簡(jiǎn)單說：是為了保證估計(jì)的無(wú)偏性!
1.總體方差為σ2,均值為μ
S=[(X1-X)^2+(X2-X)^2.+(Xn-X)^2]/(n-1)
X表示樣本均值=(X1+X2+...+Xn)/n
設(shè)A=(X1-X)^2+(X2-X)^2.+(Xn-X)^2
E(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=σ2+μ2
E(X)^2=D(X)+[E(X)]^2=σ2/n+μ2
所以E(A)=E[(X1-X)^2+(X2-X)^2.+(Xn-X)^2]
=n(σ2+μ2)-n(σ2/n+μ2)
=(n-1)σ2
故為了保證樣本方差的無(wú)偏性（即保證估計(jì)量的數(shù)學(xué)期望等于實(shí)際值,在此即要保證樣本方差的期望等于總體方差）,應(yīng)?。?br/>S=[(X1-X)^2+(X2-X)^2.+(Xn-X)^2]/(n-1)
從而保證：E(S)=E(A)/(n-1）=(n-1)σ2/(n-1)=σ2