[Math] 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

You can use the following relation

$\text{Limiting Variance} \geq \text{Asymptotic Variance} \geq CRLB_{n=1}$

Now calculate the CRLB for $n=1$ (where n is the sample size), it'll be equal to ${2σ^4}$ which is the Limiting Variance. Therefore Asymptotic Variance also equals $2\sigma^4$.