# Standard error of sample variance

We know that an unbiased estimator of the variance is: $$\hat{\sigma}^2_{unbiased} = \frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2$$

I was wondering, does it have the smallest possible standard error?

Does the biased (but consistent) estimator: $$\hat{\sigma}^2_{biased} = \frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2$$ have a lower standard error?

I know that the biased estimator of the variance is the maximum likelihood estimator (MLE) and that MLE estimators have the smallest possible variance in the set of well-behaved estimators. According to this result, what I would conclude here is that $$\sigma^2_{biased}$$ even though it is biased, has a lower standard error than the unbiased estimator.

Looking at the variance of $$\hat{\sigma}_{biased}^2$$ we have $$\begin{eqnarray*} \mathrm{Var}(\hat{\sigma}_{biased}^2) &=& \mathrm{Var} \left( \dfrac{n-1}{n} \hat{\sigma}_{unbiased}^2 \right) \\ &=& \left( \dfrac{n-1}{n} \right)^2 \mathrm{Var}(\hat{\sigma}_{unbiased}^2). \end{eqnarray*}$$ Since $$(n-1)/n < 1$$ it follows that $$\mathrm{Var}(\hat{\sigma}_{biased}^2) < \mathrm{Var}(\hat{\sigma}_{unbiased}^2)$$ so $$\mathrm{sd}(\hat{\sigma}_{biased}^2) < \mathrm{sd}(\hat{\sigma}_{unbiased}^2).$$ The maximum likelihood estimator (MLE) does indeed have a smaller standard error.
• You can make a variance estimator $a \hat{\sigma}_{unbiased}^2$ with arbitrarily low standard error by choosing some $a < 1$. For example if $a = 1/(n + 1)$ you have the minimum mean square error estimator. Smaller values of $a$ result in more bias so at some point you could say the estimator is no longer well behaved. – Estacionario Apr 12 '19 at 5:45