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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.

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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}

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  • $\begingroup$ You have found $V(X_i^2)=2 \sigma^4$, we need $V(T_n)=\frac{1}{n^2}\sum_{i=1}^{n}2 \sigma^4=\frac{2 \sigma^4 * n}{n^2}=\frac{2 \sigma^4}{n}$ $\endgroup$ – Legend Killer Oct 27 '18 at 10:10
  • $\begingroup$ thanks edited @LegendKiller $\endgroup$ – Ahmad Bazzi Oct 27 '18 at 11:00

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