Vol.07 No.05(2018), Article ID:25063,9 pages
10.12677/AAM.2018.75069

A New Method for Calculating the High Order Origin Moments of Classical Probability Models

Jie Xia, Wenqing Wu, Xiaohui Li, Zhiping Wang, Haiyang Lan

School of Science, Southwest University of Science and Technology, Mianyang Sichuan

Received: Apr. 30th, 2018; accepted: May 18th, 2018; published: May 25th, 2018

ABSTRACT

Due to the mathematical complexity of calculating the high order origin moments of classical probabilistic models directly by definition, two methods for calculating the higher order moments are given in this paper. In the method one, considering the expansion of the expression (1 + n)m+1, and applying the mathematical induction method, we obtained the recursive expression of the higher order moments. In the method two, thanks to the $\sum _{i=1}^{n}{i}^{m}$ can be represented by the representation of m + 1 polynomials, it can be further transformed into an m elements linear system. The expression of the classical model can be found by solving the system of linear equations. Finally, we use the dice test and the English alphabetic experiment as examples to show how to use the two methods.

Keywords:Classical Probability Model, High Order Origin Moment, Recursion Formula, Linear System of Equations

1. 引言

2. 问题的提出

 随机变量X的一阶原点矩：

$\begin{array}{l}E\left[X\right]=\frac{p}{n}×1+\frac{p}{n}×2+\frac{p}{n}×3+\cdots +\frac{p}{n}×n\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{p}{n}\sum _{i=1}^{n}i\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{p}{n}×\frac{n\left(n+1\right)}{2}.\end{array}$

 随机变量X的二阶原点矩：

$\begin{array}{l}E\left[{X}^{2}\right]=\frac{p}{n}×{1}^{2}+\frac{p}{n}×{2}^{2}+\frac{p}{n}×{3}^{2}+\cdots +\frac{p}{n}×{n}^{2}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{p}{n}\sum _{i=1}^{n}{i}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{p}{n}×\frac{n\left(n+1\right)\left(2n+1\right)}{6}.\end{array}$

 随机变量X的三阶原点矩：

$\begin{array}{l}E\left[{X}^{3}\right]=\frac{p}{n}×{1}^{3}+\frac{p}{n}×{2}^{3}+\frac{p}{n}×{3}^{3}+\cdots +\frac{p}{n}×{n}^{3}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{p}{n}\sum _{i=1}^{n}{i}^{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{p}{n}×{\left(\frac{n\left(n+1\right)}{2}\right)}^{2}.\end{array}$

 随机变量X的m阶原点矩：

$\begin{array}{l}E\left[{X}^{m}\right]=\frac{p}{n}×{1}^{m}+\frac{p}{n}×{2}^{m}+\frac{p}{n}×{3}^{m}+\cdots +\frac{p}{n}×{n}^{m}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{p}{n}\sum _{i=1}^{n}{i}^{m}.\end{array}$

3. 表达式 $\sum _{i=1}^{n}{i}^{m}$ 的计算

3.1. 方法1：数学归纳法

${\left(1+n\right)}^{m+1}=\left(\begin{array}{c}m+1\\ 0\end{array}\right){n}^{m+1}+\left(\begin{array}{c}m+1\\ 1\end{array}\right){n}^{m}+\left(\begin{array}{c}m+1\\ 2\end{array}\right){n}^{m-1}+\cdots +\left(\begin{array}{c}m+1\\ m+1\end{array}\right){n}^{0}$ (1)

${\left(1+n\right)}^{m+1}-{n}^{m+1}=\left(\begin{array}{c}m+1\\ 1\end{array}\right){n}^{m}+\left(\begin{array}{c}m+1\\ 2\end{array}\right){n}^{m-1}+\cdots +\left(\begin{array}{c}m+1\\ m+1\end{array}\right){n}^{0}$ (2)

$\left\{\begin{array}{l}{\left(1+n\right)}^{m+1}-{n}^{m+1}=\left(\begin{array}{c}m+1\\ 1\end{array}\right){n}^{m}+\left(\begin{array}{c}m+1\\ 2\end{array}\right){n}^{m-1}+\cdots +\left(\begin{array}{c}m+1\\ m+1\end{array}\right){n}^{0},\\ {n}^{m+1}-{\left(n-1\right)}^{m+1}=\left(\begin{array}{c}m+1\\ 1\end{array}\right){\left(n-1\right)}^{m}+\left(\begin{array}{c}m+1\\ 2\end{array}\right){\left(n-1\right)}^{m-1}+\cdots +\left(\begin{array}{c}m+1\\ m+1\end{array}\right){\left(n-1\right)}^{0},\\ {\left(n-1\right)}^{m+1}-{\left(n-2\right)}^{m+1}=\left(\begin{array}{c}m+1\\ 1\end{array}\right){\left(n-2\right)}^{m}+\left(\begin{array}{c}m+1\\ 2\end{array}\right){\left(n-2\right)}^{m-1}+\cdots +\left(\begin{array}{c}m+1\\ m+1\end{array}\right){\left(n-2\right)}^{0},\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⋮\\ {2}^{m+1}-{1}^{m+1}=\left(\begin{array}{c}m+1\\ 1\end{array}\right){1}^{m}+\left(\begin{array}{c}m+1\\ 2\end{array}\right){1}^{m-1}+\cdots +\left(\begin{array}{c}m+1\\ m+1\end{array}\right){1}^{0}.\end{array}$ (3)

${\left(1+n\right)}^{m+1}-1=\left(\begin{array}{c}m+1\\ 1\end{array}\right)\sum _{i=1}^{n}{i}^{m}+\left(\begin{array}{c}m+1\\ 2\end{array}\right)\sum _{i=1}^{n}{i}^{m-1}+\cdots +\left(\begin{array}{c}m+1\\ m+1\end{array}\right)\sum _{i=1}^{n}{i}^{0}$ (4)

$\begin{array}{l}{\left(1+n\right)}^{m+1}-1=\left(m+1\right){A}_{m}+\sum _{k=2}^{m+1}\left(\begin{array}{c}m+1\\ k\end{array}\right){A}_{m+1-k}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(m+1\right){A}_{m}+\sum _{k=0}^{m-1}\left(\begin{array}{c}m+1\\ m+1-k\end{array}\right){A}_{k}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(m+1\right){A}_{m}+\sum _{k=0}^{m-1}\left(\begin{array}{c}m+1\\ k\end{array}\right){A}_{k}\end{array}$ (5)

$\begin{array}{l}{A}_{m}=\frac{{\left(1+n\right)}^{m+1}-1-\sum _{k=0}^{m-1}\left(\begin{array}{c}m+1\\ k\end{array}\right){A}_{k}}{m+1},\\ m=1,2,�\end{array}$ (6)

$\left\{\begin{array}{l}{A}_{0}=n,\\ {A}_{1}=\sum _{i=1}^{n}i=\frac{n\left(n+1\right)}{2},\\ {A}_{2}=\sum _{i=1}^{n}{i}^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6},\\ {A}_{3}={\left(\frac{n\left(n+1\right)}{2}\right)}^{2}.\end{array}$ (7)

${A}_{m}$ 的通项公式可知，有初值和递推表达式，则 ${A}_{m}$ 的值即可求解。

3.2. 方法2：线性方程组

${\int }_{0}^{n}{x}^{2}\text{d}x<\sum _{i=1}^{n}{i}^{2}<{\int }_{0}^{n}\left({x}^{2}+1\right)\text{d}x$ (8)

$\frac{\text{1}}{\text{3}}{n}^{\text{3}}<\sum _{i=1}^{n}{i}^{\text{2}}<\frac{\text{1}}{\text{3}}{\left(n+1\right)}^{\text{3}}$ (9)

$\sum _{i=1}^{n}{i}^{2}\text{=}a{n}^{3}+b{n}^{2}+dn+d$ (10)

Figure 1. The relationship between $\sum _{i=1}^{n}{i}^{2}$ and ${\int }_{0}^{n}{x}^{2}\text{d}x$ and ${\int }_{0}^{n}{\left(x+1\right)}^{2}\text{d}x$

$\left\{\begin{array}{l}a+b+c+d=1\\ 8a+4b+2c+d=5\\ 27a+9b+3c+d=14\\ 64a+16b+4c+d=30\end{array}$ (11)

$\sum _{i=1}^{n}{i}^{2}=\frac{{n}^{3}}{3}+\frac{{n}^{2}}{2}+\frac{n}{6}$ (12)

$\frac{\text{1}}{m+1}{n}^{m+1}<\sum _{i=1}^{n}{i}^{m}<\frac{\text{1}}{m+1}{\left(n+1\right)}^{m+1}$ (13)

$\sum _{i=1}^{n}{i}^{m}\text{=}\sum _{j=0}^{m+1}{a}_{j}{n}^{j}$ (14)

$\left\{\begin{array}{c}{a}_{\text{0}}{1}^{m+1}+{a}_{\text{1}}{1}^{m}+\cdots +{a}_{m}1+{a}_{m+\text{1}}=\sum _{i=1}^{1}{i}^{m}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\\ {a}_{\text{0}}{2}^{m+1}+{a}_{\text{1}}{2}^{m}+\cdots +{a}_{m}2+{a}_{m+\text{1}}=\sum _{i=1}^{2}{i}^{m}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\\ \begin{array}{cccccccccccc}& & & & & ⋮& & & & & & \end{array}\\ {a}_{\text{0}}{\left(m+2\right)}^{m+1}+{a}_{\text{1}}{\left(m+2\right)}^{m}+\cdots +{a}_{m}\left(m+2\right)+{a}_{m+\text{1}}=\sum _{i=1}^{m+2}{i}^{m}\end{array}$ (15)

${a}_{j}=\frac{{D}_{i}}{D}，j=0,1,\cdots ,m+1.$ (16)

${D}_{j}=\left(\begin{array}{cccccc}{1}^{m+1}& {1}^{m}& \cdots & \sum _{i=1}^{1}{i}^{m}& \cdots & 1\\ {2}^{m+1}& {2}^{m}& & \sum _{i=1}^{2}{i}^{m}& & 1\\ {3}^{m+1}& {3}^{m}& & \sum _{i=1}^{3}{i}^{m}& & 1\\ ⋮& & \ddots & ⋮& \ddots & ⋮\\ {\left(m+1\right)}^{m+1}& {\left(m+1\right)}^{m}& \dots & \sum _{i=1}^{m+2}{i}^{m}& \dots & 1\end{array}\right),\text{}D=\left(\begin{array}{cccc}{1}^{m+1}& {1}^{m}& \cdots & 1\\ {2}^{m+1}& {2}^{m}& & 1\\ {3}^{m+1}& {3}^{m}& & 1\\ ⋮& & \ddots & ⋮\\ {\left(m+1\right)}^{m+1}& {\left(m+1\right)}^{m}& \cdots & 1\end{array}\right)$

4. 实例分析

1) 掷骰子实验

$\begin{array}{l}E\left[X\right]=\frac{21}{6},E\left[{X}^{2}\right]=\frac{91}{6},E\left[{X}^{3}\right]=\frac{441}{6},\\ E\left[{X}^{4}\right]=\frac{2275}{6},E\left[{X}^{5}\right]=\frac{12201}{6},E\left[{X}^{6}\right]=\frac{67171}{6}.\end{array}$

$n=6$ 时， ${A}_{1}=21，{A}_{2}=91，{A}_{3}=441$ ，可根据方法一式(6)可计算 ${A}_{4},{A}_{5},{A}_{6}$

${A}_{4}=\frac{{7}^{5}-1-\sum _{k=0}^{3}\left(\begin{array}{c}5\\ k\end{array}\right){A}_{k}}{5}=2276$

${A}_{5}=\frac{{7}^{6}-1-\sum _{k=0}^{4}\left(\begin{array}{c}6\\ k\end{array}\right){A}_{k}}{6}=12201$

${A}_{6}=\frac{{7}^{7}-1-\sum _{k=0}^{5}\left(\begin{array}{c}7\\ k\end{array}\right){A}_{k}}{7}=67171$

$\begin{array}{l}E\left[X\right]=\frac{1}{6}×{A}_{1}=\frac{21}{6},E\left[{X}^{2}\right]=\frac{1}{6}×{A}_{2}=\frac{91}{6},E\left[{X}^{3}\right]=\frac{1}{6}×{A}_{3}=\frac{441}{6},\\ E\left[{X}^{4}\right]=\frac{1}{6}×{A}_{4}=\frac{2275}{6},E\left[{X}^{5}\right]=\frac{1}{6}×{A}_{5}=\frac{12201}{6},E\left[{X}^{6}\right]=\frac{1}{6}×{A}_{6}=\frac{67171}{6}.\end{array}$

${A}_{1}=\sum _{i=1}^{n}i=\frac{{n}^{2}}{2}+\frac{n}{2},$

${A}_{2}=\sum _{i=1}^{n}{i}^{2}=\frac{{n}^{3}}{3}+\frac{{n}^{2}}{2}+\frac{n}{6},$

${A}_{3}=\sum _{i=1}^{n}{i}^{3}=\frac{{n}^{4}}{4}+\frac{{n}^{3}}{2}+\frac{{n}^{2}}{4},$

${A}_{4}=\sum _{i=1}^{n}{i}^{4}=\frac{{n}^{5}}{5}+\frac{{n}^{4}}{2}+\frac{{n}^{3}}{3}-\frac{n}{30},$

${A}_{5}=\sum _{i=1}^{n}{i}^{5}=\frac{{n}^{6}}{6}+\frac{{n}^{5}}{2}+\frac{5{n}^{4}}{12}-\frac{{n}^{2}}{12},$

${A}_{6}=\sum _{i=1}^{n}{i}^{6}=\frac{{n}^{7}}{7}+\frac{{n}^{6}}{2}+\frac{{n}^{5}}{2}-\frac{{n}^{3}}{6}+\frac{n}{42}.$

$n=6$ 时， ${A}_{1}=21，{A}_{2}=91，{A}_{3}=441,{A}_{4}=2275,{A}_{5}=12201,{A}_{6}=67171.$ 因此，可计算 $E\left[X\right]=\frac{21}{6},$ $E\left[{X}^{2}\right]=\frac{91}{6},$ $E\left[{X}^{3}\right]=\frac{441}{6},$ $E\left[{X}^{4}\right]=\frac{2275}{6},$ $E\left[{X}^{5}\right]=\frac{12201}{6},$ $E\left[{X}^{6}\right]=\frac{67171}{6}.$

2) 选英文字母实验

$\begin{array}{l}E\left[X\right]=\frac{27}{2},E\left[{X}^{2}\right]=\frac{477}{2},E\left[{X}^{3}\right]=\frac{9477}{2},\\ E\left[{X}^{4}\right]=\frac{200817}{2},E\left[{X}^{5}\right]=\frac{2216039}{2},E\left[{X}^{6}\right]=\frac{50299889}{2}.\end{array}$

$n=26$ 时， ${A}_{1}=351,{A}_{2}=6201,{A}_{3}=123201$ ，可根据方法一式(6)可计算 ${A}_{4},{A}_{5},{A}_{6}$

${A}_{4}=\frac{{27}^{5}-1-\sum _{k=0}^{3}\left(\begin{array}{c}5\\ k\end{array}\right){A}_{k}}{5}=2610621$

${A}_{5}=\frac{{27}^{6}-1-\sum _{k=0}^{4}\left(\begin{array}{c}6\\ k\end{array}\right){A}_{k}}{6}=\text{57617001}$

${A}_{6}=\frac{{27}^{7}-1-\sum _{k=0}^{5}\left(\begin{array}{c}7\\ k\end{array}\right){A}_{k}}{7}=\text{1307797101}$

$\begin{array}{l}E\left[X\right]=\frac{1}{26}×{A}_{1}=\frac{27}{2},E\left[{X}^{2}\right]=\frac{1}{26}×{A}_{2}=\frac{477}{2},E\left[{X}^{3}\right]=\frac{1}{26}×{A}_{3}=\frac{9477}{2},\\ E\left[{X}^{4}\right]=\frac{1}{26}×{A}_{4}=\frac{200817}{2},E\left[{X}^{5}\right]=\frac{1}{26}×{A}_{5}=\frac{2216039}{2},E\left[{X}^{6}\right]=\frac{1}{26}×{A}_{6}=\frac{50299889}{2}.\end{array}$

$n=26$ 时，因此 ${A}_{1}=351，$ ${A}_{2}=6201，$ ${A}_{3}=\text{123201}，$ ${A}_{4}=\text{2610621}，$ ${A}_{5}=\text{57617001}，$ ${A}_{6}=\text{1307797101}\text{.}$ ，可计算 $E\left[X\right]=\frac{27}{2},$ $E\left[{X}^{2}\right]=\frac{477}{2},$ $E\left[{X}^{3}\right]=\frac{9477}{2},$ $E\left[{X}^{4}\right]=\frac{200817}{2},$ $E\left[{X}^{5}\right]=\frac{2216039}{2},$ $E\left[{X}^{6}\right]=\frac{50299889}{2}.$

5. 结论

A New Method for Calculating the High Order Origin Moments of Classical Probability Models[J]. 应用数学进展, 2018, 07(05): 584-592. https://doi.org/10.12677/AAM.2018.75069

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