递推法求样本方差和标准差

问题来源：《算法(第4版)》习题1.2.18

#include 
class Accumulator
{
public:
    Accumulator()
    {
        m = 0.0;
        s = 0.0;
        n = 0;
    }
    void addDataValue(double x)
    {
        n++;
        s = s + 1.0 * (n - 1) / n * (x - m) * (x - m);
        m = m + (x - m) / n;
    }
    double mean()             //样本期望
    {
        return m;
    }
    double var()              //样本方差
    {
        return s / (n - 1);   //样本方差为什么除以的是n - 1而不是n请自行百度
    }
    double stddev()           //样本标准差
    {
        return sqrt(var());
    }
private:
    double m;
    double s;
    int n;
};

公式推导：
期望定义：
$$E_n=\frac{\sum _{i=1}^n{x}_i}{n}$$
期望的递推公式：
$$E_n=\frac{\sum _{i=1}^n{x}_i}{n}=\frac{\sum _{i=1}^{n-1}{x}_i+x_n}{n}=\frac{(n-1)E_{n-1}+x_n}{n}=E_{n-1}+\frac{x_n-E_{n-1}}{n}$$
方差定义：
$$S_n^2=\frac{\sum _{i=1}^n(x_i-E_n{)}^2}{n-1}$$
方差的递推公式推导：
为了方便，设
$$F_n=\sum _{i=1}^n(x_i-E_n{)}^2$$
$$\begin{array}{rl}{F}_n-F_{n-1}& =\sum _{i=1}^n(x_i-E_n{)}^2-\sum _{i=1}^{n-1}(x_i-E_{n-1}{)}^2\\ & =\sum _{i=1}^n(x_i-E_{n-1}+E_{n-1}-E_n{)}^2-\sum _{i=1}^{n-1}(x_i-E_{n-1}{)}^2\\ & =(x_n-E_{n-1}{)}^2+\sum _{i=1}^{n-1}(x_i-E_{n-1}{)}^2+2(E_{n-1}-E_n)\sum _{i=1}^n(x_i-E_{n-1})+\sum _{i=1}^n(E_{n-1}-E_n{)}^2-\sum _{i=1}^{n-1}(x_i-E_{n-1}{)}^2\\ & =(x_n-E_{n-1}{)}^2+2(E_{n-1}-E_n)(n{E}_n-n{E}_{n-1})+n(E_{n-1}-E_n{)}^2\end{array}$$
由期望的递推公式：
$$n{E}_n-n{E}_{n-1}=x_n-E_{n-1}$$
代入上面的式子里，得
$$\begin{array}{rl}{F}_n-F_{n-1}& =(x_n-E_{n-1}{)}^2+2(E_{n-1}-E_n)(x_n-E_{n-1})+n(E_{n-1}-E_n{)}^2\\ & =(x_n-E_{n-1}{)}^2+(E_{n-1}-E_n)(2x_n-2E_{n-1}+n{E}_{n-1}-E_n)\\ & =(x_n-E_{n-1}{)}^2+(E_{n-1}-E_n)(2x_n-2E_{n-1}-x_n+E_{n-1})\\ & =(x_n-E_{n-1}{)}^2+(E_{n-1}-E_n)(x_n-E_{n-1})\\ & =(x_n-E_{n-1})(x_n-E_{n-1}+E_{n-1}-E_n)\\ & =(x_n-E_{n-1})(x_n-E_n)\end{array}$$
所以，$$F_n=F_{n-1}+(x_n-E_{n-1})(x_n-E_n)$$
而
$$\begin{array}{rl}{x}_n-E_n& =x_n-(E_{n-1}+\frac{x_n-E_{n-1}}{n})\\ & =\frac{n-1}{n}(x_n-E_{n-1})\end{array}$$
最后，$$F_n=F_{n-1}+\frac{n-1}{n}(x_n-E_{n-1}{)}^2$$