为什么样本方差前是n-1

(1) $\operatorname{Var}(\bar{X})=\sigma^{2} / n$ 样本均值的方差等于总体的方差除以n
(2) $E\left(S^{2}\right)=\sigma^{2}$

(1) $\operatorname{Var}(\bar{X})=\frac{1}{n^{2}} \operatorname{Var}\left(\sum_{i=1}^{n} X_{i}\right)=\frac{n \sigma^{2}}{n^{2}}=\frac{\sigma^{2}}{n}$
(2)根据样本方差的定义:
​$S^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}=\frac{1}{n-1}\left(\sum_{i=1}^{n} X_{i}^{2}-n \bar{X}^{2}\right)$
所以:

$\begin{aligned} E\left(S^{2}\right) &=E\left(\frac{1}{n-1}\left(\sum_{i=1}^{n} X_{i}^{2}-n \bar{X}^{2}\right)\right) \\ &=\frac{1}{n-1}\left(\sum_{i=1}^{n} E\left(X_{i}^{2}\right)-{n E}\left(\bar{X}^{2}\right)\right) \end{aligned}$

根据方差性质有:$\sigma^{2}=E\left(X^{2}\right)-\mu^{2} \Rightarrow E\left(X^{2}\right)=\sigma^{2}+\mu^{2}$
同理:

$E\left(\bar{X}^{2}\right)=\operatorname{Var}(\bar{X})+[E(\bar{X})]^{2} = \frac{\sigma^{2}}{n}+\mu^{2}$

样本均值的期望等于总体的期望

$\begin{aligned} E\left(S^{2}\right) &=\frac{1}{n-1}\left(\sum_{i=1}^{n} E\left(X_{i}^{2}\right) -n E\left(\bar{X}^{2}\right)\right) \\ &=\frac{1}{n-1}\left(\sum_{i=1}^{n}\left(\sigma^{2}+\mu^{2}\right)-n\left(\frac{\sigma^{2}}{n}+\mu^{2}\right)\right) \\ &=\sigma^{2} \end{aligned}$

为什么样本方差:

$S^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}$

的分母是 $n-1$ 而不是 $n$, 这是因为:

$E\left(\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}\right)=\sigma^{2}-\frac{1}{n} \sigma^{2}$

和刚才的求样本方差的期望差不多:

$\begin{aligned} \hat S=E\left(\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}\right) &\left.=E\left(\frac{1}{n}\right.\left(\sum_{i=1}^{n} X_{i}^{2}-n \bar{X}^{2}\right)\right) \\ &=\frac{1}{n}\left(\sum_{i=1}^{n} E\left(X_{i}^{2}\right)-n E\left(\bar{X}^{2}\right)\right) \\ &=\frac{1}{n}\left(\sum_{i=1}^{n}\left(\sigma^{2}+\mu^{2}\right)-n\left(\frac{\sigma^{2}}{n}+\mu^{2}\right)\right) \\ &=\sigma^{2}-\frac{1}{n} \sigma^{2} \end{aligned}$

$S^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}=\frac{n}{n-1}\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}$

为什么这么缩放

$\frac{n-1}{n} \sigma^2 \cdot \frac{n}{n-1}=\sigma^2$

如果分母为 $n$ 的话，伩样去推算出来的方差要比真实的总体方差小 $\frac{1}{n} \sigma^{2}$ 。 不过在实践中, 如果 $n$

$\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}$

来作为样本方差, 这个时候 $n$ 和 $n-1$