定理2.5

定理2.5

简单随机样本的协方差

$S_{yx}=\frac{1}{n-1}\sum _{i=1}^n(y_i-\bar{y})(x_i-\bar{x})$

是总体协方差$S_{yx}$的无偏估计

证明：

对于总体中每个单元$Y_i$,引入示性变量$a_i$
$$a_i=\{\begin{array}{ll}1& \text{若}{Y}_i\text{入样}\\ 0& \text{若}{Y}_i\text{不入样}\end{array}$$
由引理2.2可知：$\sum _{i=1}^N{a}_i{Y}_i=\sum _{i=1}^n{y}_i,E(a_i)=\frac{n}{N},E(a_i^2)=\frac{n}{N},E(a_i{a}_j)=\frac{n(n-1)}{N(N-1)}$
$$\begin{array}{rl}E(s_{yx})& =E\left(\frac{1}{n-1}\sum _{i=1}^n(x_i-\bar{x})(y_i-\bar{y})\right)\\ & =\frac{1}{n-1}E\left(\sum _{i=1}^n(x_i{y}_i-\bar{x}{y}_i-\bar{y}{x}_i+\bar{x}\bar{y})\right)\\ & =\frac{1}{n-1}E\left(\sum _{i=1}^n{x}_i{y}_i-\bar{x}\sum _{i=1}^n{y}_i-\bar{y}\sum _{i=1}^n{x}_i+\sum _{i=1}^n\bar{x}\bar{y}\right)\\ & =\frac{1}{n-1}E\left(\sum _{i=1}^n{x}_i{y}_i-n\bar{x}\bar{y}-n\bar{x}\bar{y}+2n\bar{x}\bar{y}\right)\\ & =\frac{1}{n-1}E\left(\sum _{i=1}^n{x}_i{y}_i-n\bar{x}\bar{y}\right)\\ & =\frac{1}{n-1}E\left[\sum _{i=1}^N{X}_i{Y}_i{a}_i^2-n\left(\frac{1}{n}\sum _{i=1}^n{x}_i\right)\left(\frac{1}{n}\sum _{i=1}^n{y}_i\right)\right]\\ & =\frac{1}{n-1}E\left(\sum _{i=1}^N{X}_i{Y}_i{a}_i^2-\frac{1}{n}\sum _{i=1}^n{x}_i\sum _{i=1}^n{y}_i\right)\\ & =\frac{1}{n-1}\left[\sum _{i=1}^N{X}_i{Y}_i{a}_i^2-\frac{1}{n}\left(\sum _{i=1}^n{x}_i{y}_i+\sum _{i\ne j}^n{x}_i{y}_i\right)\right]\\ & =\frac{1}{n-1}\left[\sum _{i=1}^N{X}_i{Y}_i{a}_i^2-\frac{1}{n}\left(\sum _{i=1}^N{X}_i{Y}_i{a}_i^2+\sum _{i\ne j}^N{X}_i{Y}_i{a}_i{a}_j\right)\right]\\ & =\frac{1}{n-1}\left(E(a_i^2)\sum _{i=1}^N{X}_i{Y}_i-\frac{1}{n}E(a_i^2)\sum _{i=1}^N{X}_i{Y}_i-\frac{1}{n}E(a_i{a}_j)\sum _{i\ne j}^n{X}_i{Y}_i\right)\\ & =\frac{1}{n-1}\left(\frac{n}{N}\sum _{i=1}^N{X}_i{Y}_i-\frac{1}{N}\sum _{i=1}^N{X}_i{Y}_i-\frac{n-1}{N(N-1)}\sum _{i\ne j}^n{X}_i{Y}_i\right)\\ & =\frac{1}{n-1}\left[\frac{n-1}{N}\sum _{i=1}^N{X}_i{Y}_i-\frac{n-1}{N(N-1)}\left(\sum _{i=1}^N{X}_i\sum _{i=1}^N{Y}_i-\sum _{i=1}^N{X}_i{Y}_i\right)\right]\\ & =\frac{1}{n-1}\left[\left(\frac{n-1}{N}+\frac{n-1}{N(N-1)}\right)\sum _{i=1}^N{X}_i{Y}_i-\frac{n-1}{N(N-1)}\sum _{i=1}^N{X}_i\sum _{i=1}^N{Y}_i\right]\\ & =\frac{1}{n-1}\left(\frac{n-1}{N-1}\sum _{i=1}^N{X}_i{Y}_i-\frac{n-1}{N(N-1)}N\bar{X}\cdot N\bar{Y}\right)\\ & =\frac{1}{N-1}\left(\sum _{i=1}^N{X}_i{Y}_i-N\bar{X}\bar{Y}\right)\\ & =\frac{1}{N-1}\sum _{i=1}^N(X_i-\bar{X})(Y_i-\bar{Y})\\ & =S_{yx}\end{array}$$