Why the variance formula divides by n minus 1

Question

A bottling line fills n bottles and you record the volumes X1, \dots, Xn, an i.i.d. sample with unknown mean \mu and unknown variance \sigma^2. You estimate the process variance by averaging the squared gaps from the sample mean \bar X. If you want the estimator to be correct on average (unbiased) for \sigma^2, what number should you divide \sumi (Xi - \bar X)^2 by, and why is it not n?

Accepted Answer

The intuition first. You do not know the true mean \mu, so you center on the sample mean \bar X instead. But \bar X is pulled toward the data, so the points sit closer to \bar X than they do to the true \mu. The squared deviations are therefore too small on average, and dividing by n makes the estimate too low. You have to divide by something smaller than n to compensate. The algebra. Split each deviation around \mu: \sumi (Xi - \bar X)^2 = \sumi (Xi - \mu)^2 - n(\bar X - \mu)^2. Take expectations term by term. Each \mathbb{E}[(Xi - \mu)^2] = \sigma^2, so the first sum has expectation n\sigma…

Why the variance formula divides by n minus 1

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