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Chapter 6: Problem 5

The dog-flea model of Example (6.5.5) is a Markov chain \(X\) on the state space \(\\{0,1, \ldots, m\) ) with transition probabilities $$ p_{i, i+1}=1-\frac{i}{m}, \quad p_{i, i-1}=\frac{i}{m}, \quad \text { for } \quad 0 \leq i \leq m $$ Show that, if \(X_{0}=i\), $$ \mathrm{E}\left(X_{n}-\frac{m}{2}\right)=\left(i-\frac{m}{2}\right)\left(1-\frac{2}{m}\right)^{n} \rightarrow 0 \quad \text { as } n \rightarrow \infty $$

Short Answer

Expert verified
The expectation approaches zero as \(n\rightarrow\infty\).

Step by step solution

01

Identify Initial Definitions

Identify the definitions and initial parameters provided in the problem. Here, we have a Markov chain denoted by \(X_n\) with state space \(\{0,1,2,\ldots,m\}\). The transition probabilities are specified: \(p_{i, i+1} = 1 - \frac{i}{m}\) and \(p_{i, i-1} = \frac{i}{m}\). The expectation \(\mathrm{E}\left(X_n - \frac{m}{2}\right)\) is to be shown for given expressions.
02

Evaluate Transition Probabilities

Express the transition probabilities in terms of expected values. The equation for expectation in Markov processes depends on current state transitions. The transitions go from \(i\) to \(i+1\) with probability \(1-\frac{i}{m}\) and to \(i-1\) with probability \(\frac{i}{m}\).
03

Formulate the Expectation Equation

Develop the expectation equation based on transitions: \[ \mathrm{E}[X_{n+1} - \frac{m}{2}] = (1 - \frac{i}{m})\mathrm{E}[X_{n}] + \frac{i}{m}(\mathrm{E}[X_{n} - 1]) - \frac{m}{2}. \] Simplify the expression to track the changes due to state transitions.
04

Simplify and Solve Recurrence Relation

Solve the recurrence relation obtained for the expectation: \[\mathrm{E}(X_{n}) = i\left(1 - \frac{2}{m}\right)^{n} + \frac{m}{2}(1 - \left(1 - \frac{2}{m}\right)^{n}).\] This forms the basis for understanding long-term behavior as \(n\) increases.
05

Evaluate Long-Term Behavior

Analyze the expectation as \(n\to\infty\). Substitute the derived formula into the long-term expectation: \[ \mathrm{E}\left(X_{n} - \frac{m}{2}\right) = \left(i - \frac{m}{2}\right)\left(1 - \frac{2}{m}\right)^n. \] As \(n\rightarrow\infty\), \( \left(1 - \frac{2}{m}\right)^n \to 0\).

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Key Concepts

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

Transition Probabilities
In the context of Markov chains, transition probabilities describe the likelihood of moving from one state to another. For the dog-flea model described, these probabilities show how fleas move between dogs over time. In this model, if a current state is represented by the number of fleas on one dog, the transition probabilities determine how many fleas are likely to move to the next state:
  • If you are in state \(i\), representing \(i\) fleas, the probability of moving to state \(i+1\) (gaining a flea) is \(1 - \frac{i}{m}\).
  • Conversely, the probability of moving to state \(i-1\) (losing a flea) is \(\frac{i}{m}\).
These probabilities ensure the states鈥� progression is stochastic, meaning each next state depends solely on the current state. This concept is fundamental as it sets up all calculations related to expected values and long-term behavior.
Expectation
Expectation, in probability theory, refers to the average or mean value that you would expect from a random process after many trials. For our Markov chain, it describes the average number of fleas we expect to find on a dog after a certain number of transitions. Here, we are interested in the expectation of the difference from the midpoint, \(\frac{m}{2}\), which corresponds to half of the total fleas distributed equally over two dogs:
  • The expectation equation is structured as \(\mathrm{E}[X_{n+1} - \frac{m}{2}]\), accounting for current and next states based on transition probabilities.
Thus, the shifting of probabilities affects the outcome of this expectation, highlighting how interactions and movements within states alter the expected value over time.
Recurrence Relation
A recurrence relation is an equation that recursively defines a sequence or multidimensional array of values. In this exercise, the recurrence relation helps us to calculate expectations iteratively through each step:
  • The recurrence relation derived was \[\mathrm{E}(X_{n}) = i\left(1 - \frac{2}{m}\right)^{n} + \frac{m}{2}(1 - \left(1 - \frac{2}{m}\right)^{n})\].
  • This allows the calculation of the expectation for future states based on their previous states.
The recurrence relation remains crucial as it outlines the systematic approach needed in finding the expectation's future values, helping predict how the distribution evolves.
Long-Term Behavior
The long-term behavior of a Markov chain refers to how the system behaves as it progresses over an extended period. In this case, we analyze the expectation as \(n\) approaches infinity:
  • The given equation simplifies to \(\mathrm{E}\left(X_{n} - \frac{m}{2}\right) = \left(i - \frac{m}{2}\right)\left(1 - \frac{2}{m}\right)^n\).
  • As \(n\to\infty\), the term \(\left(1 - \frac{2}{m}\right)^{n}\) tends towards zero, indicating the expectation stabilizes and the number of fleas becomes evenly distributed between the dogs.
This convergence to a stable state is an essential characteristic of Markov chains, which often exhibit predictable behavior in the long term.

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Most popular questions from this chapter

Let \(\mathrm{P}\) be the transition matrix of a Markov chain with finite state space. Let \(\mathrm{I}\) be the identity matrix, \(\mathbf{U}\) the \(|S| \times|S|\) matrix with all entries unity, and 1 the row \(|S|\)-vector with all entries unity. Let \(\pi\) be a non-negative vector with \(\sum_{i} \pi_{i}=1\). Show that \(\pi \mathbf{P}=\pi\) if and only if \(\pi(\mathbf{I}-\mathbf{P}+\mathbf{U})=1\). Deduce that if \(\mathbf{P}\) is irreducible then \(\boldsymbol{\pi}=\mathbf{1}(\mathbf{I}-\mathbf{P}+\mathbf{U})^{-1} .\)

Let \(X\) be a Markov chain with state space \(S=(1,2,3)\) and transition matrix $$ \mathbf{P}=\left(\begin{array}{ccc} 1-p & p & 0 \\ 0 & 1-p & p \\ p & 0 & 1-p \end{array}\right) $$ where \(0

A particle performs a random walk on the vertex set of a connected graph \(G\), which for simplicity we assume to have neither loops nor multiple edges. At cach stage it moves to a ncighbour of its current position, each such neighbour being chosen with equal probability. If \(G\) has \(\eta(<\infty)\) edges, show that the stationary distribution is given by \(\pi_{v}=d_{v} /(2 \eta)\), where \(d_{0}\) is the degree of vertex \(v\).

Show that not every discrete-time Markov chain can be imbedded in a continuous-time chain. More precisely, let $$ \mathbf{P}=\left(\begin{array}{cc} \alpha & 1-\alpha \\ 1-\alpha & \alpha \end{array}\right) \quad \text { for some } 0<\alpha<1 $$ be a transition matrix. Show that there exists a uniform semigroup \(\left(\mathbf{P}_{\mathrm{t}}\right)\) of transition probabilities in, continuous time such that \(\mathbf{P}_{1}=\mathbf{P}\), if and only if \(\frac{1}{2}<\alpha<1\). In this case show that \(\left(\mathbf{P}_{f}\right)\) is unique and calculate it in terms of \(\alpha\).

Jobs arrive in a computer queue in the manner of a Poisson process with intensity \(\lambda\). The central processor handles them one by one in the order of their arrival, and each has an exponentially distributed runtime with parameter \(\mu\), the runtimes of different jobs being independent of each other and of the arrival process. Let \(X(t)\) be the number of jobs in the system (either running or waiting) at time \(t\), where \(X(0)=0 .\) Explain why \(X\) is a Markov chain, and write down its generator. Show that a stationary distribution exists if and only if \(\lambda<\mu\), and find it in this case.
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