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Chapter 6: Problem 1

Classify the states of the discrete-time Markov chains with state space \(S=(1,2,3,4)\) and transition matrices (a) \(\left(\begin{array}{llll}\frac{1}{3} & \frac{2}{3} & 0 & 0 \\\ \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ \frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{2} \\ 0 & 0 & 0 & 1\end{array}\right)\) (b) \(\left(\begin{array}{llll}0 & \frac{1}{2} & \frac{1}{2} & 0 \\\ \frac{1}{3} & 0 & 0 & \frac{2}{3} \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0\end{array}\right) .\) In case (a), calculate \(f_{34}(n)\), and deduce that the probability of ultimate absorption in state 4 , starting from 3 , equals ?. Find the mean recurrence times of the states in case (b).

Short Answer

Expert verified
In (a), states 1, 2 form a class; 3 goes to 4. Ultimate absorption from 3 to 4 is 0.5. In (b), states 1, 2, 3, 4 have different recurrence properties; 3, 4 cycle with mean 2.

Step by step solution

01

Analyze Transition Matrix (a)

The matrix is given by:\[\begin{array}{llll}\frac{1}{3} & \frac{2}{3} & 0 & 0 \\frac{1}{2} & \frac{1}{2} & 0 & 0 \\frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{2} \0 & 0 & 0 & 1\end{array}\]State 4 is an absorbing state because it can only transition to itself. States 1, 2 form a communicating class, while states 3 directly communicates to 4.
02

Find Probability Function \(f_{34}(n)\)

To calculate \(f_{34}(n)\), which represents the probability of transitioning from state 3 to state 4 in exactly \(n\) steps, observe from the transition matrix that moving from state 3 to 4 takes one step with probability \(\frac{1}{2}\). Therefore, \(f_{34}(1) = \frac{1}{2}\), and \(f_{34}(n) = 0\) for all \(n > 1\).
03

Calculate Ultimate Absorption Probability from State 3

Using the equation \( F_{34} = \sum_{n=1}^{\infty} f_{34}(n) \), we deduce that the probability of ultimately being absorbed in state 4, starting from state 3, is\[ F_{34} = \frac{1}{2} + 0 + 0 + \ldots = \frac{1}{2} \]
04

Analyze Transition Matrix (b) and State Classification

The matrix is given by:\[\begin{array}{llll}0 & \frac{1}{2} & \frac{1}{2} & 0 \\frac{1}{3} & 0 & 0 & \frac{2}{3} \1 & 0 & 0 & 0 \0 & 0 & 1 & 0\end{array}\]State 1 and 2 form a communicating class. State 3 communicates with 4, and 4 communicates with 3.
05

Calculate Mean Recurrence Times for Case (b)

For state 3, since the cycle between 3 and 4 is deterministic (3 to 4, then back to 3), the mean recurrence time for state 3 is 2 steps. For states 1 and 2, as they form part of a irreducible class, calculate mean recurrence times using the formula \( \frac{1}{\pi_i} \), where \(\pi_i\) are stationary probabilities.

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Key Concepts

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

State Classification in Markov Chains
In a discrete-time Markov chain, states can be classified into several distinct categories based on how they interact with each other within the chain. Understanding these classifications is crucial for analyzing the behavior of the chain. The classifications include:
  • Transient State: A state is transient if there is a possibility of never returning to it once it is left. This means, over time, there's a diminishing chance of re-entry into the state.
  • Recurrent State: A state is recurrent if it is guaranteed to be revisited eventually once left. This implies a 100% probability that you will come back to this state.
  • Absorbing State: A state is absorbing if, once entered, the process never leaves. This state always transitions back to itself.
  • Communicating Class: A group of states that are all reachable from each other. If you can move from state A to state B and vice versa, they belong to the same communicating class.
Your ability to identify these classifications determines how you analyze the potential outcomes and stability of a given Markov chain.
Transition Matrices
Transition matrices are square matrices used in Markov chains to display the probabilities of moving from one state to another. They are essential tools in analyzing the dynamics of the chain.
  • Each entry \(P_{ij}\) of a transition matrix represents the probability of moving from state \(i\) to state \(j\) in one step.
  • The sum of each row in a transition matrix equals 1, as all possible transitions from a given state need to account for 100% of the probabilities.
The layout of the transition matrix enables you to predict the states' future behavior. For example, in transition matrix (a) given in the exercise, state 4's row entry as \(0, 0, 0, 1\) signifies it is an absorbing state, confirming that once the chain enters state 4, it cannot leave. Transition matrices thus provide a powerful visual and mathematical tool for understanding state transitions in a Markov chain.
Absorbing States
An absorbing state is a unique classification in a Markov chain where, once reached, no further transition to any other state occurs. This has a critical impact on the long-term behavior of a Markov chain.
  • In a transition matrix, an absorbing state will have a probability of 1 on its diagonal, indicating that the chain remains in this state once entered.
  • Absorbing states determine the "end behavior" of a Markov chain since once such a state is reached, the system halts in terms of state changes, often signifying completion or failure in practical applications.
In exercise (a), state 4 is identified as absorbing because every time it is reached, it transitions directly back to itself. This implies that in any real-world process modeled by this chain, reaching state 4 means the process has concluded in the terms modeled by the chain.
Recurrence Times
Recurrence time is a critical concept in understanding the cycle of revisiting states in a Markov chain. It helps quantify the expected number of steps it takes to return to a particular state.
  • The mean recurrence time for a state is calculated as the reciprocal of its stationary probability, \( \pi_i\), which is the long-term probability of being in that state.
  • For a recurrent state, given enough time, the mean recurrence time will become evident, while for transient states, it might be infinite since the state may not be revisited.
In exercise (b), states are analyzed for their recurrence times. Particularly, the states that form communicating classes might involve complex calculations unless the transitions show patterns. In the example provided, state 3's deterministic recurrence cycle makes it a straightforward calculation, with a mean recurrence time of 2 steps, due to its direct relationship with state 4.

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Most popular questions from this chapter

Let \(\left\\{X_{r}: r \geq 1\right\\}\) be independent exponential random variables with parameter \(\lambda\), and set \(S_{n}=\) \(\sum_{r=1}^{n} X_{r .}\) Show that: (a) \(Y_{k}=S_{k} / S_{n}, 1 \leq k \leq n-1\), have the same distribution as the order statistics of independent variables \(\left|U_{k}: 1 \leq k \leq n-1\right|\) which are uniformly distributed on \((0,1)\), (b) \(Z_{k}=X_{k} / S_{n}, 1 \leq k \leq n\), have the same joint distribution as the coordinates of a point \(\left(U_{1}, \ldots, U_{n}\right)\) chosen uniformly at random on the simplex \(\sum_{r=1}^{n} u_{r}=1, u_{r} \geq 0\) for all \(r\).

Let \(X\) and \(Y\) be Markov chains on the set \(Z\) of integers. Is the sequence \(Z_{n}=X_{n}+Y_{n}\) necessarily a Markov chain?

The proof copy of a book is read by an infinite sequence of editors checking for mistakes. Each mistake is detected with probability \(p\) at each reading: between readings the printer corrects the detected mistakes but introduces a random number of new errors (errons may be introduced even if no mistakes were detected). Assuming as much independence as usual, and that the numbers of new errors after different readings are identically distributed, find an expression for the probability generating function of the stationary distribution of the number \(X_{n}\) of errors after the \(n\)th editor-printer cycle, whenever this exists. Find it explicitly when the printer introduces a Poisson-distributed number of errors at each stage.

Let \(N\) be a non-homogeneous Poisson process on \(R_{+}\)with intensity function \(\lambda\). Find the joint density of the first two inter-event times, and deduce that they are not in general independent.

Ants enter a kitchen at the instants of a Poisson process \(N\) of rate \(\lambda\); they cach visit the pantry and then the sink, and leave. The \(r\) th ant spends time \(X_{r}\) in the pantry and \(Y_{r}\) in the \(\operatorname{sink}\) (and \(X_{r}+Y_{r}\) in the kitchen altogether), where the vectors \(V_{r}=\left(X_{r}, Y_{r}\right)\) and \(V_{A}\) are independent for \(r \neq s .\) At time \(t=0\) the kitchen is free of ants. Find the joint distribution of the numbers \(A(t)\) of ants in the pantry and \(B(t)\) of ants in the \(\operatorname{sink}\) at time \(t\). Now suppose the ants arrive in pairs at the times of the Poisson process, but then separate to behave independently as above. Find the joint distribution of the numbers of ants in the two locations.
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