/*! 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 41 Let \(\pi_{i}\) denote the long-... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 41

Let \(\pi_{i}\) denote the long-run proportion of time a given irreducible 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) \(\pi_{i} P_{i j}\) represents the proportion of transitions that satisfy what property? (c) \(\sum_{i} \pi_{i} P_{i j}\) represent the proportion of transitions that satisfy what property? (d) Using the preceding explain why $$ \pi_{j}=\sum_{i} \pi_{i} P_{i j} $$

Short Answer

Expert verified
The long-run proportion of time, 蟺_i, represents the fraction of time spent in state i in the long run. It can be thought of as the steady-state probability of being in state i. Both the proportion of transitions into state i and the proportion of transitions from state i will be equal to 蟺_i in the long run. The term 蟺_i P_ij represents the fraction of transitions from state i to state j in the long run. The term 鈭慱i (蟺_i P_ij) is the long-run proportion of transitions that end up in state j. Since the long-run proportions of transitions into state j must be equal to the long-run proportions of transitions that end up in state j, the balance condition equation is: \[ \pi_{j}=\sum_{i} \pi_{i} P_{i j} \]

Step by step solution

01

(a) Proportion of transitions

The long-run proportion of time 蟺_i represents the fraction of time the Markov chain spends in state i in the long run. It can be thought of as the steady-state probability of being in state i. Since the Markov chain keeps transitioning between states, this proportion 蟺_i essentially represents the probability of being in state i at any given time in the long run. Now, since the chain is spending a proportion 蟺_i of its time in state i, it naturally implies that both the proportion of transitions into state i and the proportion of transitions from state i will be same as 蟺_i in the long run. This occurs in order to maintain that equilibrium state (steady-state) in which the proportion of time in state i remains constant over time.
02

(b) Proportion of transitions and the property

The term 蟺_i P_ij represents the proportion of transitions that have two specific properties: the chain being in state i and transitioning from state i to state j. In other words, this term represents the fraction of transitions in the Markov chain where the chain transits from state i to state j in the long run.
03

(c) Proportion of transitions and the property

The term 鈭慱i (蟺_i P_ij) sums up the product of 蟺_i and P_ij over all states i. This represents the proportion of transitions in the Markov chain where the chain ends up in state j, regardless of which state it started from. In other words, 鈭慱i (蟺_i P_ij) is the long-run proportion of transitions that end up in state j.
04

(d) Deriving the equation

Since we know that the long-run proportion of transitions into state j (蟺_j) must be equal to the long-run proportion of transitions that end up in state j (鈭慱i (蟺_i P_ij)), we can write the equation as: \[ \pi_{j}=\sum_{i} \pi_{i} P_{i j} \] This equation represents the balance condition in a Markov chain, which helps in finding the steady-state distribution of the states.

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

In Example 4.3, Gary is in a cheerful mood today. Find the expected number of days until he has been glum for three consecutive days.

A transition probability matrix \(\mathbf{P}\) is said to be doubly stochastic if the sum over each column equals one; that is, $$ \sum_{i} P_{i j}=1, \quad \text { for all } $$ If such a chain is irreducible and aperiodic and consists of \(M+1\) states \(0,1, \ldots, M\), show that the limiting probabilities are given by $$ \pi_{j}=\frac{1}{M+1}, \quad j=0,1, \ldots, M $$

Each morning an individual leaves his house and goes for a run. \(\mathrm{He}\) is equally likely to leave either from his front or back door. Upon leaving the house, he chooses a pair of running shoes (or goes running barefoot if there are no shoes at the door from which he departed). On his return he is equally likely to enter, and leave his running shoes, either by the front or back door. If he owns a total of \(k\) pairs of running shoes, what proportion of the time does he run barefooted?

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 \(i, i>0\), 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.

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

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Calculus

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.