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The Property of Transition Probability and Its Application
DOI: 10.12677/AAM.2019.89176, PDF, 下载: 1,159  浏览: 4,791

Abstract: Transition probability is an important computational tool in Markov chain. This paper derives two important properties of transition probability, and discusses the Polya jar model by using these properties, which leads to an important Recursive equation.

1. 引言

2. 定义和性质

2.1. 相关定义

$P\left({X}_{n+1}=j|{X}_{n}=i,{X}_{n-1}={i}_{n-1},\cdots ,{X}_{2}={i}_{2},{X}_{1}={i}_{1},{X}_{0}={i}_{0}\right)=P\left({X}_{n+1}=j|{X}_{n}=i\right)$ (1)

2.2. 转移概率的性质

$\begin{array}{l}P\left({X}_{n}={i}_{n},{X}_{n-1}={i}_{n-1},\cdots ,{X}_{1}={i}_{1},{X}_{0}={i}_{0}\right)\\ =P\left({X}_{n}={i}_{n}|{X}_{n-1}={i}_{n-1},\cdots ,{X}_{1}={i}_{1},{X}_{0}={i}_{0}\right)P\left({X}_{n-1}={i}_{n-1},\cdots ,{X}_{1}={i}_{1},{X}_{0}={i}_{0}\right)\\ =P\left({X}_{n}={i}_{n}|{X}_{n-1}={i}_{n-1}\right)P\left({X}_{n-1}={i}_{n-1}|{X}_{n-2}={i}_{n-2}\cdots ,{X}_{1}={i}_{1},{X}_{0}={i}_{0}\right)P\left({X}_{n-2}={i}_{n-2}\cdots ,{X}_{1}={i}_{1},{X}_{0}={i}_{0}\right)\\ =P\left({X}_{n}={i}_{n}|{X}_{n-1}={i}_{n-1}\right)P\left({X}_{n-1}={i}_{n-1}|{X}_{n-2}={i}_{n-2}\right)P\left({X}_{n-2}={i}_{n-2}\cdots ,{X}_{1}={i}_{1},{X}_{0}={i}_{0}\right)\\ =\cdots =\underset{k=1}{\overset{n}{\prod }}P\left({X}_{k}={i}_{k}|{X}_{k-1}={i}_{k-1}\right)P\left({X}_{0}={i}_{0}\right)\end{array}$

$\begin{array}{c}P\left({X}_{n+1}=j\right)=P\left\{\left({X}_{n+1}=j\right)\cap \left(\underset{k\in S}{\cup }{X}_{n}=k\right)\right\}=\underset{k\in S}{\sum }P\left\{\left({X}_{n+1}=j\right)\cap \left({X}_{n}=k\right)\right\}\\ =\underset{k\in S}{\sum }P\left({X}_{n+1}=j|{X}_{n}=k\right)P\left({X}_{n}=k\right)\end{array}$

3. 应用

Polya罐子模型是概率论中的一个重要模型， [1] 对该模型问题从古典概率角度作了详细的论述。下面，我们从转移概率的角度探讨该模型相关问题。

$P\left({X}_{n+1}=j|{X}_{n}=i\right)=\left\{\begin{array}{l}\frac{i}{b+r+nc},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}j=i+c\\ 1-\frac{i}{b+r+nc},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }j=i\\ 0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}其它\end{array}$

$P\left({X}_{n+1}=k+c|{X}_{n}=k\right)=\frac{k}{b+r+nc}$

$P\left({X}_{n}=k|{X}_{n-1}=k\right)=1-\frac{k}{b+r+\left(n-1\right)c}=\frac{b+r+\left(n-1\right)c-k}{b+r+\left(n-1\right)c}$

$P\left({X}_{n+1}=k+c|{X}_{n}=k+c\right)=1-\frac{k+c}{b+r+nc}=\frac{b+r+\left(n-1\right)c-k}{b+r+nc}$

$P\left({X}_{n}=k+c|{X}_{n-1}=k\right)=\frac{k}{b+r+\left(n-1\right)c}$

$\begin{array}{l}P\left({X}_{n+1}=k+c|{X}_{n}=k+c\right)P\left({X}_{n}=k+c|{X}_{n-1}=k\right)\\ =P\left({X}_{n+1}=k+c|{X}_{n}=k\right)P\left({X}_{n}=k|{X}_{n-1}=k\right)\end{array}$ (2)

(2)说明，罐中黑球数一定时，概率与取黑球白球的次序无关。依次类推，交换黑球白球次序，使前 ${n}_{1}$ 次全为黑球，后 ${n}_{2}$ 次为红球，则

$\begin{array}{l}P\left(前{n}_{1}次全为黑球,\text{\hspace{0.17em}}后{n}_{2}次全为红球\right)=P\left({X}_{1}=b+c\right){P}_{12}{P}_{23}\cdots {P}_{{n}_{1}-1{n}_{1}}{P}_{{n}_{1}{n}_{1}+1}\cdots {P}_{n-1n}\\ =\frac{b}{b+r}\frac{b+c}{b+r+c}\cdots \frac{b+\left({n}_{1}-1\right)c}{b+r+\left({n}_{1}-1\right)c}\frac{r}{b+r+{n}_{1}c}\cdots \frac{r+\left({n}_{2}-1\right)c}{b+r+\left(n-1\right)c}\triangleq P\end{array}$

$P\left(恰好取出{n}_{1}个黑球,\text{\hspace{0.17em}}{n}_{2}为红球\right)={C}_{n}^{{n}_{1}}P$

${P}_{k}\left(n+1\right)=P\left({X}_{n+1}=b+kc\right)$

$\begin{array}{c}{P}_{k}\left(n+1\right)=P\left({X}_{n+1}=b+kc\right)\\ =P\left({X}_{n+1}=b+kc|{X}_{n}=b+kc\right)P\left({X}_{n}=b+kc\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+P\left({X}_{n+1}=b+kc|{X}_{n}=b+\left(k-1\right)c\right)P\left({X}_{n}=b+\left(k-1\right)c\right)\\ ={P}_{k}\left(n\right)\left(1-\frac{b+kc}{b+r+nc}\right)+{P}_{k-1}\left(n\right)\frac{b+\left(k-1\right)c}{b+r+nc}\\ ={P}_{k}\left(n\right)\frac{r+\left(n-k\right)c}{b+r+nc}+{P}_{k-1}\left(n\right)\frac{b+\left(k-1\right)c}{b+r+nc}\end{array}$

 [1] 费勒. 概率论及其应用[M]. 第3版. 北京: 人民邮电出版社, 2006: 91-95. [2] 茆诗松, 程依明, 濮晓龙. 概率论与数理统计教程[M]. 第2版. 北京: 高等教育版社, 2017: 43-44. [3] Ross, S.M. 应用随机过程概率模型导论[M]. 第11版. 北京: 人民邮电出版社, 2018: 153-154. [4] 朱庆峰, 王煜. 一类古典概率问题的解法探究[J]. 高等数学研究, 2017, 20(1): 92-95.