Fisher information $I(\sigma^2)$ and Cramer-Rao of estimator $S^2$

Question

Let $X_1,...,X_n$ be a random sample with $X_1\sim N(\mu,\sigma^2)$. Compute $I(\sigma^2)$ and verify the Cramer-Rao theorem for the estimator $S^2$

Since $f_{\mu, \sigma^2} = \frac{1}{\sqrt{2\pi} \sigma} \exp({-\frac{(x-\mu)^2}{2\sigma^2}})$, we have: $$f(X|\sigma^2) = \prod_{k=1}^n\frac{1}{\sqrt{2\pi} \sigma} \exp({-\frac{(X_k-\mu)^2}{2\sigma^2}}) = (\frac{1}{\sqrt{2\pi} \sigma})^n \exp({-\frac{\sum_{k=1}^n (X_k-\mu)^2}{2\sigma^2}})$$

\begin{align*} &\log(L(X|\sigma^2)) =-n\log(\sqrt{2\pi}\sigma)-\frac{\sum_{k=1}^n (X_k - \mu)^2}{2\sigma^2} \\ &\frac{\partial}{\partial \sigma^2}\log(L(X|\sigma^2)) = -\frac{n}{2\sigma^2}+\frac{\sum_{k=1}^n (X_k - \mu)^2}{2\sigma^4} \\ &I(\sigma) = \mathbb{E}((\frac{\partial}{\partial \sigma^2}\log(L(X|\sigma^2)))^2) = \frac{n^2}{4\sigma^4} -\frac{n^2}{2\sigma^4} +\frac{n^2}{4\sigma^4} = 0 \end{align*}

I'm pretty sure it shouldn't be $0$. Where is my mistake?

And for the second part I probably have to use $S^2 = \frac{1}{n-1}\sum^n_{k=1}(X_k-\bar{X})$ and check if $Var(S^2) \geq \frac{1}{I(\sigma)}$?

Answer 1

The mistake is in the last line:

$$\mathbb{E}\left[\left(\frac{\sum_{k=1}^n (X_k - \mu)^2}{2\sigma^4}\right)^2\right]=\mathbb{E}\left[\left(\frac{\chi_n^2}{2\sigma^2}\right)^2\right]=\frac{n(n+2)}{4\sigma^4}\neq\frac{n^2}{4\sigma^4}.$$