为什么样本的方差和样本的二阶中心矩并不一样

问题引入：

很多同学会很奇怪一个问题，那就是为什么样本的二阶中心矩$m_2=\frac{1}{n}{\sum }_{i=1}^n(X_i-\overline{X}{)}^2$和样本的方差$S^2=\frac{1}{n-1}{\sum }_{i=1}^n(X_i-\overline{X}{)}^2$相差一个常数因子$\frac{n-1}{n}$，即
$$m_2=\frac{n-1}{n}{S}^2$$
这就涉及到估计量的无偏性问题。

---

1.发现问题

首先我们假设总体分布的期望和方差分别为$\mu$和${\sigma }^2$
下面我们来看看问题所在。
首先我们假设$S^2$和$m_2$一样为$\frac{1}{n}{\sum }_{i=1}^n(X_i-\overline{X}{)}^2$，那么有如下推导：
$$\begin{gathered} S^2=\frac{1}{n}\sum _{i=1}^n(X_i-\overline{X}{)}^2=\frac{1}{n}\sum _{i=1}^n(X_i-\mu +\mu -\overline{X}{)}^2 \\ =\frac{1}{n}\sum _{i=1}^n[(X_i-\mu {)}^2+2(X_i-\mu )(\mu -\overline{X})+(\mu -\overline{X}{)}^2] \\ =\frac{1}{n}\sum _{i=1}^n(X_i-\mu {)}^2+\frac{2}{n}\sum _{i=1}^n(X_i-\mu )(\mu -\overline{X})+\frac{1}{n}\sum _{i=1}^n(\mu -\overline{X}{)}^2 \\ =\frac{1}{n}\sum _{i=1}^n(X_i-\mu {)}^2+\frac{2}{n}(\mu -\overline{X})\sum _{i=1}^n(X_i-\mu )+\frac{1}{n}\sum _{i=1}^n(\mu -\overline{X}{)}^2 \end{gathered}$$
注意此时 $\because {\sum }_{i=1}^n{X}_i={\sum }_{i=1}^n\overline{X}$，$\therefore$ 继续推导有：
$$\begin{gathered} S^2=\frac{1}{n}\sum _{i=1}^n(X_i-\mu {)}^2+\frac{2}{n}(\mu -\overline{X})\sum _{i=1}^n(\overline{X}-\mu )+\frac{1}{n}\sum _{i=1}^n(\mu -\overline{X}{)}^2 \\ =\frac{1}{n}\sum _{i=1}^n(X_i-\mu {)}^2-\frac{2}{n}\sum _{i=1}^n(\mu -\overline{X}{)}^2+\frac{1}{n}\sum _{i=1}^n(\mu -\overline{X}{)}^2 \\ =\frac{1}{n}\sum _{i=1}^n(X_i-\mu {)}^2-\frac{1}{n}\sum _{i=1}^n(\mu -\overline{X}{)}^2 \\ =Var(X_i)-Var(\overline{X})={\sigma }^2-Var(\overline{X}) \end{gathered}$$
再次打断，$\because$方差有如下性质：
1.若$c$常数，则$Var(cX)=c^{2}Var(X)$[推导很简单，不解释了]
2.$Var(X_1+X_2+...+X_n)=Var(X_1)+Var(X_2)+...+Var(X_n)$
且每个样本$X_i$都可视为一个随机变量，其分布同于总体分布，因此其方差也同于总体方差
$$\begin{gathered} \therefore Var(\overline{X})=Var(\frac{1}{n}\sum _{i=1}^n{X}_i)=\frac{1}{n^2}Var(\sum _{i=1}^n{X}_i) \\ =\frac{1}{n^2}\sum _{i=1}^{n}Var(X_i)=\frac{1}{n^2}\cdot n\cdot Var(X_i)=\frac{1}{n}{\sigma }^2 \end{gathered}$$
$\therefore$ 继续推导有：
$$S^2={\sigma }^2-Var(\overline{X})={\sigma }^2-\frac{1}{n}{\sigma }^2=\frac{n-1}{n}{\sigma }^2$$
至此就会发现用$m_2$来估计${\sigma }^2$是不准确的。

---

2.证明问题

okay，那么接下来我们需要证明样本方差$S^2=\frac{1}{n-1}{\sum }_{i=1}^n(X_i-\overline{X}{)}^2$是总体分布方差${\sigma }^2$的 无偏估计。
$$\begin{gathered} \sum _{i=1}^n(X_i-\overline{X}{)}^2=\sum _{i=1}^n[(X_i-\mu )-(\overline{X}-\mu ){]}^2 \\ =\sum _{i=1}^n(X_i-\mu {)}^2-2(\overline{X}-\mu )\sum _{i=1}^n(X_i-\mu )+\sum _{i=1}^n(\overline{X}-\mu {)}^2 \\ =\sum _{i=1}^n(X_i-\mu {)}^2-2(\overline{X}-\mu )\sum _{i=1}^n(X_i-\mu )+n(\overline{X}-\mu {)}^2 \end{gathered}$$
注意此时 $\because {\sum }_{i=1}^n(X_i-\mu )=n(\overline{X}-\mu )$，$\therefore$继续有：
$$\begin{gathered} \sum _{i=1}^n(X_i-\overline{X}{)}^2=\sum _{i=1}^n(X_i-\mu {)}^2-2n(\overline{X}-\mu {)}^2+n(\overline{X}-\mu {)}^2 \\ =\sum _{i=1}^n(X_i-\mu {)}^2-n(\overline{X}-\mu {)}^2 \end{gathered}$$
$\because$ 样本均值是总体分布均值的无偏估计
$\therefore E(X_i)=E(\overline{X})=\mu$
$\therefore$有：
$$\begin{gathered} E[\sum _{i=1}^n(X_i-\mu {)}^2]=\sum _{i=1}^{n}E(X_i-\mu {)}^2=\sum _{i=1}^{n}Var(X_i)=n{\sigma }^2 \\ E[n(\overline{X}-\mu {)}^2]=nE(\overline{X}-\mu {)}^2=nVar(\overline{X})=n\cdot \frac{1}{n^2}\sum _{i=1}^{n}Var(X_i)=n\cdot \frac{n}{n^2}\cdot {\sigma }^2 \\ ={\sigma }^2 \end{gathered}$$
因此最终有：
$$\begin{gathered} E(S^2)=\frac{1}{n-1}E(\sum _{i=1}^n(X_i-\overline{X}{)}^2)=\frac{1}{n-1}E(\sum _{i=1}^n(X_i-\mu {)}^2-n(\overline{X}-\mu {)}^2) \\ =\frac{1}{n-1}(n{\sigma }^2-{\sigma }^2)={\sigma }^2 \end{gathered}$$
至此说明了样本方差$S^2=\frac{1}{n-1}{\sum }_{i=1}^n(X_i-\overline{X}{)}^2$是总体分布方差${\sigma }^2$的 无偏估计。