/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Consider a Markov chain with sta... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 27

Consider a Markov chain with states \(0,1,2,3,4\). Suppose \(P_{0,4}=1\); and suppose that when the chain is in state \(l, i>0\), the next state is equally likely to be any of the states \(0,1, \ldots, i-1\). Find the limiting probabilities of this Markov chain.

Short Answer

Expert verified
The limiting probabilities (stationary distribution) of this Markov chain are \(\pi_0 = 4/5\), \(\pi_1 = 2/5\), \(\pi_2 = 2/5\), \(\pi_3 = 2/5\), and \(\pi_4 = 3/5\). This means, in the long run, the chain will be in state 0 with probability \(4/5\), in state 1 with probability \(2/5\), in state 2 with probability \(2/5\), in state 3 with probability \(2/5\), and in state 4 with probability \(3/5\).

Step by step solution

01

Set up the transition probability matrix

Based on the given exercise, we can write out the transition probability matrix P as follows: \( P =\begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 1/2 & 0 & 1/2 & 0 & 0 \\ 1/3 & 1/3 & 0 & 1/3 & 0 \\ 1/4 &1/4 &1/4 &0 &1/4 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \)
02

Find the stationary distribution

Let \(\Pi = (\pi_0, \pi_1, \pi_2, \pi_3, \pi_4)\) be the stationary distribution (limiting probabilities) of the given Markov chain. We are looking for the vector \(\Pi\) such that: 1. \(\Pi P = \Pi\) 2. \(\sum_{i=0}^{4}\pi_i = 1\) To find the stationary distribution, we can solve the linear system of equations resulting from the conditions above.
03

Solve the linear system of equations

Condition (1) gives us the following system of linear equations: \( \begin{cases} -\pi_0 + 1/2 \pi_1 + 1/3 \pi_2 + 1/4 \pi_3 = 0 \\ \pi_0 -\pi_1 + 1/3 \pi_2 + 1/4 \pi_3 = 0 \\ 1/2 \pi_1 - \pi_2 + 1/4 \pi_3 = 0 \\ 1/3 \pi_1 + 1/3 \pi_2 - \pi_4 = 0 \end{cases} \) Using condition (2), we have: \(\pi_0 + \pi_1 + \pi_2 + \pi_3 + \pi_4 = 1\) Now we have 5 linear equations with 5 unknowns. We can solve the system of equations using any suitable method (e.g., substitution, Gaussian elimination). By solving the linear system, we get: \(\Pi = (4/5, 2/5, 2/5, 2/5, 3/5)\)
04

Interpret the results

The limiting probabilities (stationary distribution) of the given Markov chain are as follows: \( \begin{cases} \pi_0 = 4/5 \\ \pi_1 = 2/5 \\ \pi_2 = 2/5 \\ \pi_3 = 2/5 \\ \pi_4 = 3/5 \end{cases} \) This means, in the long run, the Markov chain will be in state 0 with probability 4/5, in state 1 with probability 2/5, in state 2 with probability 2/5, in state 3 with probability 2/5, and in state 4 with probability 3/5.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Markov chain is said to be a tree process if (i) \(P_{U}>0\) whenever \(P_{y}>0\). (ii) for every pair of states \(i\) and \(j, i \neq j\), there is a unique sequence of distinct states \(i=i_{0}, i_{1}, \ldots, i_{n-1}, i_{n}=j\) such that $$ P_{h \cdot h+1}>0, \quad k=0,1, \ldots, n-1 $$ In other words, a Markov chain is a tree process if for every pair of distinet states \(i\) and \(j\) there is a unique way for the process to go from \(i\) to / without reentering a state (and this path is the reverse of the unique path from \(j\) to \(l\) ). Argue that an ergodic tree process is time reversible.

Specify the classes of the following Markov chains, and determine whether they are transient or recurrent: $$ \begin{aligned} &\mathbf{P}_{1}=\left|\begin{array}{ccc} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array}\right| & \mathbf{P}_{2}=\left|\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 1 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right| \\ &\mathbf{P}_{3}=\left\|\begin{array}{lllll} \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{array}\right\| & \mathbf{P}_{4}=\mid \begin{array}{lllll} \frac{1}{2} & \frac{1}{4} & 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & \vdots & 3 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array} \| \end{aligned} $$

It follows from Theorem \(4.2\) that for a time reversible Markov chain $$ P_{U} P_{j k} P_{k i}=P_{4} P_{k /} P_{j 1}, \quad \text { for all } l, j, k $$ It turns out that if the state space is finite and \(P_{U}>0\) for all \(t, J\), then the preceding is also a sufficient condition for time reversibility. (That is, in this case, we need only check Equation \((4.26)\) for paths from \(i\) to \(i\) that have only two intermediate states.) Prove this. Fix \(i\) and show that the equations $$ \pi_{j} P_{j k}=n_{k} P_{k /} $$ are satisfied by \(\pi_{j}=c P_{U} / P_{\mu}\), where \(c\) is chosen so that \(\sum_{j} \pi_{j}=1 .\)

(a) Show that the limiting probabilities of the reversed Markov chain are the same as for the forward chain by showing that they satisfy the equations $$ \pi_{j}=\sum_{i} \pi_{i} Q_{u} $$ (b) Give an intuitive explanation for the result of part (a).

Let \(\pi_{i}\) denote the long-run proportion of time a given Markov chain is in state \(i\). (a) Explain why \(\pi_{i}\) is also the proportion of transitions that are into state \(i\) as well as being the proportion of transitions that are from state \(i .\) (b) \(\pi_{i} P_{U}\) represents the proportion of transitions that satisfy what property? (c) \(\sum_{i} \pi_{i} P_{U}\) represent the proportion of transitions that satisfy what property? (d) Using the preceding explain why $$ \pi_{j}=\sum_{i} \pi_{i} P_{U} $$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.