随机变量的方差和样本方差为什么不一样?

1个回答

  • 1.总体方差为σ2,均值为μ

    S=[(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)^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故为了保证样本方差的无偏性(即保证估计量的数学期望等于实际值,在此即要保证样本方差的期望等于总体方差),应取: