Asymptotic variance of MLE of normal distribution.

Question

I am trying to explicitly calculate (without using the theorem that the asymptotic variance of the MLE is equal to CRLB) the asymptotic variance of the MLE of variance of normal distribution, i.e.: $$\hat{\sigma}^2=\frac{1}{n}\sum_{i=1}^{n}(X_i-\hat{\mu})^2$$ I have found that: $${\rm Var}(\hat{\sigma}^2)=\frac{2\sigma^4}{n}$$ and so the limiting variance is equal to $2\sigma^4$, but how to show that the limiting variance and asymptotic variance coincide in this case?

Answer 1

From the asymptotic normality of the MLE and linearity property of the Normal r.v \begin{align} \sqrt{n}\left( \hat{\sigma}^2_n - \sigma^2 \right) \xrightarrow{D} \mathcal{N}\left(0, \ 2\sigma^4 \right) \\ \left( \hat{\sigma}^2_n - \sigma^2 \right) \xrightarrow{D} \mathcal{N}\left(0, \ \frac{2\sigma^4}{n} \right) \\ \hat{\sigma}^2_n \xrightarrow{D} \mathcal{N}\left(\sigma^2, \ \frac{2\sigma^4}{n} \right), && n\to \infty \\ & \end{align}