﻿ 计算古典概型高阶原点矩的新方法 A New Method for Calculating the High Order Origin Moments of Classical Probability Models

Advances in Applied Mathematics
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

Copyright © 2018 by authors and Hans Publishers Inc.

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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