逐次过松弛算法及块数值迭代解在马尔可夫链平稳分布中的应用

A. Sunday. O, Ayinde Semiu A
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引用次数: 1

摘要

系统的演化是由从一种状态到下一种状态的转换来表示的,系统的物理或数学行为也可以通过定义它可能处于的所有众多状态并演示它如何在它们之间移动来描述。本文研究了马尔可夫链平稳分布的迭代求解方法,该方法从解向量的初始估计开始,然后每一步或迭代使其越来越接近真实解。,还涉及矩阵运算,如与一个或多个向量的乘法,这使转换矩阵保持不变并节省时间。我们的目标是使用连续超松弛算法和块数值迭代解方法来计算解。借助一些现有的马尔可夫链定律、定理和公式,使用归一化原理和矩阵运算,如下矩阵、上矩阵和对角矩阵。通过举例得到平稳分布向量π^((k+1))={■(π_1^((k+1))&π_2^((k+1))&■(π_3^((k+1))&π_4^((k+1))&π_5^((k+1))))),k=0,1,2,…,n},取初始平稳解为π^((0))=(■(0.2&0.2&■(0.2&0.2&0.2 &0.2)))^T,观察到后续所有迭代得到的结果与π^((1))完全相同,这表明块迭代法只需一次迭代就能得到完全机器精度的解。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
On the Application of Successive Over-relaxation Algorithmic and Block Numerical Iterative Solutions for the Stationary Distribution in Markov Chain
The evolution of a system is represented by transitions from one state to the next, and the system's physical or mathematical behavior can also be depicted by defining all of the numerous states it can be in and demonstrating how it moves between them. In this study, the iterative solution methods for the stationary distribution of Markov chains were investigated, which start with an initial estimate of the solution vector and then alter it in such a way that it gets closer and closer to the genuine solution with each step or iteration., and also involved matrices operations such as multiplication with one or more vectors, which leaves the transition matrices unchanged and saves time. Our goal is to use Successive Overrelaxation Algorithmic and Block Numerical Iterative Solution Methods to compute the solutions. With the help of some existing Markov chain laws, theorems, and formulas, the normalization principle and matric operations such as lower, upper, and diagonal matrices are used. The stationary distribution vector’s π^((k+1) )={■(π_1^((k+1) )&π_2^((k+1) )&■(π_3^((k+1) )&π_4^((k+1) )&π_5^((k+1) ) )),k=0,1,2,…,n} are obtained for the illustrative examples, taken the initial stationary solution to be π^((0) )=(■(0.2&0.2&■(0.2&0.2&0.2)))^T and it was observed that all subsequent iterations yield exactly the same result as π^((1) ), and this shows that, the block iterative method requires only a single iteration to obtain the solution to full machine precision.
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