样本方差S²中为什么是乘以1/(n-1)或者说除以n-1?贝塞尔校正，无偏估计

样本方差S²中为什么是乘以1/(n-1)或者说除以n-1?贝塞尔校正，无偏估计

前言：重在记录，可能出错。

先看样本方差的公式如下：

$$\mathit{S}{}^2=\frac{\mathbf{1}}{\mathit{n}\mathbf{-}\mathbf{1}}\sum _{\mathit{i}=1}^{\mathit{n}}(\mathit{X}{}_{\mathit{i}}-\overline{\mathit{X}}){}^2=\frac{\mathbf{1}}{\mathit{n}\mathbf{-}\mathbf{1}}\left(\sum _{\mathit{i}=1}^{\text{n}}\mathit{X}{}_{\mathit{i}}{}^2-\mathit{n}\overline{\mathit{X}}{}^2\right)$$

怎么理解这个1/(n-1)?

直观思维里，样本方差应当乘以1/n，但这里并非如此。首先，这个1/(n-1)叫做“贝塞尔校正”，它的存在可以使得样本方差更接近总体方差，也就是无偏估计。那又为什么是用n-1代替n呢？

假设待定的样本方差为：

$$\mathit{s}{}^2\mathit{=}\frac{\mathit{1}}{\mathit{n}}\mathit{\sum }_{\mathit{i}\mathit{=}\mathit{1}}^{\mathit{n}}\mathit{(}\mathit{X}{}_{\mathit{i}}\mathit{-}\overline{\mathit{X}}\mathit{)}{}^{\mathrm{2}}$$

总体方差为E(s²)，很好理解，取相当多个不同的样本方差的均值最接近总体方差

$$\begin{array}{l}\mathit{E}\left(\mathit{s}{}^2\right)\\ \mathit{=}\mathit{E}\left(\frac{\mathit{1}}{\mathit{n}}\mathit{\sum }_{\mathit{i}\mathit{=}\mathit{1}}^{\mathit{n}}\mathit{(}\mathit{X}{}_{\mathit{i}}\mathit{-}\overline{\mathit{X}}\mathit{)}{}^{\mathrm{2}}\right)\\ =\frac{1}{\mathit{n}}\mathit{E}\left(\sum _{\mathit{i}=1}^{\mathit{n}}(\mathit{X}{}_{\text{i}}{}^2-2\mathit{X}{}_{\mathit{i}}\overline{\mathit{X}}+\overline{\mathit{X}}{}^2)\right)\\ =\frac{1}{\mathit{n}}\mathit{E}\left(\sum _{\mathit{i}=1}^{\mathit{n}}\mathit{X}{}_{\text{i}}{}^2-2\overline{\mathit{X}}\sum _{\mathit{i}=1}^{\mathit{n}}\mathit{X}{}_{\mathit{i}}+\sum _{\mathit{i}=1}^{\mathit{n}}\overline{\mathit{X}}{}^2\right)\text{①}\text{代}\text{入}\sum _{\mathit{i}=1}^{\mathit{n}}\mathit{X}{}_{\mathit{i}}=\mathit{n}\overline{\mathit{X}}\\ =\frac{1}{\mathit{n}}\mathit{E}\left(\sum _{\mathit{i}=1}^{\mathit{n}}\mathit{X}{}_{\text{i}}{}^2-2\mathit{n}\overline{\mathit{X}}{}^2+\mathit{n}\overline{\mathit{X}}{}^2\right)\\ =\frac{1}{\mathit{n}}\mathit{E}\left(\sum _{\mathit{i}=1}^{\mathit{n}}\mathit{X}{}_{\text{i}}{}^2-\mathit{n}\overline{\mathit{X}}{}^2\right)\\ =\frac{1}{\mathit{n}}\mathit{E}(\sum _{\mathit{i}=1}^{\mathit{n}}\mathit{X}{}_{\text{i}}{}^2)-\mathit{E}(\overline{\mathit{X}}{}^2)\\ =\mathit{E}\left(\sum _{\mathit{i}=1}^{\mathit{n}}(\mathit{X}{}_{\text{i}}{}^2\frac{1}{\mathit{n}}\right))-(\mathit{D}\left(\overline{\mathit{X}}\right)+\mathit{E}{}^2\left(\overline{\mathit{X}}\right))\text{②}\text{代}\text{入}\text{方}\text{差}\text{计}\text{算}\text{公}\text{式}\\ =\mathit{E}\left(\mathit{X}{}^2\right)-\mathit{D}\left(\overline{\mathit{X}}\right)-\mathit{E}{}^2\left(\overline{\mathit{X}}\right)\text{③}\frac{1}{\mathit{n}}\text{是}\mathit{X}{}_{\text{i}}{}^2\text{的}\text{概}\text{率}\text{，}\text{符}\text{合}\text{期}\text{望}\text{的}\text{定}\text{义}\\ =\mathit{D}\left(\mathit{X}\right)+\mathit{E}{}^2\left(\mathit{X}\right)-\mathit{D}\left(\overline{\mathit{X}}\right)-\mathit{E}{}^2\left(\overline{\mathit{X}}\right)\text{④}\mathit{X}\text{是}\text{总}\text{体}\text{，}\mathit{D}\left(\mathit{X}\right)\text{即}\text{总}\text{体}\text{方}\text{差}\\ =\mathit{D}\left(\mathit{X}\right)-\mathit{D}\left(\overline{\mathit{X}}\right)\\ =\mathit{D}\left(\mathit{X}\right)-\frac{1}{\mathit{n}}\mathit{D}\left(\mathit{X}\right)\text{⑤}\text{总}\text{体}\text{方}\text{差}=\text{n}\text{个}\text{平}\text{均}\text{值}\text{方}\text{差之和}\\ =\frac{\mathit{n}-1}{\mathit{n}}\mathit{D}\left(\mathit{X}\right)\end{array}$$

此时求得E(s²)与D(X)存在误差，通过乘以n/(n-1)来校正，得

$$\frac{\text{n}}{\mathit{n}-1}\mathit{E}\left(\mathit{s}{}^2\right)=\mathit{D}\left(\mathit{X}\right)=\mathit{E}\left(\frac{\text{n}}{\mathit{n}-1}\mathit{s}{}^2\right)$$

令S²=n/(n-1)s²，得

$$\mathit{S}{}^2\mathit{=}\frac{\mathit{1}}{\mathit{n}\mathbf{-}\mathbf{1}}\mathit{\sum }_{\mathit{i}\mathit{=}\mathit{1}}^{\mathit{n}}\mathit{(}\mathit{X}{}_{\mathit{i}}\mathit{-}\overline{\mathit{X}}\mathit{)}{}^{\mathrm{2}}$$