The expectation of the sample variance

Topic: Statistical inference · more questions on this topic
‹ Which of these are statistics? A squared t variable ›

Let X_1,\ldots,X_n be a random sample with mean \mu and variance \sigma^2. Show that \sum_i (X_i-\overline X)^2=\sum_i (X_i-\mu)^2-n(\overline X-\mu)^2, and deduce that \mathbb{E} s^2=\sigma^2.

Worked answer (click to show)

Write X_i-\overline X=(X_i-\mu)-(\overline X-\mu) and expand:

\sum_{i=1}^n(X_i-\overline X)^2=\sum_{i=1}^n(X_i-\mu)^2-2(\overline X-\mu)\sum_{i=1}^n(X_i-\mu)+n(\overline X-\mu)^2.

Since \sum(X_i-\mu)=n(\overline X-\mu), the middle term is -2n(\overline X-\mu)^2, which leaves \sum(X_i-\mu)^2-n(\overline X-\mu)^2.

Now \mathbb{E}(X_i-\mu)^2=\sigma^2 and \mathbb{E}(\overline X-\mu)^2=\operatorname{Var}\overline X=\sigma^2/n. Hence

\mathbb{E}\sum_{i=1}^n(X_i-\overline X)^2=n\sigma^2-n\cdot\frac{\sigma^2}{n}=(n-1)\sigma^2,

and \mathbb{E} s^2=\sigma^2.