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总体均值 μ = 1 N ∑ x i \mu = \frac{1}{N}\sum x_i μ=N1∑xi， 总体方差 σ 2 = 1 N ∑ i ( x i − μ ) 2 \sigma^2 = \frac{1}{N}\sum_i (x_i – \mu)^2 σ2=N1∑i(xi−μ)2
样本均值 x ˉ = 1 n ∑ x i \bar{x} = \frac{1}{n}\sum x_i xˉ=n1∑xi， 样本方差 S 2 = 1 n − 1 ∑ i ( x i − x ˉ ) 2 S^2 = \frac{1}{n-1}\sum_i (x_i – \bar{x})^2 S2=n−11∑i(xi−xˉ)2
证明：
E ( S 2 ) = E ( 1 n − 1 ∑ i = 1 n ( x i − x ˉ ) 2 ) = 1 n − 1 E ( ∑ i = 1 n ( x i − x ˉ ) 2 ) = 1 n − 1 E ( ∑ i = 1 n ( x i 2 − 2 x i x ˉ + x ˉ 2 ) ) = 1 n − 1 E ( ∑ i = 1 n x i 2 − n x ˉ 2 ) = 1 n − 1 ( ∑ i = 1 n E ( x i 2 ) − n E ( x ˉ 2 ) ) = 1 n − 1 ( ∑ i = 1 n E ( x i 2 ) − n E ( x ˉ 2 ) ) \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} E(S2)=E(n−11∑i=1n(xi−xˉ)2)=n−11E(∑i=1n(xi−xˉ)2)=n−11E(∑i=1n(xi2−2xixˉ+xˉ2))=n−11E(∑i=1nxi2−nxˉ2)=n−11(∑i=1nE(xi2)−nE(xˉ2))=n−11(∑i=1nE(xi2)−nE(xˉ2))
又
E ( x i 2 ) = D ( x i ) + E ( x i ) 2 = σ 2 + μ 2 E(x_i^2) = D(x_i) + E(x_i)^2 = \sigma^2 + \mu^2\\ E(xi2)=D(xi)+E(xi)2=σ2+μ2 E ( x ˉ 2 ) = D ( x ˉ ) + E ( x ˉ ) 2 = σ 2 n + μ 2 E(\bar{x}^2) = D(\bar{x}) + E(\bar{x})^2 = \frac{\sigma^2}{n} + \mu^2\\ E(xˉ2)=D(xˉ)+E(xˉ)2=nσ2+μ2
所以
E ( S 2 ) = 1 n − 1 ( ∑ i = 1 n E ( x i 2 ) − n E ( x ˉ 2 ) ) = 1 n − 1 ( n ( σ 2 + μ 2 ) − n ( σ 2 n + μ 2 ) ) = σ 2 \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} E(S2)=n−11(∑i=1nE(xi2)−nE(xˉ2))=n−11(n(σ2+μ2)−n(nσ2+μ2))=σ2
证毕。
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