无偏估计
f(X1,X2,……)是g(θ)的无偏估计E(f(X1,X2,……))=g(θ)f(X_1,X_2,……)是g(θ)的无偏估计\\ E(f(X_1,X_2,……))=g(θ)f(X1,X2,……)是g(θ)的无偏估计E(f(X1,X2,……))=g(θ)
样本均值方差的无偏估计
{xˉ=∑1nxin均值,E(xˉ)=μS2=∑1n(xˉ−xi)2n−1方差,E(S2)=σ2 \begin{cases} \bar{x}=\frac{ \sum_{1}^{n} x_{i} }{n}& {均值,E( \bar{x} )=μ}\\ S^2= \frac{ \sum_{1}^{n} ( \bar{x}- x_{i} )^2 }{n-1} & {方差,E( S^2 )=σ ^2} \end{cases}{xˉ=n∑1nxiS2=n−1∑1n(xˉ−xi)2均值,E(xˉ)=μ方差,E(S2)=σ2
求样本方差为什么除n-1 ?
除n−1使样本方差作为总体方差σ2的无偏话计量n−1与自由度https://baike.baidu.com/item/行对应 除n-1使样本方差作为总体方差σ^2的无偏话计量\\n-1 与自由度
https://baike.baidu.com/item/%E8%87%AA%E7%94%B1%E5%BA%A6/5936984?fr=aladdin
行对应除n−1使样本方差作为总体方差σ2的无偏话计量n−1与自由度https://baike.baidu.com/item/行对应
1n∑i=1n(Xi−Xˉ)2=1n∑i=1nXi2−Xˉ2E(1n∑i=1n(Xi−Xˉ)2)=E(1n∑i=1nXi2−Xˉ2)=E(1n∑i=1nXi2)−E(Xˉ2)=E(1n∑i=1nXi2)−(D(Xˉ)+E(Xˉ)2)=(σ2+μ2)−(σ2n+μ2)=n−1nσ2
\frac{1}{n}\sum_{i=1}^n(X_i-\bar{X})^2 =\frac{1}{n}\sum_{i=1}^{n} X_i^2-\bar{X}^2\\
E( \frac{1}{n}\sum_{i=1}^n(X_i-\bar{X})^2 )=E( \frac{1}{n}\sum_{i=1}^{n} X_i^2-\bar{X}^2 )\\
=E( \frac{1}{n}\sum_{i=1}^{n} X_i^2)-E(\bar{X}^2 )\\
=E( \frac{1}{n}\sum_{i=1}^{n} X_i^2)-(D(\bar{X} )+ E(\bar{X} )^2 )\\
=(σ ^2+μ^2)-(\frac{σ ^2}{n}+μ^2)=\frac{n-1}{n} σ ^2
n1i=1∑n(Xi−Xˉ)2=n1i=1∑nXi2−Xˉ2E(n1i=1∑n(Xi−Xˉ)2)=E(n1i=1∑nXi2−Xˉ2)=E(n1i=1∑nXi2)−E(Xˉ2)=E(n1i=1∑nXi2)−(D(Xˉ)+E(Xˉ)2)=(σ2+μ2)−(nσ2+μ2)=nn−1σ2
E(S2)=E(1n−1∑i=1n(Xi−Xˉ)2)=σ2 E(S^2)=E(\frac{1}{n-1} \sum_{i=1}^n(X_i-\bar{X})^2 )=σ ^2 E(S2)=E(n−11i=1∑n(Xi−Xˉ)2)=σ2