# Variance of the $\hat{\sigma}^2$ of a Maximum Likelihood estimator

Given some normally distributed observations $$x_1,x_2,...,x_n$$

$$\forall i\ x_i\sim\mathcal{N}(\mu, \sigma^2)$$

the ML estimator decides that the variance that maximizes the likelihood function is (see here):

$$\hat{\sigma}^2=\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^2$$

Now, I am trying to find the variance of this estimation:

$$\sigma^2_{\hat{\sigma}^2}=\operatorname{Var}[\hat{\sigma}^2]=\operatorname{Var}[\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^2]$$

If we note that: $$\hat{\sigma}^2=\frac{1}{n}\sum_{i=1}^{n}(x_i^2-2x_i\bar{x}+\bar{x}^2) \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-2\bar{x}\frac{1}{n}\sum_{i=1}^{n}x_i+\frac{1}{n}\sum_{i=1}^{n}\bar{x}^2 \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-2\bar{x}^2+\bar{x}^2 \\ =\frac{1}{n}\sum_{i=1}^{n}x_i^2-\bar{x}^2$$

we have:

$$\sigma^2_{\hat{\sigma}^2}=\operatorname{Var}[\frac{1}{n}\sum_{i=1}^{n}x_i^2-\bar{x}^2]$$

but I am stuck here since I think that $$x_i$$ and $$\bar{x}$$ are not independent in order to use the property that says that the variance of the sum is the sum of the variances.

Do you know the famous result that if $$X_1, \ldots, X_n \text{ i.i.d. } \sim N(\mu, \sigma^2)$$, then $$\frac{1}{\sigma^2}\sum_{i = 1}^n (X_i - \bar{X})^2 \sim \chi_{n - 1}^2?$$ It is also well-known that the variance of a $$\chi_k^2$$ random variable is $$2k$$.
An alternative treatment is to express the sample variance as a quadratic form of the random vector $$\mathbf{X} = (X_1, \ldots, X_n)$$ then apply the quadratic form variance formula. See this answer for a solution to a slightly more general case.
• I know about the second part about the variance of chi-squared distribution you are referring to but where does the first result come from? I mean, how can I prove it? Finally, what is the distribution of $X_i-\bar{X}$?