方差迭代计算公式

方差迭代计算过程推导

• 术语约定
• 递推公式
• 过程推导

术语约定

$E_n =\frac{1}{n} \sum_{i=1}^{n}x_i \tag{1}$

$F(n) = \sum_{i=1}^{n}{(x^2-E_n)} \tag{2}$
$V(n) = \frac{1}{n}\sum_{i=1}^{n}{(x^2-E_n)} = \frac{F(n)}{n} \tag{3}$

递推公式

$F(n) = \sum_{i=1}^ {n}{(x_i^ 2-E_n)} = \sum_{i=1}^ {n}{x_i^ 2}-2\sum_{i=1}^ {n}{x_iE_n}+nE_n^2 \\ 由E_n =\frac{1}{n} \sum_{i=1}^ {n}x_i可导出，nE_n = \sum_{i=1}^{n}x_i,故$
$F(n) = \sum_{i=1}^{n}{x_i^2}-2\sum_{i=1}^{n}{x_iE_n}+nE_n^2 = \sum_{i=1}^{n}{x_i^2} - 2nE_n^2 + nE_n^2 = \sum_{i=1}^{n}{x_i^2} - nE_n^2 \tag{4}$
另外,平均数的递推公式有
$nE_n = (n-1)E_{n-1} + x_n \tag{5}$

过程推导

$\begin{aligned} F(n)-F(n-1) &= ( \sum_{i=1}^{n}{x_i^2} - nE_n^2)-( \sum_{i=1}^{n-1}{x_i^2} -( n-1)E_{n-1}^2) \\ &=x_n^2-nE_n^2+(n-1)E_{n-1}^2 \\ \end{aligned}$
由（5）知，$nE_n = (n-1)E_{n-1} + x_n$及$(n-1)E_{n-1} = nE_n - x_n$，则有：
$\begin{aligned} F(n)-F(n-1) &=x_n^2-nE_n^2+(n-1)E_{n-1}^2 \\ &=x_n^2-E_n[(n-1)E_{n-1}+x_n]+E_{n-1}(nE_n-x_n) \\ &= x_n^2-nE_nE_{n-1}+E_nE_{n-1}-E_nx_n+nE_{n-1}E_n-E_{n-1}x_n \\ &=x_n^2+E_nE_{n-1}-E_nx_n-E_{n-1}x_n \\ &=(x_n-E_n)(x_n-E_{n-1}) \end{aligned}$
显然有$F(1)=0$

文章来源: https://blog.csdn.net/weixin_44479136/article/details/90510374