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Chapter 6: Problem 28

If \(\\{X(t)\\}\) and \(\\{Y(t)\\}\) are independent continuous-time Markov chains, both of which are time reversible, show that the process \(\\{X(t), Y(t)\\}\) is also a time reversible Markov chain.

Short Answer

Expert verified
Since both \(X(t))\) and \(Y(t)\) are Markov chains, their joint process \(\\{X(t), Y(t)\\}\) is also a Markov chain by proving the Markov property for any three times \(t_1

Step by step solution

01

Prove that the joint process is a Markov chain

To prove that the joint process \(\\{X(t), Y(t)\\}\) is a Markov chain, we need to show that for any three times \(t_1 < t_2 < t_3\) and any three pairs \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\), we have the Markov property: \(P((X(t_3), Y(t_3)) = (x_3, y_3) | (X(t_2), Y(t_2)) = (x_2, y_2), (X(t_1), Y(t_1)) = (x_1, y_1)) = P((X(t_3), Y(t_3)) = (x_3, y_3) | (X(t_2), Y(t_2)) = (x_2, y_2))\) Since we are given that \(X(t)\) and \(Y(t)\) are independent, we have \(P((X(t_3), Y(t_3)) = (x_3, y_3) | (X(t_2), Y(t_2)) = (x_2, y_2), (X(t_1), Y(t_1)) = (x_1, y_1)) = P(X(t_3) = x_3 | X(t_2) = x_2, X(t_1) = x_1)P(Y(t_3) = y_3 | Y(t_2) = y_2, Y(t_1) = y_1)\) Since both \(X(t)\) and \(Y(t)\) are Markov chains, we have \(P(X(t_3) = x_3 | X(t_2) = x_2, X(t_1) = x_1) = P(X(t_3) = x_3 | X(t_2) = x_2)\) and \(P(Y(t_3) = y_3 | Y(t_2) = y_2, Y(t_1) = y_1) = P(Y(t_3) = y_3 | Y(t_2) = y_2)\) Therefore, \(P((X(t_3), Y(t_3)) = (x_3, y_3) | (X(t_2), Y(t_2)) = (x_2, y_2), (X(t_1), Y(t_1)) = (x_1, y_1)) = P((X(t_3), Y(t_3)) = (x_3, y_3) | (X(t_2), Y(t_2)) = (x_2, y_2))\) Hence, the joint process \(\\{X(t), Y(t)\\}\) is a Markov chain.
02

Prove that the joint process is time reversible

Now, we need to show that if \(X(t)\) and \(Y(t)\) are time reversible, then so is their joint process. The joint process will be time reversible if, for any two pairs \((x_1, y_1)\) and \((x_2, y_2)\) and any time \(t\), we have \(P((X(t+s), Y(t+s)) = (x_1, y_1) | (X(t), Y(t)) = (x_2, y_2)) = P((X(s), Y(s)) = (x_2, y_2) | (X(0), Y(0)) = (x_1, y_1))\) We know that \(X(t)\) and \(Y(t)\) are time reversible, so we have \(P(X(t+s) = x_1 | X(t) = x_2) = P(X(s) = x_2 | X(0) = x_1)\) and \(P(Y(t+s) = y_1 | Y(t) = y_2) = P(Y(s) = y_2 | Y(0) = y_1)\) Using the independence of \(X(t)\) and \(Y(t)\), we have \(P((X(t+s), Y(t+s)) = (x_1, y_1) | (X(t), Y(t)) = (x_2, y_2)) = P(X(t+s) = x_1 | X(t) = x_2)P(Y(t+s) = y_1 | Y(t) = y_2) = P(X(s) = x_2 | X(0) = x_1)P(Y(s) = y_2 | Y(0) = y_1) = P((X(s), Y(s)) = (x_2, y_2) | (X(0), Y(0)) = (x_1, y_1))\) Thus, the joint process \(\\{X(t), Y(t)\\}\) is time reversible, proving the given exercise.

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Key Concepts

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

Time Reversibility
Time reversibility in a continuous-time Markov chain means that the process looks the same whether viewed forward or backward in time. Essentially, the likelihood of transitioning from state A to state B in the forward direction over a time period is the same as moving from state B to state A in the reverse. This assumption is helpful in making certain calculations and models more straightforward because it implies symmetry in the transition dynamics. In our given problem, if two independent Markov chains \(\{X(t)\}\) and \(\{Y(t)\}\) are individually time reversible, their joint process \(\{X(t), Y(t)\}\) retains this property. This can be verified by checking if the transition probabilities satisfy the time reversibility condition. Thanks to the independence of both processes, their joint transition probabilities can be easily multiplied, preserving this property.
Markov Property
The Markov Property is a defining feature of Markov chains, where the future state of the process only depends on the present state and not any past states. This is often referred to as the "memoryless" property. In other words, knowing the current state and future behavior, the past is irrelevant. For a joint process \(\{X(t), Y(t)\}\), the property is confirmed if the probability of moving to a future state depends only on the present state of \(\{X(t), Y(t)\}\), and not on how it got there. Thus, verifying that the joint process \(\{X(t), Y(t)\}\) exhibits the Markov property involves affirming that its future transition probabilities are solely influenced by its current state and independent of its past trajectory. This can be mathematically expressed by simplifying conditional probabilities, ensuring reduction to only dependence on the current state.
Independence
Independence between two processes, such as \(\{X(t)\}\) and \(\{Y(t)\}\), means that the behavior of one process does not influence the other. When two Markov chains are independent, their joint process maintains separate state-space transitions for each component, simplifying the analysis. This independence is particularly crucial when demonstrating properties like time reversibility or the Markov property in the joint process. In the given solution, the independence helps in establishing that the probability transitions for the joint process \(\{X(t), Y(t)\}\) can be expressed as products of the individual transition probabilities for each component, making the mathematical treatment more manageable.
Joint Process
A joint process in the context of Markov chains is an amalgamation of two or more processes, where states in the joint process consist of combination pairs from each individual process. For \(\{X(t), Y(t)\}\), this means that the state of the joint process at any time \(t\) is the pair \((X(t), Y(t))\). The challenge is to show that this combined process behaves like a Markov chain. By verifying that both the Markov property and time reversibility hold, we confirm that \(\{X(t), Y(t)\}\) indeed forms a valid continuous-time Markov chain. The independence of \(\{X(t)\}\) and \(\{Y(t)\}\) facilitates this by ensuring that the interactions between these two processes do not affect their intrinsic stochastic behaviors.

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Most popular questions from this chapter

Potential customers arrive at a full-service, one-pump gas station at a Poisson rate of 20 cars per hour. However, customers will only enter the station for gas if there are no more than two cars (including the one currently being attended to) at the pump. Suppose the amount of time required to service a car is exponentially distributed with a mean of five minutes. (a) What fraction of the attendant's time will be spent servicing cars? (b) What fraction of potential customers are lost?

Potential customers arrive at a single-server station in accordance with a Poisson process with rate \(\lambda .\) However, if the arrival finds \(n\) customers already in the station, then he will enter the system with probability \(\alpha_{n}\). Assuming an exponential service rate \(\mu\), set this up as a birth and death process and determine the birth and death rates.

A service center consists of two servers, each working at an exponential rate of two services per hour. If customers arrive at a Poisson rate of three per hour, then, assuming a system capacity of at most three customers, (a) what fraction of potential customers enter the system? (b) what would the value of part (a) be if there was only a single server, and his rate was twice as fast (that is, \(\mu=4)\) ?

Consider a taxi station where taxis and customers arrive in accordance with Poisson processes with respective rates of one and two per minute. A taxi will wait no matter how many other taxis are present. However, an arriving customer that does not find a taxi waiting leaves. Find (a) the average number of taxis waiting, and (b) the proportion of arriving customers that get taxis.

Consider a birth and death process with birth rates \(\lambda_{i}=(i+1) \lambda, i \geqslant 0\), and death rates \(\mu_{i}=i \mu, i \geqslant 0\) (a) Determine the expected time to go from state 0 to state 4 . (b) Determine the expected time to go from state 2 to state 5 . (c) Determine the variances in parts (a) and (b).
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