1 n ∑ i = 1 n ( X i − X ˉ ) 2 \frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2 n1∑i=1n(Xi−Xˉ)2 是有偏估计
1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 \frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2 n−11∑i=1n(Xi−Xˉ)2 是无偏估计 \\
随机变量 X X X的均值已知为 μ {\mu} μ，总体方差 σ 2 {\sigma^2} σ2未知。根据方差的定义，可知: σ 2 = 1 N ∑ i = 1 N ( X i − μ ) 2 {\sigma^2}=\frac{1}{N}\sum_{i=1}^{N}(X_i-\mu)^2 σ2=N1i=1∑N(Xi−μ)2 其中 N N N是总体个数，从总体中抽样 n n n个样本，对于不同的抽样结果，计算出的样本方差不等，但样本方差的期望等于总体方差，故有： σ 2 = E [ 1 n ∑ i = 1 n ( X i − μ ) 2 ] (1) \sigma^2=E[\frac{1}{n}\sum_{i=1}^{n}(X_i-\mu)^2]\tag1 σ2=E[n1i=1∑n(Xi−μ)2](1)
但实际情况下，总体的均值并不易得知，非常直接的想法就是用样本均值代替总体均值，那么替代后(1)式是否仍然成立呢？做如下推导： E [ 1 n ∑ i = 1 n ( X i − X ˉ ) 2 ] = E [ 1 n ∑ i = 1 n ( ( X i − μ ) − ( X ˉ − μ ) ) 2 ] = E [ 1 n ∑ i = 1 n ( ( X i − μ ) 2 − 2 ( X i − μ ) ( X ˉ − μ ) + ( X ˉ − μ ) 2 ) ] = E [ 1 n ∑ i = 1 n ( X i − μ ) 2 − 2 ( X ˉ − μ ) n ∑ i = 1 n ( X i − μ ) + ( X ˉ − μ ) 2 n ∑ i = 1 n 1 ] = E [ 1 n ∑ i = 1 n ( X i − μ ) 2 − 2 ( X ˉ − μ ) 2 + ( X ˉ − μ ) 2 ] = E [ 1 n ∑ i = 1 n ( X i − μ ) 2 ] − E [ ( X ˉ − μ ) 2 ] = σ 2 − E [ ( X ˉ − μ ) 2 ] = σ 2 − 1 n σ 2 \begin{aligned} E[\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2]&=E[\frac{1}{n}\sum_{i=1}^{n}((X_i-\mu)-(\bar{X}-\mu))^2]\\ &=E[\frac{1}{n}\sum_{i=1}^{n}((X_i-\mu)^2-2(X_i-\mu)(\bar{X}-\mu)+(\bar{X}-\mu)^2)]\\ &=E[\frac{1}{n}\sum_{i=1}^{n}(X_i-\mu)^2-\frac{2(\bar{X}-\mu)}{n}\sum_{i=1}^{n}(X_i-\mu)+\frac{(\bar{X}-\mu)^2}{n}\sum_{i=1}^{n}1]\\ &=E[\frac{1}{n}\sum_{i=1}^{n}(X_i-\mu)^2-2(\bar{X}-\mu)^2+(\bar{X}-\mu)^2]\\ &=E[\frac{1}{n}\sum_{i=1}^{n}(X_i-\mu)^2]-E[(\bar{X}-\mu)^2]\\ &=\sigma^2-E[(\bar{X}-\mu)^2]\\ &=\sigma^2-\frac{1}{n}\sigma^2 \end{aligned} E[n1i=1∑n(Xi−Xˉ)2]=E[n1i=1∑n((Xi−μ)−(Xˉ−μ))2]=E[n1i=1∑n((Xi−μ)2−2(Xi−μ)(Xˉ−μ)+(Xˉ−μ)2)]=E[n1i=1∑n(Xi−μ)2−n2(Xˉ−μ)i=1∑n(Xi−μ)+n(Xˉ−μ)2i=1∑n1]=E[n1i=1∑n(Xi−μ)2−2(Xˉ−μ)2+(Xˉ−μ)2]=E[n1i=1∑n(Xi−μ)2]−E[(Xˉ−μ)2]=σ2−E[(Xˉ−μ)2]=σ2−n1σ2 过程中，第四个等号替换是因为： X ˉ − μ = 1 n ∑ i = 1 n X i − 1 n ∑ i = 1 n μ = 1 n ∑ i = 1 n ( X i − μ ) \bar{X}-\mu=\frac{1}{n}\sum_{i=1}^{n}X_i-\frac{1}{n}\sum_{i=1}^{n}\mu=\frac{1}{n}\sum_{i=1}^{n}(X_i-\mu) Xˉ−μ=n1i=1∑nXi−n1i=1∑nμ=n1i=1∑n(Xi−μ) 倒数第二个等号是将式(1)代入；
最后一个等号是因为根据期望性质： E ( X ˉ ) = E ( 1 n ∑ i = 1 n X i ) = 1 n ∑ i = 1 n E ( X i ) = 1 n ∑ i = 1 n μ = μ E(\bar{X})=E(\frac{1}{n}\sum_{i=1}^nX_i)=\frac{1}{n}\sum_{i=1}^nE(X_i)=\frac{1}{n}\sum_{i=1}^n\mu=\mu E(Xˉ)=E(n1∑i=1nXi)=n1∑i=1nE(Xi)=n1∑i=1nμ=μ，可得：
E [ ( X ˉ − μ ) 2 ] = E [ ( X ˉ − E ( X ˉ ) ) 2 ] = D ( X ˉ ) = D ( 1 n ∑ i = 1 n X i ) = 1 n 2 D ( ∑ i = 1 n X i ) = 1 n 2 ∑ i = 1 n D ( X i ) = 1 n 2 ⋅ n ⋅ σ 2 = 1 n σ 2 \begin{aligned} E[(\bar{X}-\mu)^2]&=E[(\bar{X}-E(\bar{X}))^2]\\ &=D(\bar{X})\\ &=D(\frac{1}{n}\sum_{i=1}^nX_i)\\ &=\frac{1}{n^2}D(\sum_{i=1}^nX_i)\\ &=\frac{1}{n^2}\sum_{i=1}^nD(X_i)\\ &=\frac{1}{n^2}·n·\sigma^2\\ &=\frac{1}{n}\sigma^2 \end{aligned} E[(Xˉ−μ)2]=E[(Xˉ−E(Xˉ))2]=D(Xˉ)=D(n1i=1∑nXi)=n21D(i=1∑nXi)=n21i=1∑nD(Xi)=n21⋅n⋅σ2=n1σ2 因此， E ( S 2 ) = E [ 1 n ∑ i = 1 n ( X i − X ˉ ) 2 ] = σ 2 − 1 n σ 2 = n − 1 n σ 2 \begin{aligned} E(S^2)&=E[\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2]\\ &=\sigma^2-\frac{1}{n}\sigma^2\\ &=\frac{n-1}{n}\sigma^2 \end{aligned} E(S2)=E[n1i=1∑n(Xi−Xˉ)2]=σ2−n1σ2=nn−1σ2 进一步简单变形可得： σ 2 = n n − 1 E [ 1 n ∑ i = 1 n ( X i − X ˉ ) 2 ] = E [ 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 ] \begin{aligned} \sigma^2&=\frac{n}{n-1}E[\frac{1}{n}\sum_{i=1}^{n}(X_i-\bar{X})^2]\\ &=E[\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2]\\ \end{aligned} σ2=n−1nE[n1i=1∑n(Xi−Xˉ)2]=E[n−11i=1∑n(Xi−Xˉ)2] 因此，令样本方差 S 2 = 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2 S2=n−11∑i=1n(Xi−Xˉ)2，这样其期望才是总体方差 σ 2 \sigma^2 σ2 的无偏估计。
