/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 For the Markov chain with states... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(s_{i 3}\) 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 summary, to find the first passage probabilities \(f_{i3}\) and mean first passage times \(s_{i3}\) for \(i=1,2,3\) with given transition probability matrix \(\mathbf{P}\), we utilize the recursive formula for first passage probabilities and the definition of mean first passage time. Calculation involves iterating over the recursions until a desired level of precision is obtained.

Step by step solution

01

Calculate the first passage probabilities, \(f_{i3}\)

The first passage probabilities \(f_{ij}\) are defined as the probability that the Markov chain will first visit state \(j\) from state \(i\). They can be calculated recursively using the following formula: $$ f_{ij}^{(n)} = \sum_{k \neq j} p_{ik} f_{kj}^{(n-1)}, \quad n \geq 1, $$ where \(f_{ij}^{(n)}\) denotes the probability that the process first enters state \(j\) at step \(n\) and \(p_{ik}\) is the transition probability from state \(i\) to state \(k\). For \(i=1,2,3\), we will compute \(f_{i3}\) using the formula above: 1. When \(i = 1\), \(f_{13}^{(n)} = 0.1\cdot f_{33}^{(n-1)}+0.2\cdot f_{23}^{(n-1)}+0.4\cdot f_{13}^{(n-1)}\). 2. When \(i = 2\), \(f_{23}^{(n)} = 0.2\cdot f_{33}^{(n-1)}+0.5\cdot f_{23}^{(n-1)}+0.1\cdot f_{13}^{(n-1)}\). 3. When \(i = 3\), \(f_{33}^{(n)} = 0.2\cdot f_{33}^{(n-1)}+0.4\cdot f_{23}^{(n-1)}+0.3\cdot f_{13}^{(n-1)}\). With these recursions, we can calculate the first passage probabilities until the desired convergence is obtained.
02

Calculate the mean first passage times, \(s_{i3}\)

The mean first passage time, \(s_{ij}\), is the expected time it takes for the Markov chain to reach state \(j\) for the first time from state \(i\). It can be calculated using the first passage probabilities, \(f_{ij}^{(n)}\), as follows: $$ s_{ij} = \sum_{n=1}^{\infty} n \cdot f_{ij}^{(n)}. $$ Using the previously calculated first passage probabilities, \(f_{ij}^{(n)}\), for \(i=1,2,3\), we can now compute the mean first passage times, \(s_{i3}\), by summing the series for each \(i\). Remember that, in general, we must use numerical methods to compute both first passage probabilities and mean first passage times. By iterating over the recursions, we can obtain these values to a desired level of precision.

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Most popular questions from this chapter

Suppose that coin 1 has probability \(0.7\) of coming up heads, and \(\operatorname{coin} 2\) has probability \(0.6\) of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select coin 2 to flip tomorrow. If the coin initially flipped is equally likely to be coin 1 or coin 2 , then what is the probability that the coin flipped on the third day after the initial flip is coin \(1 ?\)

For a series of dependent trials the probability of success on any trial is \((k+1) /(k+2)\) where \(k\) is equal to the number of successes on the previous two trials. Compute \(\lim _{n \rightarrow \infty} P\) \\{success on the \(n\) th trial\\}

Consider an irreducible finite Markov chain with states \(0,1, \ldots, N\). (a) Starting in state \(i\), what is the probability the process will ever visit state \(j\) ? Explain! (b) Let \(x_{i}=P\\{\) visit state \(N\) before state \(0 \mid\) start in \(i\\}\). Compute a set of linear equations which the \(x_{i}\) satisfy, \(i=0,1, \ldots, N\). (c) If \(\sum_{j} j P_{i j}=i\) for \(i=1, \ldots, N-1\), show that \(x_{i}=i / N\) is a solution to the equations in part (b)

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Trials are performed in sequence. If the last two trials were successes, then the next trial is a success with probability \(0.8\); otherwise the next trial is a success with probability \(0.5\). In the long run, what proportion of trials are successes?

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