Why divide by n − 1 in the sample variance?

Question

Given i.i.d. observations X1, \dots, Xn with mean \mu and variance \sigma^2, the sample variance is defined as S^2 = \frac{1}{n-1}\sum{i=1}^n (Xi - \bar{X})^2. Why n-1 and not n? Prove that \mathbb{E}[S^2] = \sigma^2.

Accepted Answer

The algebra. Start from the identity \sum{i=1}^n (Xi - \bar{X})^2 = \sum{i=1}^n Xi^2 - n\bar{X}^2. Take expectations term by term, using \mathbb{E}[Z^2] = \operatorname{Var}(Z) + (\mathbb{E}Z)^2: - \mathbb{E}[Xi^2] = \sigma^2 + \mu^2, so the first sum has expectation n(\sigma^2 + \mu^2). - \mathbb{E}[\bar{X}^2] = \operatorname{Var}(\bar{X}) + \mu^2 = \sigma^2/n + \mu^2, so the second term has expectation \sigma^2 + n\mu^2. Therefore \mathbb{E}\left[\sum{i=1}^n (Xi - \bar{X})^2\right] = n\sigma^2 + n\mu^2 - \sigma^2 - n\mu^2 = (n-1)\sigma^2. Dividing by n-1 gives \mathbb{E}[S^2] = \sigma^2: un…

Why divide by n − 1 in the sample variance?

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