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。