<s>[INST] CAN and BAN estimators of $\sigma^2$ Let $X_1,X_2,..X_n \sim^{\text{i.i.d}} N(0,\sigma^2)$
Show that $T_n=\frac{1}{n}\sum_{i=1}^{n} X_i^2$ is a Consistent and Asymptotically Normal estimator(CAN) as well as the Best Asymptotic Normal Estimator(BAN) of $\sigma^2$.
Indeed, $E(T_n)=\sigma^2$ and $V(T_n)=\frac{2 \sigma^4}{n}$.
So, $T_n$ is a CAN estimator.
But to show that it is a BAN estimator , $V(T_n)$ should be equal to $n$ times the Rao Cramer lower bound.
But I am failing to show that.
 [/INST] CRB
Let's write down the PDF 
\begin{equation}
 f_{X_i}(x_i \vert \sigma^2) \sim \frac{1}{\sqrt{2\pi \sigma^2}}\exp(-\frac{x_i^2}{2\sigma^2} ) 
\end{equation}
Assuming independence, we have that 
\begin{equation}
 L(\sigma^2) = f(x_1 \ldots x_n \vert \sigma^2)
 =
 f(x_1 \vert \sigma^2) \ldots f(x_n \vert \sigma^2)
 =
 \frac{1}{(2\pi \sigma^2)^{n/2}}
 \exp(- \frac{1}{2\sigma^2}\sum_{i=1}^n x_i^2)
\end{equation}
Take the log likelihood
\begin{equation}
 l(\sigma^2)
 =
 \log L(\sigma^2)
 =
 -\frac{n}{2}
 \log(2 \pi \sigma^2)
 - \frac{1}{2\sigma^2}\sum_{i=1}^n x_i^2
\end{equation}
Deriving w.r.t $\sigma^2$, we have
\begin{equation}
 l'(\sigma^2)
 =
 -\frac{n}{2}
 \frac{1}{\sigma^2}
 + \frac{1}{2\sigma^4}\sum_{i=1}^n x_i^2
\end{equation}
Derive again
\begin{equation}
 l''(\sigma^2)
 =
 \frac{n}{2}
 \frac{1}{\sigma^4}
 - \frac{1}{\sigma^6}\sum_{i=1}^n x_i^2
\end{equation}
Take the expectation now
\begin{equation}
 E(l''(\sigma^2))
 =
 E \big(
 \frac{n}{2}
 \frac{1}{\sigma^4}
 - \frac{1}{\sigma^6}\sum_{i=1}^n x_i^2
 \big)
\end{equation}
The only random part here is $x_i^2$, hence
\begin{equation}
 E(l''(\sigma^2))
 =
 \frac{n}{2}
 \frac{1}{\sigma^4}
 - \frac{1}{\sigma^6}\sum_{i=1}^n E x_i^2
\end{equation}
But $E x_i^2 = \sigma^2$ so 
\begin{equation}
 E(l''(\sigma^2))
 =
 \frac{n}{2}
 \frac{1}{\sigma^4}
 - \frac{1}{\sigma^6}\sum_{i=1}^n \sigma^2
 =
 \frac{n}{2}
 \frac{1}{\sigma^4}
 - \frac{n}{\sigma^4}
 =
 -
 \frac{n}{2\sigma^4}
\end{equation}
So, the Fisher information is 
\begin{equation}
 F = - E(l''(\sigma^2)) = \frac{n}{2\sigma^4}
\end{equation}
Then the CRB is 
\begin{equation}
 CRB = \frac{1}{F} = \frac{2\sigma^4}{n} = V(T_n)
\end{equation}
</s>