总体方差的无偏估计

样本平均值average： $\overline{X}=\frac{1}{n}\sum_{i=1}^nx_i$

样本方差variance： $S^2$

总体均值（期望）mean： $\mu$

总体方差variance： $\sigma^2$

$\begin{array}{lcl} E((\overline{X}-\mu)^2) \\ =E((\frac{1}{n}\sum_{i=1}^nx_i-\mu)^2) \\ =E((\frac{1}{n}\sum_{i=1}^n(x_i-\mu))^2) \\ =E((\frac{1}{n}\sum_{i=1}^n(x_i-\mu))^2) \\ =\frac{1}{n}.(E(\frac{1}{n}\sum_{i=1}^n(x_i-\mu)^2)+\frac{1}{n}.E(\sum_{i\ne j}(x_i-\mu)(x_j-\mu)) \end{array}$

$\because 样本是从同一总体独立同分布抽样，协方差为0$

$\therefore E(\sum_{i\ne j}(x_i-\mu)(x_j-\mu)) = 0$

$\begin{array}{lcl} =\frac{1}{n}\sigma^2+\frac{1}{n}.E(\sum_{i\ne j}(x_i-\mu)(x_j-\mu)) \\ =\frac{1}{n}\sigma^2 \end{array}$

样本方差

$\begin{array}{lcl} E(S^2) \\ =E(\frac{1}{n}\sum_{i=1}^{n} (x_i-\overline{X})^2) \\ = E(\frac{1}{n}\sum_{i=1}^n((x_i-\mu)-(\overline{X}-\mu))^2) \\ =E(\frac{1}{n}\sum_{i=1}^n((x_i-\mu)^2-2(x_i-\mu)(\overline{X}-\mu)+(\overline{X}-\mu)^2)) \\ =E(\frac{1}{n}\sum_{i=1}^n(x_i-\mu)^2-\frac{1}{n}\sum_{i=1}^n2(x_i-\mu)(\overline{X}-\mu)+\frac{1}{n}\sum_{i=1}^n(\overline{X}-\mu)^2) \end{array}$

中心极限定理，足够样本量情况下，样本的均值趋于总体的期望
$\because$

$\begin{array}{lcl} \frac{1}{n}\sum_{i=1}^n(x_i-\mu) \\ = \frac{1}{n}\sum_{i=1}^nx_i-\mu \\ =\overline{X}-\mu \end{array}$

$E(\overline{X})=\mu$

$\therefore$

$\begin{array}{lcl} E(S^2) \\ = E(\frac{1}{n}\sum_{i=1}^n(x_i-\mu)^2-2(\overline{X}-\mu)(\overline{X}-\mu)+(\overline{X}-\mu)^2) \\ =E(\frac{1}{n}\sum_{i=1}^n(x_i-\mu)^2-(\overline{X}-\mu)^2) \\ =E(\frac{1}{n}\sum_{i=1}^n(x_i-\mu)^2)-E((\overline{X}-\mu)^2) \\ =Var(X)-Var(\overline{X}) \\ =\sigma^2-\frac{1}{n}\sigma^2 \\ = \frac{n-1}{n}\sigma^2 \end{array}$

总体方差的无偏估计

$\begin{array}{lcl} E(S^2)\\ =E(\frac{1}{n-1}\sum_{i=1}^{n} (x_i-\overline{X})^2) \\ =\frac{n}{n-1}.E(\frac{1}{n}\sum_{i=1}^{n} (x_i-\overline{X})^2) \\ =\frac{n}{n-1}.\frac{n-1}{n}\sigma^2 \\ = \sigma^2 \end{array}$

总体方差的无偏估计

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