/*! 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 3 Consider the Markov chain with g... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 3

Consider the Markov chain with general \(2 \times 2\) transition matrix $$ \mathbf{P}=\left(\begin{array}{cc} 1-a & a \\ b & 1-b \end{array}\right) $$ (a) Under what conditions is \(\mathbf{P}\) absorbing? (b) Under what conditions is \(\mathbf{P}\) ergodic but not regular? (c) Under what conditions is \(\mathbf{P}\) regular?

Short Answer

Expert verified
(a) \( a = 0 \) or \( b = 0 \). (b) \( 0 < a < 1 \) and \( 0 < b < 1 \), but cycles exist. (c) \( a > 0 \) and \( b > 0 \).

Step by step solution

01

Defining an Absorbing Markov Chain

A Markov chain is absorbing if it has at least one absorbing state, and it is possible to go from any state to at least one absorbing state.In matrix \( \mathbf{P} \), a state is absorbing if the probability of staying in that state is 1, which means the diagonal element is 1. For state 1 to be absorbing, \(1 - a = 1\) or \(a = 0\). For state 2 to be absorbing, \(1 - b = 1\) or \(b = 0\). Thus, \(\mathbf{P}\) is absorbing if \(a = 0\) or \(b = 0\).
02

Defining ErgodiCity without Regularity

A Markov chain is ergodic if it can return to any state from any state, but not in a strict cycle (therefore not regular). For \( \mathbf{P} \) to be ergodic but not regular, it should not be periodic and must have every state be possible to reach from any other state.This requires that both states can be reached from each other, but with a mixture of transitions such that the chain doesn't regularly reach all states in a consistent manner. This generally can be seen if \(0 < a < 1\) and \(0 < b < 1\) and the matrix is not primitive (not all powers have all positive entries).
03

Defining Regularity

A Markov chain is regular if there exists some power \(n\) such that each element in the transition matrix raised to \(n\) is positive. For \( \mathbf{P} \), it is regular if all elements are strictly positive after raising it to a certain power, which occurs when both \(a > 0\) and \(b > 0\) and as long as the cycling does not strictly alternate to prevent all entries being positive.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absorbing State
An "absorbing state" in a Markov chain is a special type of state with the characteristic that once entered, it cannot be left. This is because the probability of staying in this state is exactly 1. Think of it like a trap that once you get into, you can't get out of.

For the given Markov chain described by the matrix \[ \mathbf{P} = \begin{pmatrix}1-a & a \ b & 1-b\end{pmatrix} \], a state is considered absorbing if its diagonal entry (the probability of it staying in itself) is 1.
  • If state 1 is absorbing, then the probability of transitioning from state 1 to state 1 must be 1. This happens when \(1-a = 1\), leading to \(a = 0\).
  • Similarly, if state 2 is absorbing, then \(1-b = 1\), leading to \(b = 0\).
So overall, the Markov chain \(\mathbf{P}\) is said to be absorbing if either \(a = 0\) or \(b = 0\). This guarantees that once entered, either state 1 or state 2 can never be left.
Ergodic Markov Chain
An "ergodic Markov chain" is one where you can get from any state to any other state, but this does not happen in a simple or predictable cycle.

It means that over time, each state is accessible from every other state, but the movements aren't periodic or predictable in terms of timeframe.
  • For our given Markov chain, ergodicity without regularity means \(0 < a < 1\) and \(0 < b < 1\). Both these conditions ensure that you can potentially move from one state to the other, but not in a predictable way.
  • The key here is that the chain must not exhibit periodic behavior, which means it doesn't keep repeating the same trajectory in a predictable number of steps.
Thus, for the matrix \(\mathbf{P}\) to represent an ergodic chain without being regular, it requires that although states are accessible from each other, there is no fixed cycle other than a mixture of back-and-forth transitions at unequal probabilities. The matrix needs not be primitive, meaning it's not necessary for every entry in any power of the matrix to be strictly positive.
Regular Markov Chain
A "regular Markov chain" is characterized by its powerful ability to reach any state from any other state given a sufficient number of steps. This means that the chain has no periodic behavior and, given enough time, can transition to any state with positive probability.

For the matrix \[ \mathbf{P} = \begin{pmatrix}1-a & a \ b & 1-b\end{pmatrix} \] to be regular, certain conditions must be met.
  • The parameters must be such that \(a > 0\) and \(b > 0\). This ensures that transitions between states are possible in both directions in any long enough sequence of steps.
  • Regularity also involves finding some power \(n\) such that raising the transition matrix \(\mathbf{P}^n\) results in all elements being positive, reflecting the possibility of transitioning from any state to any other state without permanent barriers.
In essence, regular Markov chains are versatile in moving between states, given that over time, every entry in the matrix raised to a power will become positive, indicating that every state will eventually be accessible from any other state.

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 process moves on the integers \(1,2,3,4,\) and \(5 .\) It starts at 1 and, on each successive step, moves to an integer greater than its present position, moving with equal probability to each of the remaining larger integers. State five is an absorbing state. Find the expected number of steps to reach state five.

Define \(f(r)\) to be the smallest integer \(n\) such that for all regular Markov chains with \(r\) states, the \(n\) th power of the transition matrix has all entries positive. It has been shown, \(^{14}\) that \(f(r)=r^{2}-2 r+2\). (a) Define the transition matrix of an \(r\) -state Markov chain as follows: For states \(s_{i},\) with \(i=1,2, \ldots, r-2, \mathbf{P}(i, i+1)=1, \mathbf{P}(r-1, r)=\mathbf{P}(r-1,1)=\) \(1 / 2,\) and \(\mathbf{P}(r, 1)=1\). Show that this is a regular Markov chain. (b) For \(r=3\), verify that the fifth power is the first power that has no zeros. (c) Show that, for general \(r,\) the smallest \(n\) such that \(\mathbf{P}^{n}\) has all entries positive is \(n=f(r)\)

Let \(\mathbf{P}\) be the transition matrix of an ergodic Markov chain. Let \(\mathbf{w}\) be a fixed probability vector (i.e., \(\mathbf{w}\) is a row vector with \(\mathbf{w} \mathbf{P}=\mathbf{w}) .\) Show that if \(w_{i}=0\) and \(p_{j i}>0\) then \(w_{j}=0\). Use this to show that the fixed probability vector for an ergodic chain cannot have any 0 entries.

Find the fixed probability vector \(\mathbf{w}\) for each of the following regular matrices. (a) \(\mathbf{P}=\left(\begin{array}{cc}.75 & .25 \\ .5 & .5\end{array}\right)\). (b) \(\mathbf{P}=\left(\begin{array}{ll}.9 & .1 \\ .1 & .9\end{array}\right)\). (c) \(\mathbf{P}=\left(\begin{array}{ccc}3 / 4 & 1 / 4 & 0 \\ 0 & 2 / 3 & 1 / 3 \\ 1 / 4 & 1 / 4 & 1 / 2\end{array}\right)\)

Assume that a man's profession can be classified as professional, skilled laborer, or unskilled laborer. Assume that, of the sons of professional men, 80 percent are professional, 10 percent are skilled laborers, and 10 percent are unskilled laborers. In the case of sons of skilled laborers, 60 percent are skilled laborers, 20 percent are professional, and 20 percent are unskilled. Finally, in the case of unskilled laborers, 50 percent of the sons are unskilled laborers, and 25 percent each are in the other two categories. Assume that every man has at least one son, and form a Markov chain by following the profession of a randomly chosen son of a given family through several generations. Set up the matrix of transition probabilities. Find the probability that a randomly chosen grandson of an unskilled laborer is a professional man.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

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.