/*! 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 8 Specify the classes of the follo... [FREE SOLUTION] | 91Ó°ÊÓ

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

Short Answer

Expert verified
In summary, for the given Markov chains: 1. \(\mathbf{P}_1\) has one class that is aperiodic and recurrent. 2. \(\mathbf{P}_2\) has two classes: {0, 1, 3} aperiodic and recurrent, and {2} transient. 3. \(\mathbf{P}_3\) has two classes, both aperiodic and recurrent: {0, 1, 2} and {3, 4}. 4. \(\mathbf{P}_4\) has three classes: {0, 1} aperiodic and recurrent, {2} absorbing and recurrent, and {3, 4} aperiodic and transient.

Step by step solution

01

We can see that every state in \(\mathbf{P}_1\) can be reached from any other state in one step, so the Markov chain is irreducible. There is only one class in this case. Since the chain is irreducible, we can check the aperiodicity using any state, say state 0. The least common multiple of the numbers of steps needed to go from state 0 to state 0 is 2, so the chain is aperiodic. To determine if the states are transient or recurrent, we'll use the criterion mentioned above. \[ \sum_{n=0}^{\infty} p_{00}^{(2n)} = \sum_{n=0}^{\infty} (1)^n = \infty \] Since the sum is infinite, all the states in this Markov Chain are recurrent. #Step 2: Analyze \(\mathbf{P}_2\)

In \(\mathbf{P}_2\), the transition between states varies depending on the current state. State 0 and 1 are communicating with only state 3. State 2 is communicating with states 0, 1, and 3. Therefore, there are 2 classes: {0, 1, 3} and {2}. Since there is more than one class, the chain is reducible. We'll now analyze the aperiodicity in the first class, {0, 1, 3}. The least common multiple of the number of steps required to go from state 0 to state 0 is infinite, so this class is aperiodic. To check the transient or recurrent nature of states, since state 2 is communicating with the first class, it is a transient state. Note that, if a class has a transient state, then all other states in that class are also transient. Since states 0, 1, and 3 are not communicating with any of the other states, they are recurrent. #Step 3: Analyze \(\mathbf{P}_3\)
02

In \(\mathbf{P}_3\), the classes are very easy to observe: {0, 1, 2} and {3, 4}. Both classes are irreducible. The Markov Chain is reducible. Both classes are easily shown to be aperiodic. We can determine the transient or recurrent nature of states by inspection: The first class {0, 1, 2} will be recurrent as the system tends to stay in this class with non-zero probability. The second class {3, 4} will also be recurrent as both states have a non-zero probability of staying in the same state. #Step 4: Analyze \(\mathbf{P}_4\)

In \(\mathbf{P}_4\), states 0 and 1 are only communicating with each other. State 2 is an absorbing state (once the system gets into state 2, it stays there forever). State 3 is reaching only state 4, and state 4 is reaching only state 0. So, there are 3 classes: {0, 1}, {2}, {3, 4}. The Markov chain is reducible. States 0 and 1 are communicating with each other and are aperiodic. State 2 is an absorbing state (aperiodic, too). States 3 and 4 are also aperiodic. In terms of recurrence, since states 3 and 4 are reaching state 0 with non-zero probability, they are transient. State 2 is absorbing, so it is recurrent. States 0 and 1 will also be recurrent as they continue to communicate with each other with non-zero probabilities.

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Key Concepts

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

Irreducible Markov Chains
An irreducible Markov chain is one in which it's possible to get from any state to any other state in a finite number of steps. In simple terms, no matter where you start, you can potentially reach every other state. This property means there is only one class in the Markov chain. Since every state is interconnected, we say the chain is strongly connected.

When looking at a Markov Chain matrix like \(\mathbf{P}_1\) from the exercise, we can see that each state communicates with every other state, making it irreducible. This is useful because it simplifies many analyses, particularly when determining the nature of all states being recurrent or otherwise.
Reducible Markov Chains
A reducible Markov chain has at least two separate classes of states, meaning there exist states that cannot be reached from certain other states. This lack of full connectivity categorizes states into groups, or "classes."

For example, if you scrutinize \(\mathbf{P}_2\), you'll note that some states can only transition to specific others, creating distinct classes that don't communicate. The chain is divided into these separate sets of states, which can either be further analyzed as for their transient or recurrent nature.
Transient States
Transient states are those where it is possible to leave and never return with certainty. Think of them as temporary stops where, eventually, over time, the process will move on and not come back.

To identify transient states, see if there’s a non-zero probability that, starting from that state, the chain moves to another set of states permanently. For \(\mathbf{P}_2\), for instance, state 2 is transient because once the process leaves it, there is no mechanism in it for mandatory return.
Recurrent States
Recurrent states are ones you can return to eventually, for sure, given that you have started in it. There is always a non-zero probability of coming back to a recurrent state at some point because it's a part of a closed communication class.

Taking \(\mathbf{P}_3\) as an example, all states are recurrent because once the system reaches any state in its classes, there is a strong probability that it'll circle back around to that state again. Recurrent states essentially trap the process within their class over time.
Aperiodic Markov Chains
Aperiodicity in Markov chains means there isn't a fixed cycle or period within which a state must return to its starting point. Essentially, the state can return after varied numbers of steps each time.

An example would be when analyzing \(\mathbf{P}_1\), where the transitions do not restrict to specific periodic patterns. The least common multiple of step cycles for any state is "1," defining the chain as aperiodic. This indicates more flexibility in when state revisits can happen.

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Most popular questions from this chapter

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.

A group of \(n\) processors are arranged in an ordered list. When a job arrives, the first processor in line attempts it; if it is unsuccessful, then the next in line tries it; if it too is unsuccessful, then the next in line tries it, and so on. When the job is successfully processed or after all processors have been unsuccessful, the job leaves the system. At this point we are allowed to reorder the processors, and a new job appears. Suppose that we use the one- closer reordering rule, which moves the processor that was successful one closer to the front of the line by interchanging its position with the one in front of it. If all processors were unsuccessful (or if the processor in the first position was successful), then the ordering remains the same. Suppose that each time processor \(i\) attempts a job then, independently of anything else, it is successful with probability \(p_{i}\). (a) Define an appropriate Markov chain to analyze this model. (b) Show that this Markov chain is time reversible. (c) Find the long run probabilities.

16\. Let \(Y_{n}\) be the sum of \(n\) independent rolls of a fair dic. Find $$ \lim _{n \rightarrow \infty} P\left[Y_{n} \text { is a multiple of } 13\right] $$ Define an appropriate Markov chain and apply the results of Exercise 14.

A flea moves around the vertices of a triangle in the following manner: Whenever it is at vertex \(i\) it moves to its clockwise neighbor vertex with probability \(p_{i}\) and to the counterclockwise neighbor with probability \(q_{i}=1-p_{i}, i=1,2,3\) (a) Find the proportion of time that the flea is at each of the vertices. (b) How often does the flea make a counterclockwise move which is then followed by 5 consecutive clockwise moves?

Prove that if the number of states in a Markov chain is \(M\), and if state \(J\) can be reached from state \(i\), then it can be reached in \(M\) steps or less.

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