总体方差为σ?,均值为μS=[(X1-X)^2+(X2-X)^2....+(Xn-X)^2]/(n-1) X表示样本均值=(X1+X2+...+Xn)/n设A=(X1-X)^2+(X2-X)^2....+(Xn-X)^2E(A)=E[(X1-X)^2+(X2-X)^2....+(Xn-X)^2]=E[(X1)^2-2X*X1+X^2+(X2)^2-2X*X2+X^2+(X2-X)^2....+(Xn)^2-2X*Xn+X^2]=E[(X1)^2+(X2)^2...+(Xn)^2+nX^2-2X*(X1+X2+...+Xn)]=E[(X1)^2+(X2)^2...+(Xn)^2+nX^2-2X*(nX)]=E[(X1)^2+(X2)^2...+(Xn)^2-nX^2]而E(Xi)^2=D(Xi)+[E(Xi)]^2=σ?+μ?E(X)^2=D(X)+[E(X)]^2=σ?/n+μ?所以E(A)=E[(X1-X)^2+(X2-X)^2....+(Xn-X)^2]=n(σ?+μ?)-n(σ?/n+μ?)=(n-1)σ?所以为了保证样本方差的无偏性S=[(X1-X)^2+(X2-X)^2....+(Xn-X)^2]/(n-1) E(S)=(n-1)σ?/(n-1)=σ?
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