# Spring 2018, Math 171

#### Week 3

1. Stopping/Non-Stopping times

1. Let $$T_1, T_2$$ be stopping times for some Markov Chain $$\{X_n:n \ge 0\}$$. Which of the following will also necessarily be stopping times? Prove your claims.

1. (Discussed) $$T_3=5$$

2. $$T_4=T_1 + T_2 + 1$$

3. (Discussed) $$T_5=T_1 + T_2 - 1$$

• (Solution) $$T_5$$ will not necessarily be a stopping time. Suppose $$\{X_n:n \ge 0\}$$ is the Markov Chain corresponding to the transition matrix $P = \begin{matrix} & \mathbf 0 & \mathbf 1 \cr \mathbf 0 & 1/2 & 1/2 \cr \mathbf 1 & 0 & 1 \end{matrix}$ Suppose further that $$T_1 = \min\{n \ge 0: X_n = \mathbf 0\}$$ and $$T_2 = \min\{n \ge 0: X_n = \mathbf 1\}$$. If $$T_5$$ were a stopping time we would have $$P(T_5 = n | X_n = x_n, \dots, X_0=x_0)\in \{0, 1\} \; \forall n$$. However, by definition of $$T_5$$$P(T_5 = 0 | X_0=\mathbf 0) = P(T_1 + T_2 = 1 | X_0=\mathbf 0)$ by directly enumerating the possibilities we see $= P(T_1 = 1, T_2 = 0 | X_0=\mathbf 0)$ $+ P(T_1 = 0, T_2 = 1 | X_0=\mathbf 0)$ and now using the Multiplication Rule $= P(T_1 = 1 | T_2 = 0, X_0=\mathbf 0)P(T_2 = 0 | X_0=\mathbf 0)$ $+ P(T_2 = 1 | T_1 = 0, X_0=\mathbf 0)P(T_1 = 0 | X_0=\mathbf 0)$ since $$T_1$$ and $$T_2$$ are stopping times this simplifies to $= P(T_1 = 1 | X_0=\mathbf 0)P(T_2 = 0 | X_0=\mathbf 0)$$+ P(T_2 = 1 | X_0=\mathbf 0)P(T_1 = 0 | X_0=\mathbf 0)$ each term of which can be computed $=0 \cdot 0 + \frac 1 2 \cdot 1$$= \frac 1 2 \notin \{0, 1\}$
2. (Discussed) Solve problem 4(ii) on HW 2

• (Discussed) Show Lemma 1.3 from the textbook
2. Classification of States

1. Consider the Markov chain defined by the following transition matrix: $P = \begin{matrix} & \mathbf 1 & \mathbf 2 & \mathbf 3 & \mathbf 4 & \mathbf 5 & \mathbf 6 & \mathbf 7 & \mathbf 8 \cr \mathbf 1 & 0.5 & 0 & 0.5 & 0 & 0 & 0 & 0 & 0 \cr \mathbf 2 & 0.5 & 0.5 & 0 & 0 & 0 & 0 & 0 & 0 \cr \mathbf 3 & 0 & 0 & 0 & 0.5 & 0 & 0 & 0 & 0.5 \cr \mathbf 4 & 0 & 0 & 0.5 & 0 & 0.5 & 0 & 0 & 0 \cr \mathbf 5 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \cr \mathbf 6 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \cr \mathbf 7 & 0 & 0 & 0 & 0 & 0 & 0.5 & 0 & 0.5 \cr \mathbf 8 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{matrix}$ Identify the transient and recurrent states, and the irreducible closed sets in the Markov chain. Give reasons for your answers.

2. Consider the Markov chain defined by the following transition matrix: $P = \begin{matrix} & \mathbf 1 & \mathbf 2 & \mathbf 3 & \mathbf 4 & \mathbf 5 & \mathbf 6 \cr \mathbf 1 & 0 & 0 & 1 & 0 & 0 & 0 \cr \mathbf 2 & 0 & 0 & 0 & 0 & 0 & 1 \cr \mathbf 3 & 0 & 0 & 0 & 0 & 1 & 0 \cr \mathbf 4 & 0.25 & 0.25 & 0 & 0.5 & 0 & 0 \cr \mathbf 5 & 1 & 0 & 0 & 0 & 0 & 0 \cr \mathbf 6 & 0 & 0.5 & 0 & 0 & 0 & 0.5 \end{matrix}$ Identify the transient and recurrent states, and the irreducible closed sets in the Markov chain. Give reasons for your answers.

3. Stationary Distributions

Recall a stationary distribution is a vector $$\pi$$ satisfying: $\sum _i \pi(i)=1$ $\pi(i)\ge 0, \quad \forall i$ $\pi P = \pi$

1. Compute any and all stationary distributions of $P = \begin{bmatrix} 0 & 0 & 1 \cr 1 & 0 & 0 \cr 0 & 1 & 0 \end{bmatrix}$ If you claim $$P$$ has a unique stationary distribution, please justify.

2. (Partially Discussed) Under what circumstances is the stationary distribution of $P = \begin{bmatrix} 1-r & 0 & r \cr p & 1-p & 0 \cr 0 & q & 1-q \end{bmatrix}$ unique? Justify your answer. Compute the stationary distribution in this case.

3. Compute any and all stationary distributions of $P = \begin{bmatrix} 1 & 0 \cr 0 & 1 \end{bmatrix}$ If you claim $$P$$ has a unique stationary distribution, please justify.

4. Compute any and all stationary distributions of $P = \begin{bmatrix} P_1 & 0 \cr 0 & P_2 \end{bmatrix}$ where $$P_1$$ has a unique stationary distribution $$\pi_1$$ and $$P_2$$ has a unique stationary distribution $$\pi_2$$. If you claim $$P$$ has a unique stationary distribution, please justify.

5. Compute any and all stationary distributions of $P = \begin{bmatrix} 0 & p & 0 & 1-p \cr q & 0 & 1-q & 0 \cr 0 & 1-r & 0 & r \cr 1-s & 0 & s & 0 \end{bmatrix}$ If you claim $$P$$ has a unique stationary distribution, please justify.