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?

• This generalization, stats.stackexchange.com/q/105337/28746 may be of interest to you. – Alecos Papadopoulos Nov 27 '18 at 21:29
• Not going to lie that example made me very confused! – Olivia Roberts Nov 27 '18 at 22:33
• Why? It is the general case where the variables are not necessarily normal. Then, one just notes that the 4th central moment of the Normal is $3\sigma^4$ and arrives at the $2\sigma^4$ variance that holds for the Normal case. – Alecos Papadopoulos Nov 27 '18 at 22:56
• So in the case for i.i.d normal Xis, is the var((Xi−μ)^2) =2σ^4 – Olivia Roberts Nov 27 '18 at 22:59
• Yes. In general it is $E(x^4) - [E(x^2)]^2$. – Alecos Papadopoulos Nov 28 '18 at 0:48

$$\dfrac 1 n \sum_{i=1}^n (X_i-\mu)^2$$ 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.
• @AlecosPapadopoulos : Sorry -- I needed a fourth power there: $$\operatorname{var}((X_i-\mu)^2) = 2\sigma^4.$$ – Michael Hardy Nov 28 '18 at 1:34
• @OliviaRoberts : I should have had $\operatorname{var}((X_i - \mu)^2) = 2\sigma^4$ rather than $2\sigma^2.$ Deriving it is elementary but takes some work. Could you post a separate question on that? – Michael Hardy Nov 28 '18 at 1:42