「方差的计算公式」方差迭代计算公式

方差的计算公式

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

$\begin{matrix} (1) & E_{n} = \frac{1}{n} \sum_{i = 1}^{n} x_{i} \end{matrix} E_n =\frac{1}{n} \sum_{i=1}^{n}x_i \tag{1}$ En=n1i=1∑nxi(1)

$\begin{matrix} (2) & F (n) = \sum_{i = 1}^{n} (x^{2} - E_{n}) \end{matrix} F(n) = \sum_{i=1}^{n}{(x^2-E_n)} \tag{2}$ F(n)=i=1∑n(x2−En)(2)

$\begin{matrix} (3) & V (n) = \frac{1}{n} \sum_{i = 1}^{n} (x^{2} - E_{n}) = \frac{F (n)}{n} \end{matrix} V(n) = \frac{1}{n}\sum_{i=1}^{n}{(x^2-E_n)} = \frac{F(n)}{n} \tag{3}$ V(n)=n1i=1∑n(x2−En)=nF(n)(3)

递推公式

$F (n) = \sum_{i = 1}^{n} (x_{i}^{2} - E_{n}) = \sum_{i = 1}^{n} x_{i}^{2} - 2 \sum_{i = 1}^{n} x_{i} E_{n} + n E_{n}^{2} 由 E_{n} = \frac{1}{n} \sum_{i = 1}^{n} x_{i} 可导出， n E_{n} = \sum_{i = 1}^{n} x_{i}, 故 F(n) = \sum_{i=1}^ {n}{(x_i^ 2-E_n)} = \sum_{i=1}^ {n}{x_i^ 2}-2\sum_{i=1}^ {n}{x_iE_n}+nE_n^2 \\
由E_n =\frac{1}{n} \sum_{i=1}^ {n}x_i可导出，nE_n = \sum_{i=1}^{n}x_i,故$ F(n)=i=1∑n(xi2−En)=i=1∑nxi2−2i=1∑nxiEn+nEn2由En=n1i=1∑nxi可导出，nEn=i=1∑nxi,故

$\begin{matrix} (4) & F (n) = \sum_{i = 1}^{n} x_{i}^{2} - 2 \sum_{i = 1}^{n} x_{i} E_{n} + n E_{n}^{2} = \sum_{i = 1}^{n} x_{i}^{2} - 2 n E_{n}^{2} + n E_{n}^{2} = \sum_{i = 1}^{n} x_{i}^{2} - n E_{n}^{2} \end{matrix} F(n) = \sum_{i=1}^{n}{x_i^2}-2\sum_{i=1}^{n}{x_iE_n}+nE_n^2 = \sum_{i=1}^{n}{x_i^2} - 2nE_n^2 + nE_n^2 = \sum_{i=1}^{n}{x_i^2} - nE_n^2 \tag{4}$ F(n)=i=1∑nxi2−2i=1∑nxiEn+nEn2=i=1∑nxi2−2nEn2+nEn2=i=1∑nxi2−nEn2(4)

另外,平均数的递推公式有

$\begin{matrix} (5) & n E_{n} = (n - 1) E_{n - 1} + x_{n} \end{matrix} nE_n = (n-1)E_{n-1} + x_n \tag{5}$ nEn=(n−1)En−1+xn(5)

过程推导

$\begin{matrix} F (n) - F (n - 1) & = (\sum_{i = 1}^{n} x_{i}^{2} - n E_{n}^{2}) - (\sum_{i = 1}^{n - 1} x_{i}^{2} - (n - 1) E_{n - 1}^{2}) \\ = x_{n}^{2} - n E_{n}^{2} + (n - 1) E_{n - 1}^{2} \end{matrix} \begin{aligned}
F(n)-F(n-1) &= ( \sum_{i=1}^{n}{x_i^2} - nE_n^2)-( \sum_{i=1}^{n-1}{x_i^2} -( n-1)E_{n-1}^2) \\
&=x_n^2-nE_n^2+(n-1)E_{n-1}^2 \\
\end{aligned}$ F(n)−F(n−1)=(i=1∑nxi2−nEn2)−(i=1∑n−1xi2−(n−1)En−12)=xn2−nEn2+(n−1)En−12

由（5）知， $n E_{n} = (n - 1) E_{n - 1} + x_{n} nE_n = (n-1)E_{n-1} + x_n$ nEn=(n−1)En−1+xn及 $(n - 1) E_{n - 1} = n E_{n} - x_{n} (n-1)E_{n-1} = nE_n - x_n$ (n−1)En−1=nEn−xn，则有：

$\begin{matrix} F (n) - F (n - 1) & = x_{n}^{2} - n E_{n}^{2} + (n - 1) E_{n - 1}^{2} \\ = x_{n}^{2} - E_{n} [(n - 1) E_{n - 1} + x_{n}] + E_{n - 1} (n E_{n} - x_{n}) \\ = x_{n}^{2} - n E_{n} E_{n - 1} + E_{n} E_{n - 1} - E_{n} x_{n} + n E_{n - 1} E_{n} - E_{n - 1} x_{n} \\ = x_{n}^{2} + E_{n} E_{n - 1} - E_{n} x_{n} - E_{n - 1} x_{n} \\ = (x_{n} - E_{n}) (x_{n} - E_{n - 1}) \end{matrix} \begin{aligned}
F(n)-F(n-1) &=x_n^2-nE_n^2+(n-1)E_{n-1}^2 \\
&=x_n^2-E_n[(n-1)E_{n-1}+x_n]+E_{n-1}(nE_n-x_n) \\
&= x_n^2-nE_nE_{n-1}+E_nE_{n-1}-E_nx_n+nE_{n-1}E_n-E_{n-1}x_n \\
&=x_n^2+E_nE_{n-1}-E_nx_n-E_{n-1}x_n \\
&=(x_n-E_n)(x_n-E_{n-1})
\end{aligned}$ F(n)−F(n−1)=xn2−nEn2+(n−1)En−12=xn2−En[(n−1)En−1+xn]+En−1(nEn−xn)=xn2−nEnEn−1+EnEn−1−Enxn+nEn−1En−En−1xn=xn2+EnEn−1−Enxn−En−1xn=(xn−En)(xn−En−1)

显然有 $F (1) = 0 F(1)=0$ F(1)=0

方差迭代计算公式