1、问题描述：假设有一批独立同分布的样本{x i , i = 1 , 2 , 3... , n − 1 , n {x_i,i=1,2,3...,n-1,n} xi,i=1,2,3...,n−1,n}。
标准方差的公式为：
S 2 = 1 n ∑ i = 1 n − 1 ( x i − x ‾ ) 2 S^{2} = \frac{1}{n}\sum_{i=1}^{n-1}(x_i - \overline{x})^{2} S2=n1i=1∑n−1(xi−x)2
,其中 x ‾ = 1 n ∑ i = 1 n x i \overline{x}=\frac{1}{n}\sum_{i=1}^{n}x_i x=n1∑i=1nxi；
另外假设均值 E ( x i ) = μ E(x_i) = \mu E(xi)=μ,方差 D ( x i ) = σ 2 D(x_i) = \sigma^{2} D(xi)=σ2；
证明：标准方差是方差的无偏估计。
【分析】：要想证明标准方差是方差的无偏估计，只需证明 E ( S 2 ) = σ 2 = D ( x i ) E(S^2) = \sigma^{2} = D(x_i) E(S2)=σ2=D(xi)，其中比较麻烦的地方是 x i ∗ x ‾ x_i * \overline{x} xi∗x，所以要把 x ‾ \overline{x} x中的 x i x_i xi分离处理。
2、证明开始：
E ( S 2 ) = E ( 1 n − 1 ∑ i = 1 n ( x i − x ‾ ) 2 ) = 1 n − 1 E ( ∑ i = 1 n ( x i − 1 n x i − 1 n ∑ j ≠ i , j = 1 n x j ) ) = 1 n − 1 E ( ∑ i = 1 n ( ( n − 1 ) 2 n 2 x i 2 − 2 ( n − 1 ) n 2 x i ∗ ∑ j ≠ i , j = 1 n x j + 1 n 2 ( ∑ j ≠ i , j = 1 n x j ) 2 ) ) \begin{aligned} E(S^{2}) &= E(\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \overline{x})^{2}) \\ &= \frac{1}{n-1}E(\sum_{i=1}^{n}(x_i - \frac{1}{n}x_i - \frac{1}{n}\sum_{j\neq{i},j=1}^{n}x_j))\\ &= \frac{1}{n-1}E(\sum_{i=1}^{n}(\frac{(n-1)^2}{n^2}x_i^{2} - \frac{2(n-1)}{n^{2}}x_i *\sum_{j\neq{i},j=1}^{n}x_j + \frac{1}{n^2}(\sum_{j\neq{i},j=1}^{n}x_j)^{2})) \\ \end{aligned} E(S2)=E(n−11i=1∑n(xi−x)2)=n−11E(i=1∑n(xi−n1xi−n1j=i,j=1∑nxj))=n−11E(i=1∑n(n2(n−1)2xi2−n22(n−1)xi∗j=i,j=1∑nxj+n21(j=i,j=1∑nxj)2))
下面将 E ( ⋅ ) E(\centerdot) E(⋅)中的三部分分别计算，按照从左往右的顺序，系数 1 n − 1 \frac{1}{n-1} n−11暂时不带，最后算完结果一起带入。
第一部分：
E ( ∑ i = 1 n ( ( n − 1 ) 2 n 2 ) x i 2 ) = ( n − 1 ) 2 n 2 ∑ i = 1 n E ( x i 2 ) = ( n − 1 ) 2 n 2 ∑ i = 1 n ( E 2 ( x i ) + D ( x i ) ) = ( n − 1 ) 2 n 2 ∑ i = 1 n ( μ 2 + σ 2 ) = ( n − 1 ) 2 n ( μ 2 + σ 2 ) \begin{aligned} E(\sum_{i=1}^{n}(\frac{(n-1)^2}{n^2})x_i^{2}) &= \frac{(n-1)^2}{n^2}\sum_{i=1}^{n}E(x_i^{2}) \\ &= \frac{(n-1)^2}{n^2}\sum_{i=1}^{n}(E^2(x_i) + D(x_i)) \\ &= \frac{(n-1)^2}{n^2}\sum_{i=1}^{n}(\mu^{2} + \sigma^{2}) \\ &= \frac{(n-1)^2}{n}(\mu^{2} + \sigma^{2}) \end{aligned} E(i=1∑n(n2(n−1)2)xi2)=n2(n−1)2i=1∑nE(xi2)=n2(n−1)2i=1∑n(E2(xi)+D(xi))=n2(n−1)2i=1∑n(μ2+σ2)=n(n−1)2(μ2+σ2)
第二部分：
E ( ∑ i = 1 n ( 2 ( n − 1 ) n 2 x i ∗ ∑ j ≠ i , j = 1 n x j ) ) = 2 ( n − 1 ) n 2 ∑ i = 1 n E ( x i ∗ ∑ j ≠ i , j = 1 n x j ) ) = 2 ( n − 1 ) n 2 ∑ i = 1 n ( E ( x i ) ∗ ∑ j ≠ i , j = 1 n E ( x j ) ) = 2 ( n − 1 ) n 2 ∑ i = 1 n ( μ ∗ ( n − 1 ) μ ) = 2 ( n − 1 ) n 2 ∗ n ∗ ( n − 1 ) ∗ μ 2 = 2 ( n − 1 ) 2 ∗ μ 2 n \begin{aligned} E(\sum_{i=1}^{n}( \frac{2(n-1)}{n^{2}}x_i *\sum_{j\neq{i},j=1}^{n}x_j)) &= \frac{2(n-1)}{n^{2}}\sum_{i=1}^{n}E(x_i *\sum_{j\neq{i},j=1}^{n}x_j)) \\ &= \frac{2(n-1)}{n^{2}}\sum_{i=1}^{n}(E(x_i) * \sum_{j\neq{i},j=1}^{n}E(x_j)) \\ &= \frac{2(n-1)}{n^{2}}\sum_{i=1}^{n}(\mu * (n-1)\mu) \\ &= \frac{2(n-1)}{n^{2}} * n*(n-1)*\mu^{2} \\ & = \frac{2(n-1)^{2}*\mu^{2}}{n} \end{aligned} E(i=1∑n(n22(n−1)xi∗j=i,j=1∑nxj))=n22(n−1)i=1∑nE(xi∗j=i,j=1∑nxj))=n22(n−1)i=1∑n(E(xi)∗j=i,j=1∑nE(xj))=n22(n−1)i=1∑n(μ∗(n−1)μ)=n22(n−1)∗n∗(n−1)∗μ2=n2(n−1)2∗μ2
第三部分：
E ( ∑ i = 1 n ( 1 n 2 ( ∑ j ≠ i , j = 1 n x j ) 2 ) ) = 1 n 2 ∑ i = 1 n E ( ( ( ∑ j ≠ i , j = 1 n x j ) 2 ) ) = 1 n 2 ∑ i = 1 n ( E 2 ( ∑ j ≠ i , j = 1 n x j ) + D ( ∑ j ≠ i , j = 1 n x j ) ) = 1 n 2 ∑ i = 1 n ( ( ∑ j ≠ i , j = 1 n E 2 ( x j ) ) + ∑ j ≠ i , j = 1 n D ( x j ) ) = 1 n 2 ∑ i = 1 n ( ( ( ∑ j ≠ i , j = 1 n μ ) 2 ) + ∑ j ≠ i , j = 1 n σ 2 ) = 1 n 2 ∑ i = 1 n ( ( n − 1 ) 2 ∗ μ 2 + ( n − 1 ) ∗ σ 2 ) = ( n − 1 ) 2 n 2 ∗ n ∗ μ 2 + n − 1 n 2 ∗ n σ 2 = ( n − 1 ) 2 n ∗ μ 2 + n − 1 n σ 2 \begin{aligned} E(\sum_{i=1}^{n}(\frac{1}{n^2}(\sum_{j\neq{i},j=1}^{n}x_j)^{2})) &= \frac{1}{n^2}\sum_{i=1}^{n}E(((\sum_{j\neq{i},j=1}^{n}x_j)^{2})) \\ &= \frac{1}{n^2}\sum_{i=1}^{n}(E^{2}(\sum_{j\neq{i},j=1}^{n}x_j) +D(\sum_{j\neq{i},j=1}^{n}x_j)) \\ &= \frac{1}{n^2}\sum_{i=1}^{n}((\sum_{j\neq{i},j=1}^{n}E^{2}(x_j))+\sum_{j\neq{i},j=1}^{n}D(x_j)) \\ &= \frac{1}{n^2}\sum_{i=1}^{n}(((\sum_{j\neq{i},j=1}^{n}\mu)^{2})+\sum_{j\neq{i},j=1}^{n}\sigma^{2}) \\ &= \frac{1}{n^2}\sum_{i=1}^{n}((n-1)^{2}*\mu^2+(n-1)*\sigma^2) \\ &= \frac{(n-1)^2}{n^2} *n *\mu^2+\frac{n-1}{n^2} *n\sigma^2\\ &= \frac{(n-1)^2}{n} *\mu^2+\frac{n-1}{n} \sigma^2 \end{aligned} E(i=1∑n(n21(j=i,j=1∑nxj)2))=n21i=1∑nE(((j=i,j=1∑nxj)2))=n21i=1∑n(E2(j=i,j=1∑nxj)+D(j=i,j=1∑nxj))=n21i=1∑n((j=i,j=1∑nE2(xj))+j=i,j=1∑nD(xj))=n21i=1∑n(((j=i,j=1∑nμ)2)+j=i,j=1∑nσ2)=n21i=1∑n((n−1)2∗μ2+(n−1)∗σ2)=n2(n−1)2∗n∗μ2+n2n−1∗nσ2=n(n−1)2∗μ2+nn−1σ2
终于到了相加的时候了，将1-3部分的内容带上系数相加如下：
( n − 1 ) E ( ⋅ ) = ( n − 1 ) 2 n ( μ 2 + σ 2 ) − 2 ( n − 1 ) 2 ∗ μ 2 n + ( n − 1 ) 2 n ∗ μ 2 + n − 1 n σ 2 = ( n − 1 ) 2 n σ 2 + n − 1 n σ 2 = n 2 − n n σ 2 = ( n − 1 ) σ 2 \begin{aligned} (n-1)E(\centerdot) & =\frac{(n-1)^2}{n}(\mu^{2} + \sigma^{2}) \ - \frac{2(n-1)^{2}*\mu^{2}}{n} + \frac{(n-1)^2}{n} *\mu^2+\frac{n-1}{n} \sigma^2 \\ &= \frac{(n-1)^2}{n}\sigma^{2} + \frac{n-1}{n} \sigma^2 \\ &= \frac{n^2-n}{n}\sigma^{2} \\ &= (n-1)\sigma^{2} \end{aligned} (n−1)E(⋅)=n(n−1)2(μ2+σ2)−n2(n−1)2∗μ2+n(n−1)2∗μ2+nn−1σ2=n(n−1)2σ2+nn−1σ2=nn2−nσ2=(n−1)σ2
所以
E ( ⋅ ) = σ 2 E(\centerdot) = \sigma^2 E(⋅)=σ2
得证。
注：标准方差前面的系数 1 n − 1 \frac{1}{n-1} n−11是为了让其是方差的无偏估计而加上的。
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