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For the Markov chain with states \(1,2,3,4\) whose transition probability matrix \(\mathbf{P}\) is as specified below find \(f_{i 3}\) and \(s_{a}\) for \(i=1,2,3 .\) $$ \mathbf{P}=\left[\begin{array}{llll} 0.4 & 0.2 & 0.1 & 0.3 \\ 0.1 & 0.5 & 0.2 & 0.2 \\ 0.3 & 0.4 & 0.2 & 0.1 \\ 0 & 0 & 0 & 1 \end{array}\right] $$

Short Answer

Expert verified
In short, by solving the system of linear equations obtained from the first passage time formula, we find the first passage probabilities \(f_{13}\), \(f_{23}\) and \(f_{33}\). Next, we compute the limiting probabilities \(s_a\) by finding the left eigenvectors of the matrix P corresponding to the eigenvalue 1 and normalizing these vectors.

Step by step solution

01

Understanding the transition probability matrix

The transition probability matrix P contains the probabilities of transitioning from one state to another in the Markov chain. Each entry \(p_{ij}\) in the matrix denotes the probability of transitioning from state i to state j. Specifically, we are given the following matrix: $$ \mathbf{P}=\left[\begin{array}{llll} 0.4 & 0.2 & 0.1 & 0.3 \\\ 0.1 & 0.5 & 0.2 & 0.2 \\\ 0.3 & 0.4 & 0.2 & 0.1 \\\ 0 & 0 & 0 & 1 \end{array}\right] $$
02

Computing first passage probabilities

We will use the first passage time formula to calculate the probability of first passage to state 3 (i.e., \(f_{i3}\)). The formula for first passage probability is given by the following equation: $$ f_{ij} = p_{ij} + \sum_{k \neq j} p_{ik}f_{kj} $$ Here, \(f_{ij}\) is the first passage probability from state i to state j. In our case, j = 3. For \(i=1\): $$ f_{13} = p_{13} + \sum_{k \neq 3} p_{1k}f_{k 3} = 0.1 + 0.4f_{13} + 0.2f_{23} + 0.3f_{43} $$ For \(i=2\): $$ f_{23} = p_{23} + \sum_{k \neq 3} p_{2k}f_{k 3} = 0.2 + 0.1f_{13} + 0.5f_{23} + 0.2f_{43} $$ For \(i=3\): $$ f_{33} = p_{33} + \sum_{k \neq 3} p_{3k}f_{k 3} = 0.2 + 0.3f_{13} + 0.4f_{23} + 0.1f_{43} $$ Now, we have a system of three linear equations with three unknowns to solve. After solving it, we will get the first passage probabilities \(f_{13}\), \(f_{23}\) and \(f_{33}\).
03

Computing limiting probabilities

The limiting probabilities of a Markov chain are computed by finding the limit of the transition probability matrix as n approaches infinity. The limiting probabilities, denoted by \(s_a\), will be a probability distribution over the states. To compute \(s_a\), we can use the following equation: $$ \lim_{n \rightarrow \infty} \mathbf{P}^n \mathbf{s}_a = \mathbf{s}_a $$ Here, \(\mathbf{P}^n\) is the matrix P raised to the power n. The limiting probabilities \(\mathbf{s}_a\) are the left eigenvectors of the matrix \(\mathbf{P}\) corresponding to the eigenvalue 1. To find the limiting probabilities, you will need to compute the left eigenvectors of the transition probability matrix P and normalize these vectors so that they sum to 1. After finding the limiting probabilities, you will have the values for \(s_a\) for each state in the Markov chain.

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Key Concepts

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

Transition Probability Matrix
A transition probability matrix is at the heart of any Markov chain analysis. This matrix, often denoted as \( \mathbf{P} \), consists of entries that represent the likelihood of moving from one state to another within a system. Imagine the matrix as a roadmap of possibilities, where each row corresponds to a current state and each column to a potential next state. The entry at \( p_{ij} \) tells us the probability of transitioning from state \( i \) to state \( j \) in one step. According to our exercise, the matrix is structured as follows:

\[ \mathbf{P}=\begin{bmatrix} 0.4 & 0.2 & 0.1 & 0.3 \ 0.1 & 0.5 & 0.2 & 0.2 \ 0.3 & 0.4 & 0.2 & 0.1 \ 0 & 0 & 0 & 1 \end{bmatrix} \]
The sum of probabilities in any row must equal 1 since one of the moves must take place. For effective learning, students should become comfortable with reading and interpreting these matrices, as they show the complete set of transitions at a glance. It's critical to note any patterns, such as states that do not lead to other states (e.g., the last row in our matrix), which can significantly impact the behavior of the chain.
First Passage Probabilities
First passage probabilities are a fascinating area within Markov chains, telling us the odds of reaching a specific state for the first time, starting from another state. In other words, it measures the likelihood that the process will enter state \( j \) for the first time, starting from state \( i \), in any future step. Our exercise gives us a formula to find this probability:

\[ f_{ij} = p_{ij} + \sum_{k eq j} p_{ik}f_{kj} \]
Here, \( f_{ij} \) represents the first passage probability from state \( i \) to state \( j \) where \( p_{ik} \) is the transition probability from state \( i \) to an intermediary state \( k \) and \( f_{kj} \) is the probability of first passage from this state \( k \) to the desired state \( j \). Such calculations often involve solving a system of linear equations, which can be both challenging and enlightening. The math sheds light not just on the probability of eventual arrival, but also on the interconnected nature of states within the chain.

Students are encouraged to understand this concept thoroughly, as it provides deeper insights into the dynamics of stochastic processes, and strengthens one's ability to predict the behavior of complex systems.
Limiting Probabilities
In exploring Markov chains, we often seek to understand their long-term behavior, and limiting probabilities serve this purpose splendidly. These probabilities, when they exist, describe the steady-state distribution of the states, after many transitions have been made — essentially, they tell us where the system is likely to settle in the long run. But how do we find these elusive probabilities? One way is by examining the powers of the transition probability matrix \( \mathbf{P} \), considering an infinite number of steps:

\[ \lim_{n \to \infty} \mathbf{P}^n \mathbf{s}_a = \mathbf{s}_a \]
In this equation, if the matrix raised to higher and higher powers converges to a stable matrix, the rows of that matrix give us the limiting probabilities for each state. Alternatively, we look for the left eigenvectors associated with the eigenvalue 1. As students work through these problems, they must understand that not all Markov chains have limiting distributions, and those that do, reach a sort of equilibrium. Digesting this concept is essential for using Markov chains to predict long-term outcomes in various fields, from population genetics to queueing theory and finance. By mastering this, students can appreciate the power of the Markovian approach to understanding systems that evolve randomly over time.

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Most popular questions from this chapter

(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 $$ \pi_{j}=\sum_{i} \pi_{i} Q_{u} $$ (b) Give an intuitive explanation for the result of part (a).

A flea moves around the vertices of a triangle in the following manner: Whenever it is at vertex \(i\) it moves to its clockwise neighbor vertex with probability \(p_{i}\) and to the counterclockwise neighbor with probability \(q_{i}=1-p_{i}, i=1,2,3\) (a) Find the proportion of time that the flea is at each of the vertices. (b) How often does the flea make a counterclockwise move which is then followed by 5 consecutive clockwise moves?

In the gambler's ruin problem of Section \(4.5 .1\), suppose the gambler's fortune is presently \(i\), and suppose that we know that the gambler's fortune will eventually reach \(N\) (before it goes to 0). Oiven this information, show that the probability he wins the next gamble is $$ \begin{aligned} &\frac{p\left[1-(q / p)^{1+1}\right]}{1-(q / p)^{l}}, & \text { if } p \neq \frac{1}{2} \\ &\frac{i+1}{2 i}, \quad \text { if } p=\frac{1}{2} \end{aligned} $$ The probability we want is $$ \begin{aligned} P\left(X_{n+1}\right.&\left.=i+1 \mid X_{n}=i, \lim _{m \rightarrow \infty} X_{m}=N\right] \\ &=\frac{P\left[X_{n+1}=i+1, \lim _{m} X_{m}=N \mid X_{n}=i\right)}{P\left[\lim _{m} X_{m}=N \mid X_{n}=i\right]} \end{aligned} $$

Suppose that on each play of the game a gambler either wins 1 with probability \(p\) or loses 1 with probability \(1-p\). The gambler continues betting until she or he is either winning \(n\) or losing \(m\). What is the probability that the gambler quits a winner?

Each day, one of \(n\) possible elements is requested, the ith one with probability \(P_{i}, i \geq 1, \sum_{1}^{n} P_{i}=1 .\) These elements are at all times arranged in an ordered list which 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 \(\pi\left(i_{1}, \ldots, t_{n}\right)\) 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 \(l_{3}\), and so on. Hence, it appears intuitive that $$ \pi\left(i_{1}, \ldots, i_{n}\right)=P_{t_{1}} \frac{P_{i_{2}}}{1-P_{i_{1}}} \frac{P_{h_{1}}}{1-P_{i_{1}}-P_{h_{2}}} \cdots \frac{P_{i_{m-1}}}{1-P_{i_{1}}-\cdots-P_{i_{\alpha}-2}} $$ Verify when \(n=3\) that the above are indeed the limiting probabilities.

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