样本方差是总体方差的无偏估计

总体均值 $\mu = \frac{1}{N}\sum x_i$， 总体方差 $\sigma^2 = \frac{1}{N}\sum_i (x_i - \mu)^2$

样本均值 $\bar{x} = \frac{1}{n}\sum x_i$， 样本方差 $S^2 = \frac{1}{n-1}\sum_i (x_i - \bar{x})^2$

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证明：
$\begin{array}{ll} E(S^2) &= E\left(\frac{1}{n-1}\sum_{i=1}^n (x_i - \bar{x})^2 \right) \\\\ &=\frac{1}{n-1} E\left(\sum_{i=1}^n (x_i - \bar{x})^2 \right) \\\\ &= \frac{1}{n-1} E\left(\sum_{i=1}^n (x_i^2 - 2x_i\bar{x} + \bar{x}^2) \right) \\\\ &= \frac{1}{n-1} E\left(\sum_{i=1}^n x_i^2 - n\bar{x}^2 \right) \\\\ &= \frac{1}{n-1} \left(\sum_{i=1}^n E(x_i^2) - nE(\bar{x}^2) \right) \\\\ &= \frac{1}{n-1} \left(\sum_{i=1}^n E(x_i^2) - nE(\bar{x}^2) \right) \\\\ \end{array}$

又
$E(x_i^2) = D(x_i) + E(x_i)^2 = \sigma^2 + \mu^2\\$ $E(\bar{x}^2) = D(\bar{x}) + E(\bar{x})^2 = \frac{\sigma^2}{n} + \mu^2\\$

所以
$\begin{array}{ll} E(S^2) &= \frac{1}{n-1} \left(\sum_{i=1}^n E(x_i^2) - nE(\bar{x}^2) \right) \\\\ &= \frac{1}{n-1} \left(n (\sigma^2 + \mu^2) - n(\frac{\sigma^2}{n} + \mu^2) \right) \\\\ &=\sigma^2 \end{array}$
证毕。