If the mean is known, do you still divide by n − 1?

Question

Suppose the true mean \mu is known (not estimated from the data), and you estimate the variance from i.i.d. observations X1, \dots, Xn. What is the right divisor now, n or n - 1? Prove that your choice is unbiased.

Accepted Answer

With \mu known, use the divisor n: \hat\sigma^2 = \frac{1}{n}\sum{i=1}^n (Xi - \mu)^2. Proof of unbiasedness. Each term has expectation \mathbb{E}[(Xi - \mu)^2] = \operatorname{Var}(Xi) = \sigma^2 exactly, because \mu is the true center. So \mathbb{E}\left[\sum{i=1}^n (Xi - \mu)^2\right] = \sum{i=1}^n \sigma^2 = n\sigma^2, and dividing by n gives \mathbb{E}[\hat\sigma^2] = \sigma^2. Why no correction. The n - 1 divisor exists only because the usual sample variance measures deviations around the estimated mean \bar X, which is fitted to the data and hugs it too closely. Knowing \mu costs no de…

If the mean is known, do you still divide by n − 1?

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