I am given $X_1,\ldots,X_n$ I.I.D random variables, where $X_i\sim N(\theta,\theta ^2)$. I am asked to find the loglikelihood function of this a long with the first two derivates of said function, as well as the Fisher Information $i(\theta)$.
So far what I have got is the following. Seeing as the density for the simultaneous outcome is given as the product of the marginal densitites, all alike, I have;
$L_X(\theta)=\frac{1}{\left(2\pi \theta^2\right)^{n/2}}e^{-\frac{1}{2\theta^2}\sum_{i=1}^n (x_i- \theta)^2}$
And thus the loglikelihoodfunction; (in my course we do $-log(L_X(\theta))$ )
$\ell_X(\theta)=\frac{n}{2}log(\sqrt{2\pi})+\frac{n}{2}log(\theta^2)+\frac{1}{2\theta^2}\sum_{i=1}^n (x_i- \theta)^2$
And the first two derivates
$D\ell_X(\theta)=\frac{n}{\theta}-\frac{1}{\theta^2}\sum_{i=1}^n x_i-\frac{1}{\theta^3}\sum_{i=1}^n x_i^2$
$D^2\ell_X(\theta)=-\frac{n}{\theta^2}+\frac{2}{\theta^3}\sum_{i=1}^n x_i+\frac{3}{\theta^4}\sum_{i=1}^n x_i^2$
But now I need to decide $i(\theta)=E_{\theta}D^2\ell_X(\theta)$, but that involves saying something about the mean of the independent random variables $X_i^2$, and I am stuggling to find the distribution of these. I know that if $X_i\sim N(0,1)$, $X_i^2$ would be Chi-squared distributed with 1 degree of freedom, but that is not the case here.
The whole point is that I need to decide the Maximum Likelihood Estimator for $\theta$, seeing as the mean and variance of my random variables are so closely and obviously related, and find its asymptotic distribution (smells like Cramér).
You've almost got it. The Fischer Information is a function of $\theta$, the expectation is on $X$, not $\theta$, in the FI formula. The FI formula can be written as
$$ i(\theta)=-E_X\left[\frac{\delta^2}{\delta\theta^2}\ln(f(x;\theta))|\theta\right] $$