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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

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 .

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state \(i, i=0,1,2,3\), if the first urn contains \(i\) white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left\\{X_{n}, n=0,1,2, \ldots\right\\}\) is a Markov chain and calculate its transition probability matrix.

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?

A Markov chain is said to be a tree process if (i) \(\quad P_{i j}>0\) whenever \(P_{j i}>0\), (ii) for every pair of states \(i\) and \(j, i \neq j\), there is a unique sequence of distinct states \(i=i_{0}, i_{1}, \ldots, i_{n-1}, i_{n}=j\) such that $$ P_{i_{k}, i_{k+1}}>0, \quad k=0,1, \ldots, n-1 $$ In other words, a Markov chain is a tree process if for every pair of distinct states \(i\) and \(j\) there is a unique way for the process to go from \(i\) to \(j\) without reentering a state (and this path is the reverse of the unique path from \(j\) to \(i\) ). Argue that an ergodic tree process is time reversible.

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.

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