/*! 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 72 For a time reversible Markov cha... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 72

For a time reversible Markov chain, argue that the rate at which transitions from \(i\) to \(j\) to \(k\) occur must equal the rate at which transitions from \(k\) to \(j\) to \(i\) occur.

Short Answer

Expert verified
For a time-reversible Markov chain, the detailed balance condition states that \(\pi_i P_{ij} = \pi_j P_{ji}\). Using this property, we can express the transition rate from i to j to k as \(\pi_i P_{ij} P_{jk}\) and from k to j to i as \(\pi_k P_{kj} P_{ji}\). Applying detailed balance, we find that these two rates are equal, confirming the statement for the time-reversible Markov chain.

Step by step solution

01

Understanding Time Reversibility

In a Markov chain, time reversibility means that the process behaves identically when the time direction is reversed. Mathematically, this can be expressed using the concept of detailed balance: \(\pi_i P_{ij} = \pi_j P_{ji}\) where \(\pi_i\) represents the stationary distribution (i.e., the probability of being in state i in the long run), \(P_{ij}\) refers to the probability of transitioning from state i to state j, and \(P_{ji}\) refers to the probability of transitioning from state j to state i. Now we want to show that the rate of transitions from i to j to k is equal to the rate of transitions from k to j to i.
02

Transition Rate from i to j to k

First, let's calculate the rate of transition transitions from i to j to k. Since we're dealing with discrete states, the transition rate can be defined as a product of transition probabilities and the probabilities to be in the corresponding state: Rate(i to j to k) = \(\pi_i P_{ij} P_{jk}\)
03

Transition Rate from k to j to i

Similarly, let's calculate the rate of transition transitions from k to j to i: Rate(k to j to i) = \(\pi_k P_{kj} P_{ji}\)
04

Applying Time Reversibility

Using the detailed balance condition explained in Step 1, we can substitute transition probabilities in the rate expressions: Rate(i to j to k) = \(\pi_i \frac{\pi_j P_{ji}}{\pi_i} \frac{\pi_k P_{jk}}{\pi_j} = \pi_k P_{kj} P_{ji}\) Comparing the expressions for the transition rate from i to j to k and the transition rate from k to j to i: \(\pi_k P_{kj} P_{ji} = \pi_k P_{kj} P_{ji}\) which shows that the rate of transitions from i to j to k is indeed equal to the rate of transitions from k to j to i, confirming the statement for the time-reversible Markov chain.

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

Let the transition probability matrix of a two-state Markov chain be given, as in Example 4.2, by $$ \mathbf{P}=\left\|\begin{array}{cc} p & 1-p \\ 1-p & p \end{array}\right\| $$ Show by mathematical induction that $$ \mathbf{P}^{(n)}=\left\|\begin{array}{|ll} \frac{1}{2}+\frac{1}{2}(2 p-1)^{n} & \frac{1}{2}-\frac{1}{2}(2 p-1)^{n} \\ \frac{1}{2}-\frac{1}{2}(2 p-1)^{n} & \frac{1}{2}+\frac{1}{2}(2 p-1)^{n} \end{array}\right\| $$

On a chessboard compute the expected number of plays it takes a knight, starting in one of the four corners of the chessboard, to return to its initial position if we assume that at each play it is equally likely to choose any of its legal moves. (No other pieces are on the board.) Hint: Make use of Example 4.36.

A taxi driver provides service in two zones of a city. Fares picked up in zone \(A\) will have destinations in zone \(A\) with probability \(0.6\) or in zone \(B\) with probability \(0.4\). Fares picked up in zone \(B\) will have destinations in zone \(A\) with probability \(0.3\) or in zone \(B\) with probability \(0.7 .\) The driver's expected profit for a trip entirely in zone \(A\) is 6 ; for a trip entirely in zone \(B\) is \(8 ;\) and for a trip that involves both zones is 12 . Find the taxi driver's average profit per trip.

Coin 1 comes up heads with probability \(0.6\) and \(\operatorname{coin} 2\) with probability \(0.5 . \mathrm{A}\) coin is continually flipped until it comes up tails, at which time that coin is put aside and we start flipping the other one. (a) What proportion of flips use coin 1? (b) If we start the process with \(\operatorname{coin} 1\) what is the probability that \(\operatorname{coin} 2\) is used on the fifth flip?

Suppose that coin 1 has probability \(0.7\) of coming up heads, and \(\operatorname{coin} 2\) has probability \(0.6\) of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select \(\operatorname{coin} 2\) to flip tomorrow. If the coin initially flipped is equally likely to be \(\operatorname{coin} 1\) or \(\operatorname{coin} 2\), then what is the probability that the coin flipped on the third day after the initial flip is coin 1? Suppose that the coin flipped on Monday comes up heads. What is the probability that the coin flipped on Friday of the same week also comes up heads?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

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.