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A particle moves on a circle through points which have been markéd \(0,1,2,3,4\) (in a clockwise order). At each step it has a probability \(p\) of moving to the right (clockwise) and \(1-p\) to the left (counterclockwise). Let \(X_{n}\) denote its location on the circle after the \(n\) th step. The process \(\left[X_{n}, n \geqslant 0\right\\}\) is a Markov chain. (a) Find the transition probability matrix. (b) Calculate the limiting probabilities.

Short Answer

Expert verified
The transition probability matrix for the given Markov chain is: P = \( \begin{bmatrix} 0 & p & 0 & 0 & 1-p \\ 1-p & 0 & p & 0 & 0 \\ 0 & 1-p & 0 & p & 0 \\ 0 & 0 & 1-p & 0 & p \\ p & 0 & 0 & 1-p & 0 \\ \end{bmatrix} \) The limiting probabilities for all states are: \(\pi_0 = \pi_1 = \pi_2 = \pi_3 = \pi_4 = \frac{1}{5}\)

Step by step solution

01

Identify the states of the Markov chain

The given process \(\{X_n, n \geqslant 0\}\) is a Markov chain, where the states are the positions of the particle on the circle. There are five different states: 0, 1, 2, 3, and 4.
02

Determine the transition probabilities

Now, let's determine the probabilities of transitioning between states. The transition probabilities are the probabilities for the particle to move either clockwise or counterclockwise on the circle at each step. We know that the probability to go clockwise is p and counterclockwise is 1-p.
03

Set up the transition probability matrix

The transition probability matrix, denoted as P, of a Markov chain is a matrix that stores the probabilities of going from one state to another (in a single step). Let's set up the matrix for the given process. The matrix will have the form: P = \( \begin{bmatrix} P_{00}& P_{01}& P_{02}& P_{03}& P_{04} \\ P_{10}& P_{11}& P_{12}& P_{13}& P_{14} \\ P_{20}& P_{21}& P_{22}& P_{23}& P_{24} \\ P_{30}& P_{31}& P_{32}& P_{33}& P_{34} \\ P_{40}& P_{41}& P_{42}& P_{43}& P_{44} \\ \end{bmatrix} \) Each entry \(P_{ij}\) represents the probability of going from state i to state j in one step. For example, the particle in state 0 has a probability p to go to state 1 (clockwise) and probability 1-p to go to state 4 (counter-clockwise). So, \(P_{0,1}=p\) and \(P_{0,4}=1-p\). All other row entries are 0. Applying the same logic to the rest of the states, we obtain the transition probability matrix: P = \( \begin{bmatrix} 0 & p & 0 & 0 & 1-p \\ 1-p & 0 & p & 0 & 0 \\ 0 & 1-p & 0 & p & 0 \\ 0 & 0 & 1-p & 0 & p \\ p & 0 & 0 & 1-p & 0 \\ \end{bmatrix} \)
04

Calculate the limiting probabilities

To calculate the limiting probabilities of the Markov chain, we would need to find a probability vector \(\pi = (\pi_0, \pi_1, \pi_2, \pi_3, \pi_4)\) such that \(\pi P = \pi\), where \(\sum_{i=0}^{4} \pi_i = 1\). However, since the chain is symmetric (the probability of moving clockwise and counterclockwise is the same for all states), we can deduce that the limiting probabilities will be the same for each state, i.e.,: \(\pi_0=\pi_1=\pi_2=\pi_3=\pi_4\) Since \(\sum_{i=0}^{4} \pi_i = 1\), we can easily see that: \(\pi_0 = \pi_1 = \pi_2 = \pi_3 = \pi_4 = \frac{1}{5}\) #Output# Therefore, the transition probability matrix is: P = \( \begin{bmatrix} 0 & p & 0 & 0 & 1-p \\ 1-p & 0 & p & 0 & 0 \\ 0 & 1-p & 0 & p & 0 \\ 0 & 0 & 1-p & 0 & p \\ p & 0 & 0 & 1-p & 0 \\ \end{bmatrix} \) The limiting probabilities for the given Markov chain are: \(\pi_0 = \pi_1 = \pi_2 = \pi_3 = \pi_4 = \frac{1}{5}\)

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Key Concepts

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

Transition Probability Matrix
A transition probability matrix is a fundamental component in the study of Markov chains, serving as a compass that guides the probability of transitioning from one state to another over time. In our case, a particle moving on a circle marked with points labeled 0 through 4 illustrates a clear example. The transition probability matrix, denoted with a capital P, encapsulates the chances of this particle advancing in either a clockwise or counter-clockwise direction.

Imagine this matrix as a grid, where rows and columns represent the current state and the next state, respectively. Constructing such a matrix requires a methodical approach - analyzing potential movements and their associated probabilities. For the given problem, whenever the particle is at one of the points, it has only two potential movements: one step to the right (with probability p) or one step to the left (with probability 1-p). Therefore, the transitions from a state to itself are impossible except in loops, translating to zero probabilities except when the current state equals the next state, resulting in a diagonal of non-zero values.

By filling out the matrix with these probabilities, we establish a visual representation of the particle's journey around the circle, where the matrix's specific entries, labeled as P_{ij}, reflect the likelihood of transitioning from state i to state j.
Limiting Probabilities
When delving into Markov chains, we often seek to understand the concept of 'limiting probabilities.' These probabilities provide insight into the chain's long-term behavior and revealing a state's potential ‘residence’ as time extends indefinitely. In essence, limiting probabilities represent a state of equilibrium for the Markov chain after countless transitions have occurred.

For the particle circling through points 0 to 4, we initially observe a period of variability where its location after each step is unpredictable. However, as many steps are taken, the Markov chain may reveal a pattern of stabilizing probabilities for each state. This is where the magic of limiting probabilities comes into play. They are found by solving the equation \(\pi P = \pi\), under the condition that the sum of the probabilities must equal 1, adhering to the principles of probability.

In the case of our symmetrical (equal clockwise and counterclockwise probabilities) Markov chain, each state eventually carries the same weight in terms of limiting probability. This equal distribution signifies that, regardless of the starting point, the particle has an equal chance of being at any of the five marked points over a long period, resulting in the conclusion that the limiting probability for each state is \(\frac{1}{5}\).
Probability Models
Probability models are essentially blueprints for predicting the likelihood of various outcomes within random processes. Markov chains themselves form a class of probability models characterized by the 'memoryless' property - the future state depends only on the present state and not on the sequence of events that preceded it.

For educational purposes, the particle on the circle scenario serves as an exemplary probability model. In this construct, the model quantifies the particle's behavior in stepping from one marked point to another, with the transition governed by probabilities p and 1-p. The continuity and predictability of the process align with the model's design, making the Markov chain a powerful mathematical tool.

Probability models like Markov chains find applications across a diverse range of fields such as finance, computer science, and even biology, equipping researchers and analysts with a method to forecast future events based on current information. They enable the crafting of sophisticated simulations that offer a glimpse into the potential evolution of complex systems, enhancing our understanding and ability to make informed decisions.

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Most popular questions from this chapter

Suppose that a population consists of a fixed number, say, \(m\), of genes in any generation. Each gene is one of two possible genetic types. If any generation has exactly \(i\) (of its \(m\) ) genes being type 1 , then the next generation will have \(j\) type 1 (and \(m-j\) type 2 ) genes with probability $$ \left(\begin{array}{c} m \\ j \end{array}\right)\left(\frac{i}{m}\right)^{j}\left(\frac{m-i}{m}\right)^{m-j}, \quad j=0,1, \ldots, m $$ Let \(X_{n}\) denote the number of type 1 genes in the \(n\) th generation, and assume that \(X_{0}=i\). (a) Find \(E\left[X_{n}\right]\). (b) What is the probability that eventually all the genes will be type \(1 ?\)

44\. Suppose that a population consists of a fixed number, say, \(m\), of genes in any generation. Each gene is one of two possible genetic types. If any generation Consider an irreducible finite Markov chain with states \(0,1, \ldots, N\). (a) Starting in state \(i\), what is the probability the process will ever visit state \(j\) ? Explaia! (b) Let \(x_{i}=P\) \\{visit state \(N\) before state 0|start in \(\left.i\right\\}\). Compute a set of linear equations which the \(x_{i}\) satisfy, \(i=0,1, \ldots, N\). (c) If \(\sum_{j} j P_{i j}=i\) for \(i=1, \ldots, N-1\), show that \(x_{i}=i / N\) is a solution to the equations in part (b).

Let the transition probability matrix of a two-state Markov chain be given, as in Example 4.2, by $$ \mathbf{P}=\left\|\begin{array}{cc} p & 1-p \\ 1-p & p \end{array}\right\| $$ Show by mathematical induction that $$ \mathbf{P}^{(n)}=\left\|\begin{array}{ll} \frac{1}{2}+\frac{1}{2}(2 p-1)^{n} & \frac{1}{2}-\frac{1}{2}(2 p-1)^{n} \\ \frac{1}{2}-\frac{1}{2}(2 p-1)^{n} & \frac{1}{2}+\frac{1}{2}(2 p-1)^{n} \end{array}\right\| $$

Trials are performed in sequence. If the last two trials were successes, then the next trial is a success with probability \(0.8\); otherwise the next trial is a success with probability 0.5. In the long run, what proportion of trials are successes?

Prove that if the number of states in a Markov chain is \(M\), and if state \(j\) can be reached from state \(i\), then it can be reached in \(M\) steps or less.

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