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Chapter 3: Problem 1

Consider the following random walk: $$ \begin{array}{llll} P_{i, i+1} & =p & \text { with } \quad 0

Short Answer

Expert verified
The closed-form expression for the expected time to absorption, \(d(k)\), given the initial state k and transition probabilities p and q, is: \(d(k) = \frac{k}{p} \cdot \frac{r - k}{r - 1} + \frac{k (k - 1)}{2 (r - 1)}\)

Step by step solution

01

Define the Waiting Time Problem

We want to find the expected time to get absorbed into states 0 or r, given an initial state k. We can denote this expected time by \(d(k)\), where \(k = 1, 2, ..., r-1\).
02

Set Up Recursive Equations

In order to find the expected time to absorption, we consider all possible paths that would lead to absorption into states 0 or r. At each state k, there are two possible transitions: moving right to state \(k+1\) with probability p, or moving left to state \(k-1\) with probability q. For each transition, we need its respective waiting time until absorption, and we will average these times according to their probabilities to derive an equation for \(d(k)\). The recursive equation can be written as: \(d(k) = 1 + p \cdot d(k + 1) + q \cdot d(k - 1)\) This equation can be interpreted as: the expected waiting time at state k is equal to 1 (the current step) plus the expected waiting time for moving right or left, weighted by their respective probabilities.
03

Solve the Equations

We can solve this recursive equation for \(d(k)\) iteratively (by programming or other means), considering also that \(d(0) = 0\) and \(d(r) = 0\) since these are absorbing states. Alternatively, we can find a closed-form expression for \(d(k)\): Rearrange the equation: \(d(k + 1) - d(k) = \frac{1 - d(k - 1) }{p}\) Keep unrolling the recursive equation until absorption states to arrive at: \(d(k) = \frac{1}{p} \sum_{j=1}^k ( d(j) - d(j - 1) )\) Using the fact that \(d(0) = 0\), we can rewrite the sum as: \(d(k) = \frac{1}{p} \sum_{j=1}^k d(j)\) At this point, you can either keep this expression or solve it explicitly for any value of r and k: \(d(k) = \frac{k}{p} \cdot \frac{r - k}{r - 1} + \frac{k (k - 1)}{2 (r - 1)}\)
04

Find the Expected Time to Absorption

Finally, we can use the expression for the expected time to absorption, \(d(k)\), to calculate the expected time to absorption for any initial state k using the closed-form expression found in Step 3 (depending on what level of explicitness you want). To do this, we can plug in the given values of p, q, r, and the initial state k into the \(d(k)\) expression, and calculate the expected time to absorption.

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Most popular questions from this chapter

Consider a Markov chain with the \(N+1\) states \(0,1, \ldots, N\) and transition I robabilities $$ \begin{aligned} &P_{i j}=\left(\begin{array}{l} N \\ j \end{array}\right) \pi \dot{i}\left(1-\pi_{i}\right)^{N-j}, \quad 0 \leq i, \quad j \leq N \\ &\pi_{i}=\frac{1-e^{-2 a l / N}}{1-e^{-2 a}}, \quad a>0 \end{aligned} $$ Note that 0 and \(N\) are absorbing states. Verify that \(\exp \left(-2 a X_{t}\right)\) is a martingale [or, what is equivalent, prove the identity \(E\left(\exp \left(-2 a X_{t+1}\right)\left[X_{t}\right)=\right.\) \(\left.\exp \left(-2 a X_{t}\right)\right]\), where \(X_{t}\) is the state at time \(t(t=0,1,2, \ldots)\). Using this property show that the probability \(P_{N}(k)\) of absorption into state \(N\) starting nt state \(k\) is given by $$ P_{N}(k)=\frac{1-e^{-2 a k}}{1-e^{-2 a N}} $$

Consider a discrete time Markov chain with states \(0,1, \ldots, N\) whose matrix has elements $$ P_{i j}=\left\\{\begin{array}{cc} \mu_{i}, & j=i-1 \\ \lambda_{i}, & j=i+1 ; \quad i, j=0,1, \ldots, N . \\ 1-\lambda_{i}-\mu_{i}, & j=i \\ 0, & |j-i|>1 \end{array}\right. $$ Suppose that \(\mu_{0}=\lambda_{0}=\mu_{N}=\lambda_{N}=0\), and all other \(\mu_{i}^{\prime} s\) and \(\lambda_{i}\) 's are positive, and that the initial state of the process is \(k\). Determine the absorption probabilities at 0 and \(N\).

If \(\mathbf{P}\) is a finite Markov matrix, we define \(\mu(\mathbf{P})=\max _{i_{1}, i_{2} j}\left(P_{i_{1}, j}-P_{I_{2}, j}\right)\). Suppose \(P_{1}, P_{2}, \ldots, P_{k}\) are \(3 \times 3\) transition matrices of irreducible aperiodic Markov chains. Asbume furthermore that for any set of integers \(\alpha_{i}\left(1 \leq \alpha_{l} \leq k\right)\), \(i=1,2, \ldots, m, \Pi_{i=1}^{m_{1}} \mathbf{P}_{\alpha_{i}}\) is also the matrix of an aperiodic irreducible Markov chain. Prove that, for every \(\varepsilon>0\), there exists an \(M(\varepsilon)\) such that \(m>M\) implies $$ \mu\left(\prod_{i=1}^{m} \mathbf{P}_{\alpha_{i}}\right)<\varepsilon \quad \text { for any set } \alpha_{i}\left(1 \leq \alpha_{i} \leq k\right) \quad i=1,2, \ldots, m $$

Consider an irreducible Markov chain with a finite set of states \(\\{1,2, \ldots, N\\}\). Let \(\left\|P_{i j}\right\|\) be the transition probability matrix of the Markov chain and denote by \(\left\\{\pi_{j}\right\\}\) the stationary distribution of the process. Let \(\left\|P_{i j}^{(m)}\right\|\) denote the \(m\)-step transition probability matrix. Let \(\varphi(x)\) be a concave function on \(x \geq 0\) and define $$ E_{m}=\sum_{j=1}^{N} \pi_{j} \varphi\left(P_{j t}^{(m)}\right) \quad \text { with } l \text { fixed. } $$ Prove that \(E_{m}\) is a nondecreasing function of \(m\), i.e., \(E_{m+1} \geq E_{m}\) for all \(m \geq 1\)

Let \(\\{X(t), t \geq 0\\}\) and \(\\{Y(t), t \geq 0\\}\) be two independent Poisson processes with parameters \(\lambda_{1}\) and \(\lambda_{2}\), respectively. Define $$ Z(t)=X(t)-Y(t), \quad t \geq 0 $$ This is a stochastic process whose state space consists of all the integers. (positive, negative, and zero). Let $$ P_{n}(t)=\operatorname{Pr}\\{Z(t)=n\\}, \quad n=0, \pm 1, \pm 2, \ldots $$ Establish the formula $$ \sum_{n=-\infty}^{\infty} P_{n}(t) z^{n}=\exp \left(-\left(\lambda_{1}+\lambda_{2}\right) t\right) \exp \left(\lambda_{1} z t+\left(\lambda_{2} / z\right) t\right), \quad|z| \neq 0 $$ Compute \(E(Z(t))\) and \(E\left(Z(t)^{2}\right)\)
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