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

Short Answer

Expert verified
Matrix P1 has one recurrent class {1, 2, 3}. Matrix P2 has one transient class {1, 2} and one recurrent class {3, 4}. Matrix P3 has two recurrent classes {1, 2, 3} and {4, 5}. Matrix P4 has two recurrent classes {1, 2, 5} and {3, 4}.

Step by step solution

01

We will analyze the structure of matrix P1 to identify the classes. A class is a set of states that are interconnected (i.e., can reach each other). Matrix P1: \[ \left( \begin{array}{ccc} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0\end{array} \right) \] From the matrix, we can see that states 1, 2, and 3 are interconnected as there are nonzero probabilities of transitioning between them. Thus, there is one class containing all states {1, 2, 3}. #Step 2: Determine if P1 is transient or recurrent#

Now, we will determine if the class in P1 is transient or recurrent. A class is transient if it's likely that once you leave the class, you will never return to it; it's recurrent if you will eventually return to the class. States 1, 2, and 3 form a closed class, meaning there are no outgoing connections from this class to other states. This closed class represents a recurrent class since once in this class, you will always stay in it. #Step 3: Identify the classes for matrix P2#
02

We will now analyze the structure of matrix P2 to identify the classes. Matrix P2: \[ \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0\end{array} \right) \] From the matrix, we can see that the states can be grouped into two classes: - Class 1: {1, 2} - Class 2: {3, 4} #Step 4: Determine if P2 is transient or recurrent#

Now, we will determine if the classes in P2 are transient or recurrent. - Class 1: {1, 2} is transient because once state 4 is reached from state 1 or 2, there is no way to return to either state 1 or 2. - Class 2: {3, 4} is recurrent because it is a closed class. Once in this class, you will always stay in it. #Step 5: Identify the classes for matrix P3#
03

We will now analyze the structure of matrix P3 to identify the classes. Matrix P3: \[ \left( \begin{array}{ccccc} \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2}\end{array} \right) \] From the matrix, we can see that the states can be grouped into two classes: - Class 1: {1, 2, 3} - Class 2: {4, 5} #Step 6: Determine if P3 is transient or recurrent#

Now, we will determine if the classes in P3 are transient or recurrent. - Class 1: {1, 2, 3} is recurrent because it is a closed class. Once in this class, you will always stay in it. - Class 2: {4, 5} is recurrent because it is also a closed class. Once in this class, you will always stay in it. #Step 7: Identify the classes for matrix P4#
04

We will now analyze the structure of matrix P4 to identify the classes. Matrix P4: \[ \left( \begin{array}{ccccc} \frac{1}{4} & \frac{3}{4} & 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 \\ 1 & 0 & 0 & 0 & 0\end{array} \right) \] From the matrix, we can see that the states can be grouped into two classes: - Class 1: {1, 2, 5} - Class 2: {3, 4} #Step 8: Determine if P4 is transient or recurrent#

Now, we will determine if the classes in P4 are transient or recurrent. - Class 1: {1, 2, 5} is recurrent because it is a closed class. Once in this class, you will always stay in it. - Class 2: {3, 4} is recurrent because it is also a closed class. Once in this class, you will always stay in it. #Conclusion# For matrix P1, there is one recurrent class containing states {1, 2, 3}. For matrix P2, there is one transient class containing states {1, 2} and one recurrent class containing states {3, 4}. For matrix P3, there are two recurrent classes containing states {1, 2, 3} and {4, 5}, respectively. For matrix P4, there are two recurrent classes containing states {1, 2, 5} and {3, 4}, respectively.

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Most popular questions from this chapter

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.

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 $$ \lim _{n \rightarrow \infty} P\left\\{Y_{n}=(i, j)\right\\} $$

Let \(\mathrm{P}\) be the transition probability matrix of a Markov chain. Argue that if for some positive integer \(r, \mathrm{P}^{r}\) has all positive entries, then so does \(\mathrm{P}^{n}\), for all integers \(n \geqslant r\).

Consider a process \(\left\\{X_{n}, n=0,1, \ldots\right\\}\), which takes on the values 0,1 , or 2 . Suppose $$ \begin{aligned} &P\left\\{X_{n+1}=j \mid X_{n}=i, X_{n-1}=i_{n-1}, \ldots, X_{0}=i_{0}\right\\} \\ &\quad=\left\\{\begin{array}{ll} P_{i j}^{\mathrm{I}}, & \text { when } n \text { is even } \\ P_{i i}^{\mathrm{II}}, & \text { when } n \text { is odd } \end{array}\right. \end{aligned} $$ where \(\sum_{j=0}^{2} P_{i j}^{\mathrm{I}}=\sum_{j=0}^{2} P_{i j}^{\mathrm{II}}=1, i=0,1,2 .\) Is \(\left\\{X_{n}, n \geqslant 0\right\\}\) a Markov chain? If not, then show how, by enlarging the state space, we may transform it into a Markov chain.

Recall that state \(i\) is said to be positive recurrent if \(m_{i, i}<\infty\), where \(m_{i, i}\) is the expected number of transitions until the Markov chain, starting in state \(i\), makes a transition back into that state. Because \(\pi_{i}\), the long-run proportion of time the Markov chain, starting in state \(i\), spends in state \(i\), satisfies $$ \pi_{i}=\frac{1}{m_{i, i}} $$ it follows that state \(i\) is positive recurrent if and only if \(\pi_{i}>0\). Suppose that state \(i\) is positive recurrent and that state \(i\) communicates with state \(j .\) Show that state \(j\) is also positive recurrent by arguing that there is an integer \(n\) such that $$ \pi_{j} \geqslant \pi_{i} P_{i, j}^{n}>0 $$

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