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