# When estimating variance, why do unbiased estimators divide by n-1 yet maximum likelihood estimates divide by n?

I am totally confused: On the one hand you can read all kinds of explanations why you have to divide by n-1 to get an unbiased estimator for the (unknown) population variance (degrees of freedom, not defined for sample size 1 etc.) - see e.g. here or here.

On the other hand when it comes to variance estimation of a supposed normal distribution all of this doesn't seem to be true anymore. There it is said that the maximum likelihood estimator for variance includes only a division by n - see e.g. here.

Now, can anyone please enlighten me why it is true here but not there? I mean normality is what most models boil down to (not least due to the CLT). So is the choice "division by n" yet the best choice for finding the best estimation for the true population variance after all?!?

Thank you!

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The answer to your question is contained within your question.

When choosing an estimator for a parameter, you should ask yourself, what property would you like your estimator to have:

• Robustness
• Unbiasedness
• Have the distributional properties of a MLE
• Consistency
• Asymptotically normal
• You know the population mean, but the variance is unknown

If your estimator is the one that is divided by (n-1), then you want an unbiased estimtor of the variance. If your estimator is the one that is divided by n, then you have an MLE estimator. Of course, when n is large; dividing by either (n-1) or n will give you approximately the same results and the estimator will be approximately unbiased and have the properties of all MLE estimators.

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...where practice is large $n$. –  mbq May 5 '11 at 9:07