Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Show that if state \(i\) is recurrent and state \(i\) does not communicate with state \(j\), then \(P_{i j}=0 .\) This implies that once a process enters a recurrent class of states it can never leave that class. For this reason, a recurrent class is often referred to as a closed class.

Short Answer

Expert verified
Since state i is recurrent and does not communicate with state j, there do not exist positive integers n and m such that \(P_{ij}^{(n)}>0\) and \(P_{ji}^{(m)}>0\). This implies that \(P_{ij}=0\), meaning once a process enters a recurrent state (a closed class), it can never leave that class.

Step by step solution

01

State the Definitions

Before we begin, let's go over few important definitions: 1. Recurrent State: State i is recurrent if and only if the expected number of times it is visited is infinite, i.e., once a process enters a recurrent state, there is a 100% chance that it will return to that state eventually (and infinitely many times). 2. Communicating States: States i and j are said to communicate if there exist positive integers n and m such that the probabilities of transitioning from state i to state j in n steps and from state j to state i in m steps are both strictly positive. 3. Transition Probabilities: \(P_{ij}\) denotes the probability that a process transitions from state i to state j in one step.
02

Show That Transition Probabilities are Non-Negative

Recall that the transition probability \(P_{ij}\) is the probability that a process transitions from state i to state j. Since probabilities are always non-negative, we have that \(0 \le P_{ij} \le 1\).
03

Use the Definitions to Derive the Result

Now, we will use the given conditions: state i is recurrent and i does not communicate with state j. 1. Since state i is recurrent, there is a 100% chance that it will return to state i eventually. Mathematically, this means that the sum of probabilities of transitioning from state i to state i in any number of steps is equal to 1. 2. Since state i does not communicate with state j, there do not exist positive integers n and m such that \(P_{ij}^{(n)}>0\) and \(P_{ji}^{(m)}>0\) (where \(P_{ij}^{(n)}\) and \(P_{ji}^{(m)}\) denote the probabilities of transitioning from state i to state j in n steps and from state j to state i in m steps, respectively). In particular, \(P_{ij}^{(1)}=P_{ij}=0\).
04

Conclude

From step 3, we found that \(P_{ij}=0\), as there is no positive integer n such that the probability of transitioning from state i to state j in n steps is strictly positive. This means that once a process enters a recurrent state, it can never leave the class of recurrent states (since it can never transition to a non-communicating state). Thus, a recurrent class is often referred to as a closed class, and we have proved the given proposition.

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 Vaia!

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

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?

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

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state \(i, i=0,1,2,3\), if the first urn contains \(i\) white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left\\{X_{n}, n=0,1,2, \ldots\right\\}\) is a Markov chain and calculate its transition probability matrix.

Consider a process \(\left\\{X_{n}, n=0,1, \ldots\right\\}\), which takes on the values 0,1 , or 2 . Suppose $$ \begin{aligned} &P\left\\{X_{n+1}=j \mid X_{n}=i, X_{n-1}=i_{n-1}, \ldots, X_{0}=i_{0}\right\\} \\ &\quad=\left\\{\begin{array}{ll} P_{i j}^{\mathrm{I}}, & \text { when } n \text { is even } \\ P_{i i}^{\mathrm{II}}, & \text { when } n \text { is odd } \end{array}\right. \end{aligned} $$ where \(\sum_{j=0}^{2} P_{i j}^{\mathrm{I}}=\sum_{j=0}^{2} P_{i j}^{\mathrm{II}}=1, i=0,1,2 .\) Is \(\left\\{X_{n}, n \geqslant 0\right\\}\) a Markov chain? If not, then show how, by enlarging the state space, we may transform it into a Markov chain.

Suppose that a population consists of a fixed number, say, \(m\), of genes in any generation. Each gene is one of two possible genetic types. If exactly \(i\) (of the \(m\) ) genes of any generation are of type 1 , then the next generation will have \(j\) type 1 (and \(m-j\) type 2 ) genes with probability $$ \left(\begin{array}{c} m \\ j \end{array}\right)\left(\frac{i}{m}\right)^{j}\left(\frac{m-i}{m}\right)^{m-j}, \quad j=0,1, \ldots, m $$ Let \(X_{n}\) denote the number of type 1 genes in the \(n\) th generation, and assume that \(X_{0}=i\) (a) Find \(E\left[X_{n}\right]\). (b) What is the probability that eventually all the genes will be type \(1 ?\)

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free