對(duì)總體X進(jìn)行n次抽樣,得到X1,X2,……,Xn
平均值X`=(X1+X2+...+Xn)/n
X方差的無(wú)偏估計(jì)量為：
S(n-1) = [(X1-X`)^2+(X2-X`)^2+...+(Xn-X`)^2]/(n-1)
證明如下：
E[Xi^2] = [EX]^2 + DX
E[X`] = EX D[X`] = DX/n
E[X`^2] = [EX]^2 + DX/n
E[Xi·X`] = E[Xi^2]/n + (n-1)[EX]^2/n
E[S(n-1)] = [ 1/(n-1) ] · { nE[Xi^2] - 2nE[X`·Xi] + nE[X`^2] }
= [ 1/(n-1) ] · n · [ (n-1)DX/n ]
= DX