# Variance of variance MLE estimator of a normal distribution

The MLE estimator of the variance of a normal distribution is $\hat \sigma^2 = \frac{1}{N} \sum_{i=1}^{N}(X_i - \bar X)^2$, where $\bar X$ is the sample mean and $X_i \sim^{iid} \mathcal{N}(\mu,\sigma^2)$ . I'm curious because I've seen (e.g. http://www.stat.ufl.edu/~winner/sta6208/allreg.pdf , p.20) that the variance of this estimator is equal to $\frac{2\sigma ^4}{N}$, but I find something different:

Since $\frac{1}{\sigma ^2} \sum_{i=1}^{N}(X_i-\bar X)^2 \sim \chi^2_{N-1}$, we have that

\begin{align} var\big(\frac{N\hat \sigma^2}{\sigma^2}\big) = var(\chi^2_{N-1}) = 2(N-1) \\ \implies var(\hat \sigma ^2) = \frac{2(N-1) \sigma^4}{N^2} \end{align}

What am I missing?

The variance of the estimator in the course notes is based on maximum likelihood estimation which typically results in biased estimators. The second variance calculation has a "correction" term that makes the estimator unbiased. You have likely seen this phenomenon with the unbiased estimator for the sample mean, i.e., dividing by $n-1$ instead of $n$.
• So, if I don't divide by $n-1$, the bias is not 0 and the variance will be $\frac{2(n−1)\sigma^4}{n^2}$, so the bias and the variance are greater than in the case where I divide by $n-1$? – Amanjo Sep 23 '17 at 1:54
• The estimators themselves are random and have distributions. The bias of an estimator quantifies how far its mean over all samples (usually of a particular size, e.g., $n$) in the population is from the (true/actual but unknown) value of the parameter you are estimating. Similarly, an estimator has its own variance which roughly conveys how far (on average) an estimate based on a particular sample can be from the value of the parameter you are estimating. It might be confusing because you are estimating a variance and both estimators (notes & yours) of the variance have their own variances. – shoda Sep 23 '17 at 2:01
• I got it. The variance in the no bias case (i.e. dividing by $n-1$) is greater than in the bias case (i.e. dividing by $n$). – Amanjo Sep 23 '17 at 2:09