. The exact distribution of the random variable $\ \frac{n\thest}{\theta}\ $ is chi-squared with $\ n\ $ degrees of freedom. Therefore, if we find $\ x_\alpha\ $ that satisfies
$$
1-\alpha=\chi_n^2\left(\frac{n}{1-x_\alpha}\right)-\chi_n^2\left(\frac{n}{1+x_\alpha}\right)\ ,
$$
then
\begin{align}
P\big(\thest\big(1-x_\alpha\big)\le\,&\theta\le\thest\big(1+x_\alpha\big)\big)\\
&=P\left(\frac{1}{1+x_\alpha}\le\frac{\thest}{\theta}
\le \frac{1}{1-x_\alpha}\right)\\
&=1-\alpha\ .
\end{align}
