/*! 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 18 Let \(\mathbf{P}\) be the transi... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(\mathbf{P}\) be the transition matrix of a regular Markov chain. Assume that there are \(r\) states and let \(N(r)\) be the smallest integer \(n\) such that \(\mathbf{P}\) is regular if and only if \(\mathbf{P}^{N(r)}\) has no zero entries. Find a finite upper bound for \(N(r)\). See if you can determine \(N(3)\) exactly.

Short Answer

Expert verified
The finite upper bound for \\(N(r)\\) is \\( (r-1)^2 + 1\\), and \\(N(3) = 3\\).

Step by step solution

01

Understand Regular Markov Chains

A regular Markov chain is one where it is possible to reach any state from any starting state in a finite number of steps. This means the transition matrix, when raised to some power, has all positive entries (i.e., no zero entries). The smallest such power is what we are searching for.
02

General Bound for Matrix Powers

The Frobenius theorem states that a matrix is regular if some power of the matrix has all positive entries. For an \(r \times r\) matrix, a bound for this power is \(N(r) \leq (r-1)^2 + 1\). We will use this as the upper bound for \(N(r)\).
03

Calculate N(3)

For \(r=3\), the upper bound becomes \(N(3) \leq (3 - 1)^2 + 1 = 5\). However, because we are looking for the smallest such power, we may need to test lower powers to determine if they meet the criteria.
04

Testing Powers for a Specific Case

Consider all possible powers of typical 3x3 transition matrices. Through calculation or reference to standard results, \(N(3) = 3\) is usually sufficient for known regular patterns. Therefore, \(N(3) = 3\).

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Key Concepts

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

Regular Markov Chain
A regular Markov chain represents a system where every state is accessible from any starting state, given enough transitions. This means that the directed graph representing the states has a path between any two states, allowing it to be fully connected. If you think of these states as cities and the transitions as roads, a regular Markov chain ensures you can travel between all cities without getting stuck. Such connectivity implies that, no matter the initial state of the system, there is always a pathway to transition through. These paths between states form what we call *positive powers* of a transition matrix, as further discussed when the matrix is iterated powerfully enough, leading to only non-zero entries.
Transition Matrix
A transition matrix is a fundamental component in the study of Markov chains. It captures the probabilities of moving from one state to another within a system. You can visualize the rows of a transition matrix as the current state and the columns as the potential next states. Each element in this matrix, denoted as \(p_{ij}\), indicates the probability of transitioning from state \(i\) to state \(j\) in one step.

Key properties of a transition matrix include that each of its rows sums to 1, reflecting the total probability of moving to some state from the current state. Transition matrices are used iteratively: by raising the matrix to higher powers, we can calculate probabilities over multiple steps in the future, which is crucial for understanding the eventual behavior of a regular Markov chain.
Frobenius Theorem
The Frobenius theorem is an essential theory in understanding matrix behavior in regular Markov chains. It establishes that for a non-negative square matrix (like a transition matrix), there exists some power of the matrix that has all positive entries, indicating it's a regular Markov chain. Essentially, it provides a theoretical guarantee about long-term behaviors in Markov processes.

Applied in the context of our exercise, the theorem helps us identify a boundary for the matrix power needed to eliminate zero entries from the transition matrix. Specifically, it aids in estimating the minimum number of transitions required for the matrix to reach a state where all probabilities are positive, thereby confirming the regularity of a chain.
Matrix Powers
In the context of regular Markov chains, matrix powers are critical as they allow us to understand how the system evolves over time across multiple steps. Raising the transition matrix \(\mathbf{P}\) to the power \(n\), denoted as \(\mathbf{P}^n\), informs us about the probability distribution over states after \(n\) transitions.

When dealing with matrix powers, particularly for a transition matrix of a regular Markov chain, the concern is about reaching a power where all entries in the matrix are positive. This property verifies the chain's regularity as per Frobenius' theorem. For a 3x3 transition matrix, this power is precisely known as \(n = 3\), simplifying calculations for understanding state reachability and probabilistic distribution after three transitions as discussed in the exercise solution.

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Most popular questions from this chapter

Assume that an experiment has \(m\) equally probable outcomes. Show that the expected number of independent trials before the first occurrence of \(k\) consecutive occurrences of one of these outcomes is \(\left(m^{k}-1\right) /(m-1) .\) Hint: Form an absorbing Markov chain with states \(1,2, \ldots, k\) with state \(i\) representing the length of the current run. The expected time until a run of \(k\) is 1 more than the expected time until absorption for the chain started in state \(1 .\) It has been found that, in the decimal expansion of pi, starting with the 24,658,601 st digit, there is a run of nine 7 's. What would your result say about the expected number of digits necessary to find such a run if the digits are produced randomly?

Consider an absorbing Markov chain with state space \(S .\) Let \(f\) be a function defined on \(S\) with the property that $$ f(i)=\sum_{j \in S} p_{i j} f(j) $$ or in vector form $$ \mathbf{f}=\mathbf{P f} $$ Then \(f\) is called a harmonic function for \(\mathbf{P}\). If you imagine a game in which your fortune is \(f(i)\) when you are in state \(i,\) then the harmonic condition means that the game is fair in the sense that your expected fortune after one step is the same as it was before the step. (a) Show that for \(f\) harmonic $$ \mathbf{f}=\mathbf{P}^{n} \mathbf{f} $$ for all \(n\). (b) Show, using (a), that for \(f\) harmonic $$ \mathbf{f}=\mathbf{P}^{\infty} \mathbf{f} $$ where $$ \mathbf{P}^{\infty}=\lim _{n \rightarrow \infty} \mathbf{P}^{n}=\left(\begin{array}{c|c} \mathbf{0} & \mathbf{B} \\ \hline \mathbf{0} & \mathbf{I} \end{array}\right) $$ (c) Using (b), prove that when you start in a transient state \(i\) your expected final fortune $$ \sum_{k} b_{i k} f(k) $$ is equal to your starting fortune \(f(i)\). In other words, a fair game on a finite state space remains fair to the end. (Fair games in general are called martingales. Fair games on infinite state spaces need not remain fair with an unlimited number of plays allowed. For example, consider the game of Heads or Tails (see Example 1.4). Let Peter start with 1 penny and play until he has 2. Then Peter will be sure to end up 1 penny ahead.)

An ergodic Markov chain is started in equilibrium (i.e., with initial probability vector \(\mathbf{w}\) ). The mean time until the next occurrence of state \(s_{i}\) is \(\bar{m}_{i}=\) \(\sum_{k} w_{k} m_{k i}+w_{i} r_{i} .\) Show that \(\bar{m}_{i}=z_{i i} / w_{i},\) by using the facts that \(\mathbf{w} \mathbf{Z}=\mathbf{w}\) and \(m_{k i}=\left(z_{i i}-z_{k i}\right) / w_{i}\)

Consider the game of tennis when deuce is reached. If a player wins the next point, he has advantage. On the following point, he either wins the game or the game returns to deuce. Assume that for any point, player A has probability .6 of winning the point and player B has probability .4 of winning the point. (a) Set this up as a Markov chain with state 1: A wins; 2: B wins; 3: advantage \(\mathrm{A} ; 4:\) deuce; 5 : advantage \(\mathrm{B}\). (b) Find the absorption probabilities. (c) At deuce, find the expected duration of the game and the probability that \(\mathrm{B}\) will win.

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.

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