/*! 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 57 A particle moves among \(n+1\) v... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle moves among \(n+1\) vertices that are situated on a circle in the following manner. At each step it moves one step either in the clockwise direction with probability \(p\) or the counterclockwise direction with probability \(q=1-p\). Starting at a specified state, call it state 0 , let \(T\) be the time of the first return to state \(0 .\) Find the probability that all states have been visited by time \(T\).

Short Answer

Expert verified
The probability that all states have been visited by time T is given by \(\frac{p - pq^n - q^n + q^{n+1}}{1 - q^n}\), where the particle moves clockwise with probability p and counterclockwise with probability q.

Step by step solution

01

Condition on the Initial Transition

To begin with, let us consider two cases: the first case where the particle moves clockwise initially, and the second case when the particle moves counterclockwise initially. For the first case, the probability of moving clockwise is p; for the second case, the probability of moving counterclockwise is q.
02

Utilize Gambler's Ruin Problem

Now, let's use the results from the gambler's ruin problem. Let P_i be the probability of visiting all the remaining states before returning to state 0, starting from state i. Using the gambler's ruin problem, we know that the probability of ruin is given by: \(P_i = \frac{1 - \left(\frac{q}{p}\right)^i}{1 - \left(\frac{q}{p}\right)^{n}}\) Now consider our two cases mentioned in Step 1, and calculate the probability using the formula above, for each case.
03

Probability in First Case (Clockwise Initial Movement)

In the first case, the particle moves clockwise initially. So, it starts from state 1. We can calculate P_1 using the formula above: \(P_1 = \frac{1 - \left(\frac{q}{p}\right)^1}{1 - \left(\frac{q}{p}\right)^{n}}\)
04

Probability in Second Case (Counterclockwise Initial Movement)

In the second case, the particle moves counterclockwise initially. So, it starts from state n. We can calculate P_n using the formula above: \(P_n = \frac{1 - \left(\frac{q}{p}\right)^n}{1 - \left(\frac{q}{p}\right)^{n}}\)
05

Combine the Two Cases

Now that we have calculated the probabilities for both cases, we need to find the overall probability of visiting all vertices before returning to state 0. We can do this by multiplying each case's probability by its respective initial transition probability and summing the two results: Overall Probability = \(p \times P_1 + q \times P_n\)
06

Simplify the Overall Probability

Substitute the expressions for P_1 and P_n from Steps 3 and 4 into the overall probability: Overall Probability = \(p \times \frac{1 - \frac{q}{p}}{1 - \left(\frac{q}{p}\right)^{n}} + q \times \frac{1 - \left(\frac{q}{p}\right)^n}{1 - \left(\frac{q}{p}\right)^{n}}\) Now, simplify the expression to get the final answer: Overall Probability = \(\frac{p - pq^n - q^n + q^{n+1}}{1 - q^n}\) So, the probability that all states have been visited by time T is \(\frac{p - pq^n - q^n + q^{n+1}}{1 - q^n}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gambler's Ruin Problem
The Gambler's Ruin problem is a classic problem in probability theory and stochastic processes. It examines a scenario where a gambler places bets in a series of fair or biased games. The game typically ends when the gambler either goes broke or reaches a target fortune. In a mathematical context, this problem helps us determine the likelihood of a player achieving their desired fortune before losing everything, starting from a certain amount.
In our specific exercise, we relate this problem to a particle moving among vertices on a circle. The particle's movement is akin to a gambler trying to "ruin" by moving to different states. The use of probabilities in both clockwise and counterclockwise directions gives this problem a twist, as it takes into consideration the biased movement akin to winning or losing bets in the gambler’s ruin scenario.
  • The calculation formula for the probability in Gambler's Ruin problem is given by: \(P_i = \frac{1 - \left(\frac{q}{p}\right)^i}{1 - \left(\frac{q}{p}\right)^{n}}\).
  • "Ruin" in this context means reaching back to the starting state without visiting all states, which is calculated using probabilities of moving in either direction.
Understanding this problem helps us mathematically grasp complex systems' behavior with an element of random "chance," very useful in fields like finance, physics, and more.
Stochastic Processes
Stochastic processes represent a collection of random variables that are indexed by time or space, offering a way to model randomly changing systems or phenomena. They are used to model situations where outcomes are uncertain and vary over time. In our particle movement problem, the positions of the particle over time align well with the concept of a stochastic process.
The movement of the particle is contingent upon the probability of moving either clockwise with probability \(p\), or counterclockwise with \(q = 1-p\). Here, each move is a step in our stochastic process, and every vertex or state the particle lands on represents a possible outcome.
  • A fundamental property of stochastic processes is that they can be Markovian, meaning the future state depends only on the current state, not on the series of events that preceded it.
  • In our exercise, the position is updated each step and this movement is influenced by inherent randomness quantified by \(p\) and \(q\).
Modelling problems like our particle movements with stochastic processes helps forecast system behavior under uncertainty, an instrumental tool in sectors like weather prediction, stock markets, and population dynamics.
Probability Theory
Probability theory is essential for analyzing and interpreting problems involving random processes or events. It provides the mathematical foundation on which we examine the likelihood of different outcomes, as seen when calculating the probability of the particle visiting all states in the circle.
In this exercise, probability theory helps us structure and solve the problem regarding the particle's travel from one vertex to another. By calculating the probability formulas for moving clockwise or counterclockwise initially, and determining the probability that all states have been visited, we make informed predictions about the system.
  • Probability in basic terms quantifies how likely an event is to occur, with important properties such as additivity and independence.
  • In the context of our problem, probability theory underpins all calculations, from the initial setup of \(p\) and \(q\), to using those probabilities in the Gambler's Ruin problem model.
Probability theory offers a robust framework to rigorously tackle ambiguity and prediction in real-world problems, setting the stage for decision-making in uncertain scenarios.

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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 \(\alpha^{n}\) ). 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 \(j\) un\(\operatorname{der} \beta\). (b) Show that $$ \begin{aligned} \sum_{j} \sum_{a} y_{j a} &=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a} &=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \end{aligned} $$ (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 \(\boldsymbol{\beta}\), 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 $$ \begin{array}{ll} \text { maximize } & \sum_{j} \sum_{a} y_{j a} R(j, a), \\ \text { such that } & \sum_{j} \sum_{a} y_{j a}=\frac{1}{1-\alpha}, \\ \sum_{a} y_{j a}=b_{j}+\alpha \sum_{i} \sum_{a} y_{i a} P_{i j}(a) \\ y_{j a} \geqslant 0, \quad \text { all } j, a \end{array} $$ and then defining the policy \(\beta^{*}\) by $$ \beta_{i}^{*}(a)=\frac{y_{i a}^{*}}{\sum_{a} y_{i a}^{*}} $$ where the \(y_{i a}^{*}\) are the solutions of the linear program.

It follows from the argument made in Exercise 38 that state \(i\) is null recurrent if it is recurrent and \(\pi_{i}=0 .\) Consider the one-dimensional symmetric random walk of Example \(4.15\). (a) Argue that \(\pi_{i}=\pi_{0}\) for all \(i\). (b) Argue that all states are null recurrent.

A certain town never has two sunny days in a row. Each day is classified as being either sunny, cloudy (but dry), or rainy. If it is sunny one day, then it is equally likely to be either cloudy or rainy the next day. If it is rainy or cloudy one day, then there is one chance in two that it will be the same the next day, and if it changes then it is equally likely to be either of the other two possibilities. In the long run, what proportion of days are sunny? What proportion are cloudy?

Suppose that on each play of the game a gambler either wins 1 with probability \(p\) or loses 1 with probability \(1-p .\) The gambler continues betting until she or he is either winning \(n\) or losing \(m\). What is the probability that the gambler quits a winner?

Let \(A\) be a set of states, and let \(A^{c}\) be the remaining states. (a) What is the interpretation of $$ \sum_{i \in A} \sum_{j \in A^{c}} \pi_{i} P_{i j} ? $$ (b) What is the interpretation of $$ \sum_{i \in A^{c}} \sum_{j \in A} \pi_{i} P_{i j} ? $$ (c) Explain the identity $$ \sum_{i \in A} \sum_{j \in A^{c}} \pi_{i} P_{i j}=\sum_{i \in A^{c}} \sum_{j \in A} \pi_{i} P_{i j} $$

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