n阶中心距和原点矩¶

1 定义¶

n阶原点矩可以由1到n阶中心距和1阶原点矩导出 $\mu_n=E(X^n) \\=E(((X-E(X)+E(X))^n) \\ = C_{n}^0E((X-E(X))^n)+C_{n}^1E((X-E(X))^{n-1})E(X)\\+C_{n}^2E((X-E(X))^{n-2})E(X)^2+\dots+C_{n}^nE(X)^n \\ = \sum_{i=0}^nC_{n}^i v_{i}\mu_1^{n-i} \label{eq_1}$
n阶中心距可以由1到n阶原点距导出 $v_n=E((X-E(X))^n) \\ = \sum_{i=0}^n C_{n}^i(-E(X))^iE(X^{n-i}) \\ = \sum_{i=0}^n C_{n}^i (-\mu_1)^i\mu_{n-i}$

概率密度函数(pdf)， $x\sim N(\mu,\sigma^2)$ ： $p(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\exp{\left(\frac{(x-\mu)^2}{2\sigma^2}\right)}$
k阶中心距： $v_k =\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}(x-\mu)^k\exp{\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)}dx \\ = \frac{\sigma^k}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(x-\mu)^k\exp{\left(-\frac{1}{2}\frac{(x-\mu)}{\sigma}^2\right)}d\frac{(x-\mu)}{\sigma} \\ = \frac{\sigma^k}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u^k\exp{\left(-\frac{1}{2}u^2\right)}du,\because u = \frac{(x-\mu)}{\sigma}$
如果 $k$ 是奇数，上式是奇函数，则 $v_k=0$
如果 $k$ 是偶数，则上市继续化简 $v_k = \frac{\sigma^k}{\sqrt{2\pi}}\int_{-\infty}^{\infty}u^{k-1}\exp{\left(-\frac{1}{2}u^2\right)}d\frac{u^2}{2} \\ = \frac{\sqrt{2}\sigma^k}{\sqrt{\pi}}\int_{0}^{\infty}u^{k-1}\exp{\left(-\frac{1}{2}u^2\right)}d\frac{u^2}{2} \\ = \frac{\sqrt{2}\sigma^k}{\sqrt{\pi}}2^{(k-1)/2}\int_{0}^{\infty}z^{(k-1)/2}\exp{(-z)}dz \\ = \frac{\sqrt{2}\sigma^k}{\sqrt{\pi}}2^{(k-1)/2} \Gamma(\frac{k+1}{2}),(gamma函数) \\ = \sqrt{\frac{2}{\pi}}\sigma^k 2^{(k-1)/2} \frac{k-1}{2}\frac{k-3}{2}\dots\frac{1}{2}\sqrt{\pi} \\ = (k-1)!! \sigma^k$
k阶原点矩：根据 $\eqref{eq_1}$ , 可以推得原点矩