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Chapter 4: Problem 68

(a) Show that the limiting probabilities of the reversed Markov chain are the same as for the forward chain by showing that they satisfy the equations $$ \pi_{j}=\sum_{i} \pi_{i} Q_{i j} $$ (b) Give an intuitive explanation for the result of part (a).

Short Answer

Expert verified
In summary, the limiting probabilities of the reversed Markov chain are the same as those for the forward chain because they both satisfy the equation \(\pi_j = \sum_{i} \pi_i Q_{ij}\). Intuitively, this is because the long-term behavior of the Markov chain is not affected by the direction in which it is traversed, and the limiting probabilities represent the proportion of time spent in each state, which is independent of the direction of the chain.

Step by step solution

01

Understanding the Forward Markov Chain

For a (discrete-time) forward Markov chain with state space \(S\), we have a transition probability matrix \(P\), where the element \(P_{ij}\) is the probability of transitioning from state \(i\) to state \(j\). The limiting probabilities \(\pi = (\pi_i)_{i\in S}\) are the long-term probabilities that the chain is in each state \(i\), known as the stationary distribution, which satisfy \(\pi_i \geq 0\) for all \(i\), \(\sum_{i} \pi_i = 1\), and \[\pi = \pi P .\]
02

Understanding the Reversed Markov Chain

For a reversed Markov chain, we still have the state space \(S\) and stationary distribution \(\pi\). We define the transition probability matrix \(Q\) for the reversed chain by: \[Q_{ij} = \frac{\pi_j}{\pi_i} P_{ji},\] which represents the probability of transitioning from state \(j\) to state \(i\), reversing the direction of the forward chain.
03

Show that the Limiting Probabilities Satisfy the Equations

Now we need to show that the limiting probabilities of the reversed Markov chain, represented by the vector \(\pi\), satisfy the given equation: \[\pi_j = \sum_{i} \pi_i Q_{ij} .\] Let's compute the right-hand side of this equation and show that it indeed equals \(\pi_j\): \[\sum_{i} \pi_i Q_{ij} = \sum_{i} \pi_i \left(\frac{\pi_j}{\pi_i}P_{ji}\right) = \pi_j \sum_{i} P_{ji} .\] Recall that each row in the transition probability matrix \(P\) must sum to 1, so we get \[\pi_j \sum_{i} P_{ji} = \pi_j (1) = \pi_j .\] Thus, the limiting probabilities of the reversed Markov chain are the same as those for the forward chain.
04

Intuitive Explanation

The reversed Markov chain is essentially "playing the forward Markov chain backward". Since the stationary distribution reflects long-term behavior, it is reasonable that the limiting probabilities should be the same for both the forward and reversed chains. At a high level, this is because the long-term behavior of the chain is not affected by the direction in which it is traversed. The limiting probabilities represent the proportion of time spent in each state over the long run, and these proportions are independent of the direction of the chain.

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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.

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.

Each day, one of \(n\) possible elements is requested, the ith one with probability \(P_{i}, i \geqslant 1, \sum_{1}^{n} P_{i}=1\). These elements are at all times arranged in an ordered list that is revised as follows: The element selected is moved to the front of the list with the relative positions of all the other elements remaining unchanged. Define the state at any time to be the list ordering at that time and note that there are \(n !\) possible states. (a) Argue that the preceding is a Markov chain. (b) For any state \(i_{1}, \ldots, i_{n}\) (which is a permutation of \(\left.1,2, \ldots, n\right)\), let \(\pi\left(i_{1}, \ldots, i_{n}\right)\) denote the limiting probability. In order for the state to be \(i_{1}, \ldots, i_{n}\), it is necessary for the last request to be for \(i_{1}\), the last non- \(i_{1}\) request for \(i_{2}\), the last non- \(i_{1}\) or \(i_{2}\) request for \(i_{3}\), and so on. Hence, it appears intuitive that $$ \pi\left(i_{1}, \ldots, i_{n}\right)=P_{i_{1}} \frac{P_{i_{2}}}{1-P_{i_{1}}} \frac{P_{i_{3}}}{1-P_{i_{1}}-P_{i_{2}}} \cdots \frac{P_{i_{n-1}}}{1-P_{i_{1}}-\cdots-P_{i_{n-2}}} $$ Verify when \(n=3\) that the preceding are indeed the limiting probabilities.

Consider a branching process having \(\mu<1\). Show that if \(X_{0}=1\), then the expected number of individuals that ever exist in this population is given by \(1 /(1-\mu)\). What if \(X_{0}=n ?\)

On a chessboard compute the expected number of plays it takes a knight, starting in one of the four corners of the chessboard, to return to its initial position if we assume that at each play it is equally likely to choose any of its legal moves. (No other pieces are on the board.) Hint: Make use of Example 4.36.
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