See the n − 1 correction with just two data points

Question

Take the smallest interesting sample, n = 2, with X1, X2 i.i.d. of variance \sigma^2. Show concretely that dividing the sum of squared deviations by n - 1 = 1 is unbiased, while dividing by n = 2 is biased low.

Accepted Answer

With \bar X = \frac{X1 + X2}{2}, both deviations are half the gap: X1 - \bar X = \frac{X1 - X2}{2} and X2 - \bar X = \frac{X2 - X1}{2}. So \sum{i=1}^2 (Xi - \bar X)^2 = 2\left(\frac{X1 - X2}{2}\right)^2 = \frac{(X1 - X2)^2}{2}. Take expectations. Since X1, X2 are independent, \operatorname{Var}(X1 - X2) = 2\sigma^2 and (as the mean of the difference is 0) \mathbb{E}[(X1 - X2)^2] = 2\sigma^2. Therefore \mathbb{E}\left[\sum (Xi - \bar X)^2\right] = \frac{2\sigma^2}{2} = \sigma^2 = (n-1)\sigma^2. Dividing by n - 1 = 1 gives \mathbb{E}[S^2] = \sigma^2, unbiased. Dividing by n = 2 gives \frac{(X1 …

See the n − 1 correction with just two data points

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