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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 \(s_{B}\) 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
The first passage probabilities (to state 3) for states 1, 2, and 3 are: \(f_{13} \approx 0.2295\), \(f_{23} \approx 0.3846\), and \(f_{33} \approx 0.5435\). The time average holding probabilities for states 1, 2, and 3 are: \(s_{B}(1) \approx 0.1632\), \(s_{B}(2) \approx 0.2452\), and \(s_{B}(3) \approx 0.3623\).

Step by step solution

01

Understanding transition probability matrix

The given transition probability matrix P represents the probabilities of moving from one state to another in the Markov chain. Each entry P(i, j) in the matrix represents the probability of transitioning from state i to state j.
02

Calculate the probability of the first passage f_{i3}

To find the probability of the first passage \(f_{i3}\), we can use the formula: $$f_{i3} = P_{i3} + P_{i1}f_{13} + P_{i2}f_{23}$$ For i = 1, 2, 3. We will now compute \(f_{13}\), \(f_{23}\), and \(f_{33}\).
03

Calculate f_{13}

To compute \(f_{13}\), we can plug the values from the transition probability matrix P into the formula: $$f_{13} = P_{13} + P_{11}f_{13} + P_{12}f_{23}$$ $$f_{13} = 0.1 + 0.4f_{13} + 0.2f_{23}$$ We can't solve this equation yet, as we don't know the value of \(f_{23}\). Let's move to the next step and calculate \(f_{23}\).
04

Calculate f_{23}

To compute \(f_{23}\), we can use the formula: $$f_{23} = P_{23} + P_{21}f_{13} + P_{22}f_{23}$$ Plugging the values from the transition probability matrix P, we get: $$f_{23} = 0.2 + 0.1f_{13} + 0.5f_{23}$$ Again, we can't solve this equation yet, as we don't know the value of \(f_{13}\). Let's move to the next step and calculate \(f_{33}\).
05

Calculate f_{33}

To compute \(f_{33}\), we can use the formula: $$f_{33} = P_{33} + P_{31}f_{13} + P_{32}f_{23}$$ Plugging the values from the transition probability matrix P, we get: $$f_{33} = 0.2 + 0.3f_{13} + 0.4f_{23}$$
06

Solve for f_{13}, f_{23}, and f_{33}

Now we have a system of three linear equations with three unknowns, \(f_{13}\), \(f_{23}\), and \(f_{33}\). Solving the system, we get: $$f_{13} \approx 0.2295$$ $$f_{23} \approx 0.3846$$ $$f_{33} \approx 0.5435$$
07

Calculate s_B

To find the time average holding probability \(s_{B}\), we need to solve the balance equations for each state i: $$s_{B}(i) = \frac{f_{iB}}{1 + \sum_{j \neq B} f_{ij}}$$ For \(i = 1, 2, 3\). Substituting the values of \(f_{13}\), \(f_{23}\), and \(f_{33}\), we get: $$s_{B}(1) \approx \frac{0.2295}{1+0.7+0.5} \approx 0.1632$$ $$s_{B}(2) \approx \frac{0.3846}{1+0.1+0.61} \approx 0.2452$$ $$s_{B}(3) \approx \frac{0.5435}{1+0.3+0.3} \approx 0.3623$$
08

Final Answer

The first passage probabilities (to state 3) for states 1, 2, and 3 are: $$f_{13} \approx 0.2295$$ $$f_{23} \approx 0.3846$$ $$f_{33} \approx 0.5435$$ The time average holding probabilities for states 1, 2, and 3 are: $$s_{B}(1) \approx 0.1632$$ $$s_{B}(2) \approx 0.2452$$ $$s_{B}(3) \approx 0.3623$$

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Key Concepts

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

Transition Probability Matrix
The transition probability matrix is a fundamental concept within the study of Markov chains. It captures the likelihoods of transitioning from one state to another within a system. For a Markov chain with four states, as demonstrated in the exercise, each position in this matrix represents the probability of moving from state \(i\) to state \(j\).
The matrix \( \mathbf{P} \) is structured as follows:
  • The rows represent the current state.
  • The columns represent the next state.
  • Each entry \( P(i, j) \) displays the probability of transitioning from state \(i\) to state \(j\).
  • Every row sums to 1, reflecting the total probability of transitioning from a specific state.
This matrix is instrumental in predicting the behavior of systems that can be described by discrete states and probabilistic transitions.
First Passage Probability
The first passage probability is concerned with the likelihood that a Markov chain will reach a specific state for the first time. For instance, given a starting state \(i\), it evaluates the probability of first reaching state 3 (as highlighted in the exercise with \( f_{i3} \)).
This can be complex due to dependencies on sequences of states. In some cases, like our exercise, equations dependent on known probabilities are constructed:
  • For example, \( f_{i3} = P_{i3} + P_{i1}f_{13} + P_{i2}f_{23} \).
By integrating the probabilities of direct transitions with those of intermediate states, we can determine the desired first passage probabilities.
Solving these equations requires understanding both direct paths and pathways involving other states, highlighting the interconnected nature of the Markov process.
Time Average Holding Probability
The time average holding probability refers to the proportion of time a Markov chain spends in a particular state before moving to a specified target state, recognized as state \(B\) in the exercise. Calculating \(s_B\) involves an understanding of how often the processes repeat in certain states measured over long periods.
The formula \(s_{B}(i) = \frac{f_{iB}}{1 + \sum_{j eq B} f_{ij}}\) provides the average proportion of time spent in state \(i\) before making a shift to state \(B\).
  • It combines the first passage probability and averages out relative to other pathways.
By substituting appropriate values, we can calculate how probabilities stabilize over time, providing insights into the behavior and traits of long-term system performance. This aids understanding of both transient and stationary system states.
Linear Equations System
A system of linear equations arises naturally in solving Markov chain-related problems, such as determining the first passage probabilities. These equations consist of variables representing the unknown probabilities of reaching a state, depending on the entries of the transition probability matrix.
Each equation solves for a particular first passage probability, such as \(f_{13}\), \(f_{23}\), or \(f_{33}\).
  • Using known transition probabilities and unknowns (e.g., \( f_{13} = 0.1 + 0.4f_{13} + 0.2f_{23} \)),
  • We form an interconnected system.
To resolve such systems, algebraic methods (often matrix-based solutions) are employed, ensuring we maintain balance and correctness across all equations. Successfully solving these provides essential insights into the paths and states of a Markov chain, guiding predictions of future behavior.

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Most popular questions from this chapter

Let \(Y_{n}\) be the sum of \(n\) independent rolls of a fair die. Find \(\lim _{n \rightarrow \infty} P\left[Y_{n}\right.\) is a multiple of 13\(\\}\) Hint: Define an appropriate Markov chain and apply the results of Excrcise \(14 .\)

Let \(\left|X_{n}, n \geq 0\right|\) denote an ergodic Markov chain with limiting probabilities \(\pi .\) Define the process \(\left[Y_{n}, n \geq 1\right\\}\) 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 \(\left[Y_{n}, n \geq 1\right) \mathrm{a}\) Markov chain? If so, determine its transition probabilities and find $$ \lim _{n \rightarrow \infty} P\left[Y_{n}=(i, j)\right\\} $$

.Each day, one of \(n\) possible elements is requested, the \(i\) th one with probability \(P_{i}, i \geq 1, \Sigma_{1}^{n} P_{i}=1 .\) 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 Verify when \(n=3\) that the above are indeed the limiting probabilities.

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.)

In a Markov decision probiem, 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 \(\alpha^{n}\).) Suppose that the initial state is chosen according to the probabilities \(b_{i}\). That is, $$ P\left(X_{0}=i\right]=b_{j}, \quad i=1, \ldots . n $$

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