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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.

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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.

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  • $\begingroup$ So I conclude that among all well behaved estimators, the MLE has the lowest standard error. Also, I was wondering, are they other unbiased variance estimators beside the above formula? If they are, I think it would be interesting to prove that the above formula gives the minimum variance unbiased estimator. $\endgroup$
    – Victor
    Apr 11, 2019 at 20:05
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    $\begingroup$ 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. $\endgroup$ Apr 12, 2019 at 5:45
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    $\begingroup$ I checked the properties of MLE more thoroughly: the MLE has the smallest asymptotic variance. Here, both the unbiased and the biased estimators have the same asymptotic variance, they are both efficient or asymptotically optimal. $\endgroup$
    – Victor
    Apr 14, 2019 at 0:20

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