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Chapter 4: Problem 47

Let \(\left\\{X_{n}, n \geqslant 0\right\\}\) denote an ergodic Markov chain with limiting probabilities \(\pi_{i} .\) Define the process \(\left\\{Y_{n}, n \geqslant 1\right\\}\) by \(Y_{n}=\left(X_{n-1}, X_{n}\right)\). That is, \(Y_{n}\) keeps track of the last two states of the original chain. Is \(\left\\{Y_{n}, n \geqslant 1\right\\}\) a Markov chain? If so, determine its transition probabilities and find $$ \lim _{n \rightarrow \infty} P\left\\{Y_{n}=(i, j)\right\\} $$

Short Answer

Expert verified
The process \(\{Y_n, n\geqslant 1\}\) is a Markov chain, with transition probabilities given by \(p_{j k}\), which are the transition probabilities of the original chain \(\{X_n, n\geqslant 0\}\). The limiting probability of the \(\{Y_n, n\geqslant 1\}\) chain is given by \(\lim_{n\rightarrow\infty} P\{Y_n = (i, j)\} = \pi_i \cdot p_{i j}\).

Step by step solution

01

Verify the Markov property

For the process \(\{Y_n, n\geqslant 1\}\) to be a Markov chain, it must satisfy the Markov property, which states that: $$ P(Y_{n+1} = (j, k) \mid Y_n = (i, j), Y_{n-1} = (h, i), \ldots, Y_1 = (a, b)) = P(Y_{n+1} = (j, k) \mid Y_n = (i, j)) $$ Notice that \(Y_{n+1} = (X_n, X_{n+1})\). Since \(\{X_n, n \geqslant 0\}\) is a Markov chain, we know that: $$ P(X_{n+1} = k \mid X_n = j, X_{n-1} = i, \ldots, X_0 = a) = P(X_{n+1} = k \mid X_n = j) $$ Given that \(Y_n = (X_{n-1}, X_n)\) and the original chain satisfies the Markov property, we can rewrite the property for the \(Y_n\) chain as: $$ P(Y_{n+1} = (j, k) \mid Y_n = (i, j)) = P(X_{n+1} = k \mid X_n = j) $$ This shows that the process \(\{Y_n, n\geqslant 1\}\) does satisfy the Markov property and is thus a Markov chain.
02

Determine the transition probabilities

To find the transition probabilities of the \(\{Y_n, n\geqslant 1\}\) chain, we need to calculate: $$ P(Y_{n+1} = (j, k) \mid Y_n = (i, j)) $$ Using the Markov property we've shown in Step 1, we can write this as: $$ P(Y_{n+1} = (j, k) \mid Y_n = (i, j)) = P(X_{n+1} = k \mid X_n = j) = p_{j k} $$ So, the transition probabilities for the \(\{Y_n, n\geqslant 1\}\) chain are given by \(p_{j k}\), which are the transition probabilities of the original chain \(\{X_n, n\geqslant 0\}\).
03

Calculate the limiting probability

To find the limiting probability, we want to calculate: $$ \lim_{n\rightarrow\infty} P\{Y_n = (i, j)\} $$ Since we know that \(\{X_n, n \geqslant 0\}\) is an ergodic Markov chain with limiting probabilities \(\pi_i\), we have: $$ \lim_{n\rightarrow\infty} P(X_n = i) = \pi_i $$ and $$ \lim_{n\rightarrow\infty} P(X_{n-1} = i) = \pi_i $$ Now, we can use these limiting probabilities along with the transition probabilities calculated in Step 2: $$ \lim_{n\rightarrow\infty} P\{Y_n = (i, j)\} = \lim_{n\rightarrow\infty} P(X_{n-1} = i \text{ and } X_n = j) = \lim_{n\rightarrow\infty} P(X_{n-1} = i) \cdot P(X_n = j \mid X_{n-1} = i) $$ Using the limiting probabilities \(\pi_i\) and transition probabilities \(p_{i j}\), we get: $$ \lim_{n\rightarrow\infty} P\{Y_n = (i, j)\} = \pi_i \cdot p_{i j} $$ Therefore, the limiting probability of the \(\{Y_n, n\geqslant 1\}\) chain is given by \(\pi_i \cdot p_{i j}\).

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Key Concepts

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

Ergodic Markov Chain
A Markov chain is a mathematical model describing a sequence of possible events in which the probability of each event depends only on the state attained in the previous event. An ergodic Markov chain is a special type of Markov chain that is recurrent, aperiodic, and positive. This means it has the property that from any starting state, it is possible to eventually reach all other states with a non-zero probability, and every state has a chance of being revisited in a non-fixed number of steps. A key feature of ergodic Markov chains is that they have limiting probabilities, which are the long-term, steady-state probabilities of being in any given state.

In the context of the exercise provided, your understanding of the ergodic nature of the Markov chain is crucial. It guarantees that no matter which state the process starts from, it will eventually reach a steady-state distribution. This property is essential to solve for the limiting probabilities of the process \(\{Y_n, n \geqslant 1\}\).
Limiting Probabilities
The concept of limiting probabilities relates to the long-run behavior of a Markov chain. After a large number of steps, the probability of being in a specific state will stabilize and no longer change significantly as the chain progresses. These are the probabilities that a Markov chain will be in a particular state in the long-term, and are independent of the initial state. This is particularly important in the provided exercise, where you鈥檙e asked to find the limiting probability for the process \(\{Y_n, n \geqslant 1\}\).

Exercise Improvement Tip:

When approaching an exercise like this, consider reinforcing the concept of limiting probabilities by mapping out the transition diagram of the Markov chain. By understanding how one state leads to another and applying the ergodic property, you'll more readily see why the states eventually stabilize, allowing for the calculation of limiting probabilities.
Transition Probabilities
At the core of any Markov chain model are the transition probabilities, which are the probabilities of moving from one state to another. They are represented by \(p_{ij}\) which is the probability of transitioning from state \(i\) to state \(j\). Understanding these probabilities is essential for modeling the behavior of the chain over time and for computing other values such as limiting probabilities.

In the exercise, the transition probabilities for the new process \(\{Y_n, n \geqslant 1\}\) are derived from the transition probabilities of the original ergodic Markov chain \(\{X_n, n \geqslant 0\}\). It's important to note that these probabilities dictate how likely it is to move between various states and ultimately drive the results when computing the long-term behavior of the chain.

Exercise Improvement Tip:

A great way to fully comprehend and visualize transition probabilities is by creating a matrix or a graph that represents all possible transitions. This approach simplifies complex chains by breaking them down into individual movements between states and can significantly improve conceptual understanding.

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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 certain town never has two sunny days in a row. Each day is classified as being either sunny, cloudy (but dry), or rainy. If it is sunny one day, then it is equally likely to be either cloudy or rainy the next day. If it is rainy or cloudy one day, then there is one chance in two that it will be the same the next day, and if it changes then it is equally likely to be either of the other two possibilities. In the long run, what proportion of days are sunny? What proportion are cloudy?

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

Consider a Markov chain in steady state. Say that a \(k\) length run of zeroes ends at time \(m\) if $$ X_{m-k-1} \neq 0, \quad X_{m-k}=X_{m-k+1}=\ldots=X_{m-1}=0, X_{m} \neq 0 $$ Show that the probability of this event is \(\pi_{0}\left(P_{0,0}\right)^{k-1}\left(1-P_{0,0}\right)^{2}\), where \(\pi_{0}\) is the limiting probability of state 0 .

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 \(X_{0}=n ?\)
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