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Chapter 4: Problem 20

A transition probability matrix \(\mathbf{P}\) is said to be doubly stochastic if the sum over each column equals one; that is, $$ \sum_{i} P_{i j}=1, \quad \text { for all } j $$ 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 $$ \pi_{j}=\frac{1}{M+1}, \quad j=0,1, \ldots, M $$

Short Answer

Expert verified
The limiting probabilities for the given irreducible, aperiodic, and doubly stochastic Markov chain with \(M+1\) states are \(\pi_j = \frac{1}{M+1}\) for \(j = 0,1, \ldots, M\).

Step by step solution

01

Understanding the properties of the given Markov chain

It is given that the transition probability matrix \(\mathbf{P}\) is doubly stochastic, which means the sum of probabilities in each column equals 1. Also, the chain is irreducible and aperiodic, implying that it is possible to reach any state from any state in a finite number of steps, and there is no fixed period for which the chain returns to the same state.
02

Writing the balance equations

In order to find the limiting probabilities \(\pi_j\), we need the balance equations, which are given by: $$ \pi_j = \sum_{i=0}^M \pi_i P_{ij}, \quad j = 0,1, \ldots, M $$
03

Proving that all limiting probabilities are equal

Since the matrix is doubly stochastic, we know that: $$ \sum_{i=0}^M P_{ij} = 1, \quad j = 0,1, \ldots, M $$ Substitute the balance equations into this equation: $$ \sum_{i=0}^M \pi_i P_{ij} = \sum_{i=0}^M \pi_i = 1, \quad j = 0,1, \ldots, M $$ Notice that the sum of the limiting probabilities is the same for each column, indicating that all the limiting probabilities are equal. To represent this, we can write: $$ \pi_j = \pi, \quad j = 0,1, \ldots, M $$
04

Finding the limiting probabilities

Since we know all limiting probabilities are equal, we can sum up all the probabilities and set them equal to 1. This will give us the value of \(\pi\): $$ (M+1) \pi = 1 $$ Solve for \(\pi\): $$ \pi = \frac{1}{M+1} $$ Therefore, the limiting probabilities are given by: $$ \pi_j = \frac{1}{M+1}, \quad j = 0,1, \ldots, M $$

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Most popular questions from this chapter

Let \(P^{(1)}\) and \(P^{(2)}\) denote transition probability matrices for ergodic Markov chains having the same state space. Let \(\pi^{1}\) and \(\pi^{2}\) denote the stationary (limiting) probability vectors for the two chains. Consider a process defined as follows: (a) \(X_{0}=1 .\) A coin is then flipped and if it comes up heads, then the remaining states \(X_{1}, \ldots\) are obtained from the transition probability matrix \(P^{(1)}\) and if tails from the matrix \(P^{(2)} .\) Is \(\left\\{X_{n}, n \geqslant 0\right\\}\) a Markov chain? If \(p=\) \(P\left\\{\right.\) coin comes up heads\\}, what is \(\lim _{n \rightarrow \infty} P\left(X_{n}=i\right) ?\) (b) \(X_{0}=1\). At each stage the coin is flipped and if it comes up heads, then the next state is chosen according to \(P^{(1)}\) and if tails comes up, then it is chosen according to \(P^{(2)} .\) In this case do the successive states constitute a Markov chain? If so, determine the transition probabilities. Show by a counterexample that the limiting probabilities are not the same as in part (a).

Find the average premium received per policyholder of the insurance company of Example \(4.27\) if \(\lambda=1 / 4\) for one-third of its clients, and \(\lambda=1 / 2\) for two-thirds of its clients.

Consider the following approach to shuffling a deck of \(n\) cards. Starting with any initial ordering of the cards, one of the numbers \(1,2, \ldots, n\) is randomly chosen in such a manner that each one is equally likely to be selected. If number \(i\) is chosen, then we take the card that is in position \(i\) and put it on top of the deck-that is, we put that card in position 1 . We then repeatedly perform the same operation. Show that, in the limit, the deck is perfectly shuffled in the sense that the resultant ordering is equally likely to be any of the \(n !\) possible orderings.

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

An individual possesses \(r\) umbrellas that he employs in going from his home to office, and vice versa. If he is at home (the office) at the beginning (end) of a day and it is raining, then he will take an umbrella with him to the office (home), provided there is one to be taken. If it is not raining, then he never takes an umbrella. Assume that, independent of the past, it rains at the beginning (end) of a day with probability \(p\). (a) Define a Markov chain with \(r+1\) states, which will help us to determine the proportion of time that our man gets wet. (Note: He gets wet if it is raining, and all umbrellas are at his other location.) (b) Show that the limiting probabilities are given by $$ \pi_{i}=\left\\{\begin{array}{ll} \frac{q}{r+q}, & \text { if } i=0 \\ \frac{1}{r+q}, & \text { if } i=1, \ldots, r \end{array} \quad \text { where } q=1-p\right. $$ (c) What fraction of time does our man get wet? (d) When \(r=3\), what value of \(p\) maximizes the fraction of time he gets wet
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