Proving Asymptotic distribution of $\sqrt n( \widehat\sigma^2 -\sigma^2)$ I am looking at trying to derive an expression for the asymptotic distribution. We have $X_1,\ldots, X_n$ i.i.d $N(\mu, σ^2)$.
So we have defined $\hat \sigma^2 = \frac 1n \sum_{i=1}^n(X_i-\mu)^2$. (MLE of $σ^2$)
If further we are told that $\hat\sigma^2$ is a sample mean, how would I go about deriving an expression for the asymptotic distribution for:
$$\sqrt n( \hat\sigma^2 -σ^2).$$
I initially had the thought of using a chi squared argument and ultimately proving that $\sqrt n( \hat\sigma^2 -σ^2)$ -> $N(0, 2σ^4)$. However, I now think that my argument is flawed.
How would you guys go about it?
 A: $\dfrac 1 n \sum_{i=1}^n (X_i-\mu)^2$ is the MLE of $\sigma^2$ only if $\mu$ is known.
You have
$$
\operatorname E((X_i-\mu)^2) = \sigma^2
$$
and
$$
\operatorname{var}((X_i-\mu)^2) = 2\sigma^4
$$
and $(X_i-\mu)^2,\,\, i = 1,\ldots,n$ are independent.
Therefore the central limit theorem is applicable and you get
$$
\frac{\left( \sum_{i=1}^n (X_i - \mu)^2 \right) - n\sigma^2}{\sqrt2\,\,\sigma^2\cdot\sqrt n} \overset{L}{\longrightarrow} N(0,1)
$$
or in other words
$$
\frac{\left( \frac 1 n \sum_{i=1}^n (X_i-\mu)^2\right) - \sigma^2}{\sqrt2\,\,\sigma^2/\sqrt n} \overset{L}{\longrightarrow} N(0,1).
$$
If $\mu$ is not known, you have a more complicated problem since $(X_i-\overline X)^2,\,\,i=1,\ldots,n$ are not independent. But it can be shown that $\frac 1 {\sigma^2} \sum_{i=1}^n (X_i - \overline X)^2 \sim \chi^2_{n-1},$ and thus it has the same distribution as $Z_1^2+\cdots + Z_{n-1}^2$ where $Z_i,\,\,i=1,\ldots,n-1$ are i.i.d. $N(0,1),$ and then the central limit theorem can again be applied.
