定理2.2
对于简单随机抽样， y ˉ \bar{y} yˉ 的方差
V ( y ˉ ) = 1 − f N S 2 V(\bar{y})=\frac{1-f}{N}S^2 V(yˉ)=N1−fS2
证明：
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\bar{y}=\frac{1}{n}\displaystyle\sum^{N}_{i=1}{a_iY_i}
yˉ=n1i=1∑NaiYi
这里证明时需要使用方差的性质，已知两个随机变量和的方差 V ( X ± Y ) = V ( X ) + V ( Y ) ± 2 C o v ( X , Y ) V(X±Y)=V(X)+V(Y)±2Cov(X,Y) V(X±Y)=V(X)+V(Y)±2Cov(X,Y)
下面将其扩展到三个随机变量和的方差
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\begin{aligned} V(X_1+X_2+X_3)&=V(X_1)+V(X_2+X_3)+2Cov(X_1,(X_2+X_3))\\ &=V(X_1)+V(X_2+X_3)+2E(X_1(X_2+X_3))-2E(X_1)E(X_2+X_3)\\ &=V(X_1)+V(X_2+X_3)+2E(X_1X_2)+2E(X_1X_3)-2E(X_1)E(X_2)-2E(X_1)E(X_3)\\ &=V(X_1)+V(X_2)+V(X_3)+2Cov(X_1,X_2)+2Cov(X_1,X_3)+2Cov(X_2,X_3) \end{aligned}
V(X1+X2+X3)=V(X1)+V(X2+X3)+2Cov(X1,(X2+X3))=V(X1)+V(X2+X3)+2E(X1(X2+X3))−2E(X1)E(X2+X3)=V(X1)+V(X2+X3)+2E(X1X2)+2E(X1X3)−2E(X1)E(X2)−2E(X1)E(X3)=V(X1)+V(X2)+V(X3)+2Cov(X1,X2)+2Cov(X1,X3)+2Cov(X2,X3)
将其推广到
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V(\displaystyle\sum^{n}_{i=1}{X_i})=\displaystyle\sum^{n}_{i=1}{V(X_i)}+\displaystyle\sum^{n}_{i<j}{X_i,X_j}
V(i=1∑nXi)=i=1∑nV(Xi)+i<j∑nXi,Xj
所以有：
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\begin{aligned} V(\bar{y})&=V\left(\frac{1}{n}\displaystyle\sum^{N}_{i=1}{a_iY_i}\right)\\ &=\frac{1}{n^2}\displaystyle\sum^{N}_{i=1}{Y_i^2V(a_i)}+\frac{1}{n^2}[Y_1Y_2Cov(Y_1Y_2)+Y_1Y_3Cov(Y_1Y_3)+…+Y_{N-1}Y_NCov(Y_{N-1}Y_N)]\\ &=\frac{1}{n^2}\left(\displaystyle\sum^{N}_{i=1}{Y_i^2V(a_i)}+2\displaystyle\sum^{n}_{i<j}{X_i,X_j}Cov(Y_iY_j)\right)\\ &=\frac{1}{n^2}\left(\displaystyle\sum^{N}_{i=1}{Y_i^2\frac{n}{N}(1-f)}-2\displaystyle\sum^{n}_{i<j}{X_iX_j}\frac{n(1-f)}{N(N-1)}\right)\\ &=\frac{1}{n^2}\left(\frac{n(1-f)}{N}\displaystyle\sum^{N}_{i=1}{Y_i^2}-\frac{2n(1-f)}{N(N-1)}\displaystyle\sum^{n}_{i<j}{X_iX_j}\right)\\ &=\frac{1}{n^2}·\frac{n(1-f)}{N}\left(\displaystyle\sum^{N}_{i=1}{Y_i^2}-\frac{2}{N-1}\displaystyle\sum^{N}_{i<j}{Y_iY_j}\right)\\ &=\frac{1-f}{nN}\left(\frac{N}{N-1}\displaystyle\sum^{N}_{i=1}{Y_i^2}-\frac{1}{N-1}\displaystyle\sum^{N}_{i=1}{Y_i^2}-\frac{2}{N-1}\displaystyle\sum^{N}_{i<j}{Y_iY_j}\right)\\ &=\frac{1-f}{nN}\left[\frac{N}{N-1}\displaystyle\sum^{N}_{i=1}{Y_i^2}-\frac{1}{N-1}\left(\displaystyle\sum^{N}_{i=1}{Y_i^2}+2\displaystyle\sum^{N}_{i<j}{Y_iY_j}\right)\right]\\ &=\frac{1-f}{nN}·\frac{N}{N-1}\left(\displaystyle\sum^{N}_{i=1}{Y_i^2}-\frac{1}{N}\left(\displaystyle\sum^{N}_{i=1}{Y_i^2}\right)^2\right)\\ &=\frac{1-f}{n(N-1)}\left(\displaystyle\sum^{N}_{i=1}{Y_i^2}-N\left(\frac{1}{N}\displaystyle\sum^{N}_{i=1}{Y_i^2}\right)^2\right)\\ &=\frac{1-f}{n(N-1)}\left(\displaystyle\sum^{N}_{i=1}{Y_i^2}-N\bar{Y}^2\right)\\ &=\frac{1-f}{n}·\frac{1}{N-1}\displaystyle\sum^{N}_{i=1}{(Y_i-\bar{Y})^2}\\ &=\frac{1-f}{n}S^2\\ \end{aligned}
V(yˉ)=V(n1i=1∑NaiYi)=n21i=1∑NYi2V(ai)+n21[Y1Y2Cov(Y1Y2)+Y1Y3Cov(Y1Y3)+…+YN−1YNCov(YN−1YN)]=n21(i=1∑NYi2V(ai)+2i<j∑nXi,XjCov(YiYj))=n21(i=1∑NYi2Nn(1−f)−2i<j∑nXiXjN(N−1)n(1−f))=n21(Nn(1−f)i=1∑NYi2−N(N−1)2n(1−f)i<j∑nXiXj)=n21⋅Nn(1−f)(i=1∑NYi2−N−12i<j∑NYiYj)=nN1−f(N−1Ni=1∑NYi2−N−11i=1∑NYi2−N−12i<j∑NYiYj)=nN1−f[N−1Ni=1∑NYi2−N−11(i=1∑NYi2+2i<j∑NYiYj)]=nN1−f⋅N−1N⎝⎛i=1∑NYi2−N1(i=1∑NYi2)2⎠⎞=n(N−1)1−f⎝⎛i=1∑NYi2−N(N1i=1∑NYi2)2⎠⎞=n(N−1)1−f(i=1∑NYi2−NYˉ2)=n1−f⋅N−11i=1∑N(Yi−Yˉ)2=n1−fS2
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