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It follows from Theorem \(4.2\) that for a time reversible Markov chain $$ P_{i j} P_{j k} P_{k i}=P_{i k} P_{k j} P_{j i}, \quad \text { for all } i, j, k $$ It turns out that if the state space is finite and \(P_{i j}>0\) for all \(i, j\), then the preceding is also a sufficient condition for time reversibility. (That is, in this case, we need only check Equation \((4.26)\) for paths from \(i\) to \(i\) that have only two intermediate states.) Prove this. Hint: Fix \(i\) and show that the equations $$ \pi_{j} P_{j k}=\pi_{k} P_{k j} $$ are satisfied by \(\pi_{j}=c P_{i j} / P_{j i}\), where \(c\) is chosen so that \(\sum_{j} \pi_{j}=1\)

Short Answer

Expert verified
In conclusion, we have proved that the given sufficient condition for time reversibility is true for finite state spaces when \(P_{ij}>0\) for all \(i,j\). We found a valid expression for \(\pi_j\) and \(\pi_k\) as \(\pi_j=c\frac{P_{ij}}{P_{ji}}\) and \(\pi_k=c\frac{P_{ik}}{P_{ki}}\), where \(c\) is chosen to make the sum of all \(\pi_j\) equal to 1. These expressions satisfy the detailed balance equations and, considering the given relation from Theorem 4.2, further validate the claim for time reversibility under the given conditions.

Step by step solution

01

Use Theorem 4.2 and the sufficient condition

Theorem 4.2 states that for a time reversible Markov chain, we have: \[ P_{ij}P_{jk}P_{ki} = P_{ik}P_{kj}P_{ji} \quad \text{for all } i,j,k \] Given the sufficient condition for time reversibility, if the state space is finite and \(P_{ij}>0\) for all \(i,j\), then we need to check the equality for paths from \(i\) to \(i\) that have only two intermediate states, \(j\) and \(k\). Taking this into account, we need to prove that the detailed balance equations, \[ \pi_j P_{jk} = \pi_k P_{kj} \] are satisfied by \(\pi_j=c\frac{P_{ij}}{P_{ji}}\), where \(c\) is chosen so that \(\sum_j \pi_j=1\).
02

Find the relationship between \(\pi_j\) and \(\pi_k\)

Using the given expression for \(\pi_j\), we have: \[ \pi_j = c\frac{P_{ij}}{P_{ji}} \] Now, we need to find the relationship between \(\pi_j\) and \(\pi_k\). We can do this by expressing \(\pi_k\) in terms of \(\pi_j\). Using the same expression for \(\pi_k\), we have: \[ \pi_k = c\frac{P_{ik}}{P_{ki}} \]
03

Check if the detailed balance equation is satisfied by the given expression

We need to check if our given expression for \(\pi_j\) and \(\pi_k\) satisfies the detailed balance equation: \[ \pi_j P_{jk} = \pi_k P_{kj} \] Plug in our expressions for \(\pi_j\) and \(\pi_k\): \[ c\frac{P_{ij}}{P_{ji}}P_{jk} = c\frac{P_{ik}}{P_{ki}}P_{kj} \] Notice that the constant \(c\) can be cancelled from both sides of the equation. Now, we need to check if the remaining equality holds true for all states \(i, j, k\), given the sufficient condition for time reversibility from Theorem 4.2: \[ P_{ij}P_{jk}P_{ki} = P_{ik}P_{kj}P_{ji} \] Comparing the remaining equation from Step 3 and Theorem 4.2, we see that they are indeed the same, proving that the detailed balance equation is satisfied by the given expression: \[ \frac{P_{ij}}{P_{ji}}P_{jk} = \frac{P_{ik}}{P_{ki}}P_{kj} \]

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Key Concepts

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

Detailed Balance Equations
In understanding Markov chains, the concept of detailed balance equations is fundamental. These equations provide the conditions under which the flow of probability between any two states is balanced, meaning the system is in equilibrium. For a Markov chain, the detailed balance conditions can be written as:

\[\begin{equation}\pi_j P_{jk} = \pi_k P_{kj}\end{equation}\]
where \(\pi_j\) and \(\pi_k\) are the stationary probabilities of states \(j\) and \(k\) respectively, and \(P_{jk}\) and \(P_{kj}\) are the transition probabilities from state \(j\) to \(k\), and vice versa.
The detailed balance equations are not just abstract mathematical constructs—they are crucial for practical applications such as analyzing the long-run behavior of Markovian systems used in queueing theory, population genetics, and thermodynamics, to name a few. When these equations hold, they ensure that the Markov chain is time reversible, implying that observing the process backwards in time would yield the same transition probabilities as forwards.
State Space
The state space of a Markov chain is basically the set of all possible states that the process can occupy. It's like the board in a board game wherein each spot marks a state that players could land on. For Markov chains, the state space can have finite or infinite members. Let's consider a Markov chain with a finite state space, which is particularly relevant for computationally modeling many real-world processes. Here, every state can be listed out and the transition probabilities between them quantified.

In mathematical terms, if a Markov chain has \(n\) states, the state space can be represented as \(\{1, 2, ..., n\}\). The transition probabilities associated with this state space are represented by a matrix, which gives the probability of moving from any given state to another in one step. In this matrix, rows typically represent the current state, while columns represent the next state.
Theorem 4.2
The backbone of understanding the time reversibility of a Markov chain is encapsulated in Theorem 4.2. This theorem establishes a criteria that defines when a Markov chain can be considered time reversible. The time reversible property is an elegant aspect of a Markov process that tells us that navigating through the state space backwards is as probable as navigating forwards.

Specifically, Theorem 4.2 dictates that for a Markov chain to be time reversible, it must satisfy the following condition for all states \(i\), \(j\), and \(k\):\[\begin{equation}P_{ij}P_{jk}P_{ki} = P_{ik}P_{kj}P_{ji}\end{equation}\]
This equation implies a balance for transitions between three states in both directions. It serves as a key to unlocking many otherwise difficult to solve problems about the long-term behavior and structure of Markovian systems.
Markov Chain Time Reversibility
Let's focus on the concept of time reversibility in Markov chains, which essentially determines whether the process looks the same when observed in forward or reverse time. If you think of a movie of a physical process, time reversibility suggests that playing the movie backwards wouldn't make a difference to the transitions being observed. Mathematically, this is a powerful property because it simplifies a lot of the analysis and opens the door to using powerful tools from equilibrium statistical mechanics.

For a Markov chain to be time reversible, it has to adhere to not just the detailed balance equations but also the conditions laid out in Theorem 4.2. Together, these form a set of criteria that, when met, indicate that the process does not distinguish between the direction of time. Time reversibility is an attractive property when modeling systems in statistical physics or finance because it implies a kind of symmetry and equilibrium that can be very helpful for analyses.

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Most popular questions from this chapter

In a Markov decision problem, another criterion often used, different than the expected average return per unit time, is that of the expected discounted return. In this criterion we choose a number \(\alpha, 0<\alpha<1\), and try to choose a policy so as to maximize \(E\left[\sum_{i=0}^{\infty} \alpha^{i} R\left(X_{i}, a_{i}\right)\right]\) (that is, rewards at time \(n\) are discounted at rate \(\left.\alpha^{n}\right)\) Suppose that the initial state is chosen according to the probabilities \(b_{i} .\) That is, $$ P\left\\{X_{0}=i\right\\}=b_{i}, \quad i=1, \ldots, n $$ For a given policy \(\beta\) let \(y_{j a}\) denote the expected discounted time that the process is in state \(j\) and action \(a\) is chosen. That is, $$ y_{j a}=E_{\beta}\left[\sum_{n=0}^{\infty} \alpha^{n} I_{\left[X_{n}=j, a_{n}=a\right\\}}\right] $$ where for any event \(A\) the indicator variable \(I_{A}\) is defined by $$ I_{A}=\left\\{\begin{array}{ll} 1, & \text { if } A \text { occurs } \\ 0, & \text { otherwise } \end{array}\right. $$ (a) Show that $$ \sum_{a} y_{j a}=E\left[\sum_{n=0}^{\infty} \alpha^{n} I_{\left\\{X_{n}=j\right\\}}\right] $$ or, in other words, \(\sum_{a} y_{j a}\) is the expected discounted time in state \(j\) under \(\beta\). (b) Show that $$ \begin{aligned} \sum_{j} \sum_{a} y_{j a} &=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a} &=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \end{aligned} $$ Hint: For the second equation, use the identity $$ I_{\left\\{X_{n+1}=j\right\\}}=\sum_{i} \sum_{a} I_{\left\\{X_{n-i}, a_{n-a}\right\rangle} I_{\left\\{X_{n+1}=j\right\\}} $$ Take expectations of the preceding to obtain $$ E\left[I_{\left.X_{n+1}=j\right\\}}\right]=\sum_{i} \sum_{a} E\left[I_{\left\\{X_{n-i}, a_{n-a}\right\\}}\right] P_{i j}(a) $$ (c) Let \(\left\\{y_{j a}\right\\}\) be a set of numbers satisfying $$ \begin{aligned} \sum_{j} \sum_{a} y_{j a} &=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a} &=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \end{aligned} $$ Argue that \(y_{j a}\) can be interpreted as the expected discounted time that the process is in state \(j\) and action \(a\) is chosen when the initial state is chosen according to the probabilities \(b_{j}\) and the policy \(\beta\), given by $$ \beta_{i}(a)=\frac{y_{i a}}{\sum_{a} y_{i a}} $$ is employed. Hint: Derive a set of equations for the expected discounted times when policy \(\beta\) is used and show that they are equivalent to Equation \((4.38) .\) (d) Argue that an optimal policy with respect to the expected discounted return criterion can be obtained by first solving the linear program $$ \begin{array}{ll} \operatorname{maximize} & \sum_{j} \sum_{a} y_{j a} R(j, a), \\ \text { such that } & \sum_{j} \sum_{a} y_{j a}=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a}=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \\ y_{j a} \geqslant 0, \quad \text { all } j, a \end{array} $$ and then defining the policy \(\beta^{*}\) by $$ \beta_{i}^{*}(a)=\frac{y_{i a}^{*}}{\sum_{a} y_{i a}^{*}} $$ where the \(y_{j a}^{*}\) are the solutions of the linear program.

In a good weather year the number of storms is Poisson distributed with mean \(1 ;\) in a bad year it is Poisson distributed with mean 3. Suppose that any year's weather conditions depends on past years only through the previous year's condition. Suppose that a good year is equally likely to be followed by either a good or a bad year, and that a bad year is twice as likely to be followed by a bad year as by a good year. Suppose that last year-call it year 0 -was a good year. (a) Find the expected total number of storms in the next two years (that is, in years 1 and 2 ). (b) Find the probability there are no storms in year 3 . (c) Find the long-run average number of storms per year.

Coin 1 comes up heads with probability \(0.6\) and \(\operatorname{coin} 2\) with probability \(0.5 . \mathrm{A}\) coin is continually flipped until it comes up tails, at which time that coin is put aside and we start flipping the other one. (a) What proportion of flips use coin 1? (b) If we start the process with \(\operatorname{coin} 1\) what is the probability that \(\operatorname{coin} 2\) is used on the fifth flip?

Show that the stationary probabilities for the Markov chain having transition probabilities \(P_{i, j}\) are also the stationary probabilities for the Markov chain whose transition probabilities \(Q_{i, j}\) are given by $$ Q_{i, j}=P_{i, j}^{k} $$ for any specified positive integer \(k\).

Find the average premium received per policyholder of the insurance company of Example \(4.27\) if \(\lambda=1 / 4\) for one-third of its clients, and \(\lambda=1 / 2\) for two-thirds of its clients.

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