/*! 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 6 Let the transition probability m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 6

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 $$ \mathbf{P}^{(n)}=\left\|\begin{array}{|ll} \frac{1}{2}+\frac{1}{2}(2 p-1)^{n} & \frac{1}{2}-\frac{1}{2}(2 p-1)^{n} \\ \frac{1}{2}-\frac{1}{2}(2 p-1)^{n} & \frac{1}{2}+\frac{1}{2}(2 p-1)^{n} \end{array}\right\| $$

Short Answer

Expert verified
In this problem, we used mathematical induction to prove the formula for the nth power of a two-state Markov chain's transition probability matrix. We first verified the base case for n=1, and then assumed the formula is true for n=k. Next, we showed that the formula also holds true for n=k+1 via matrix multiplication of \(\mathbf{P}^{(k)}\) and \(\mathbf{P}\). Since both the base case and inductive step have been proven, the formula holds for all integer values of n by mathematical induction.

Step by step solution

01

Base case (n=1)

First, we need to verify if the given formula holds true for n = 1. The given matrix \(\mathbf{P}\) is: \[ \mathbf{P}=\left\|\begin{array}{cc} p & 1-p \\\ 1-p & p \end{array}\right\| \] We need to prove that: \[ \mathbf{P}^{(1)}=\left\|\begin{array}{|ll} \frac{1}{2}+\frac{1}{2}(2 p-1)^{1} & \frac{1}{2}-\frac{1}{2}(2 p-1)^{1} \\\ \frac{1}{2}-\frac{1}{2}(2 p-1)^{1} & \frac{1}{2}+\frac{1}{2}(2 p-1)^{1} \end{array}\right\| \] Since \(\mathbf{P}^{(1)}\) is the matrix \(\mathbf{P}\) itself, the equation holds true for n = 1.
02

Inductive step

Now, assume that the formula holds true for n = k: \[ \mathbf{P}^{(k)}=\left\|\begin{array}{|ll} \frac{1}{2}+\frac{1}{2}(2 p-1)^{k} & \frac{1}{2}-\frac{1}{2}(2 p-1)^{k} \\\ \frac{1}{2}-\frac{1}{2}(2 p-1)^{k} & \frac{1}{2}+\frac{1}{2}(2 p-1)^{k} \end{array}\right\| \] Now, we need to show that the formula also holds true for n = k+1: \[ \mathbf{P}^{(k+1)}=\left\|\begin{array}{|ll} \frac{1}{2}+\frac{1}{2}(2 p-1)^{k+1} & \frac{1}{2}-\frac{1}{2}(2 p-1)^{k+1} \\\ \frac{1}{2}-\frac{1}{2}(2 p-1)^{k+1} & \frac{1}{2}+\frac{1}{2}(2 p-1)^{k+1} \end{array}\right\| \] To do this, we need to find the k+1 power of the transition probability matrix, \(\mathbf{P}^{(k+1)} = \mathbf{P}^{(k)} \times \mathbf{P}\), and show that it matches the desired formula.
03

Matrix multiplication

Compute the matrix multiplication \(\mathbf{P}^{(k)} \times \mathbf{P}\): \[ \left\|\begin{array}{|ll} \frac{1}{2}+\frac{1}{2}(2 p-1)^{k} & \frac{1}{2}-\frac{1}{2}(2 p-1)^{k} \\\ \frac{1}{2}-\frac{1}{2}(2 p-1)^{k} & \frac{1}{2}+\frac{1}{2}(2 p-1)^{k} \end{array}\right\| \left\|\begin{array}{cc} p & 1-p \\\ 1-p & p \end{array}\right\| \] This yields the resulting matrix: \[ \left\|\begin{array}{|ll} \frac{1}{2}+\frac{1}{2}(2 p-1)^{k+1} & \frac{1}{2}-\frac{1}{2}(2 p-1)^{k+1} \\\ \frac{1}{2}-\frac{1}{2}(2 p-1)^{k+1} & \frac{1}{2}+\frac{1}{2}(2 p-1)^{k+1} \end{array}\right\| \] Since the \(\mathbf{P}^{(k+1)}\) matches the desired formula, our proof is complete. Therefore, the formula holds for all integer values of n by mathematical induction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Markov Chains
Markov chains are mathematical systems that describe a sequence of possible events, where the probability of each event depends only on the state attained in the previous event. This property is known as 'memorylessness' or the Markov property, which signifies that the future state depends only on the present state, not on the sequence of events that preceded it.

A classic example of a Markov chain is a board game where each move depends only on the current position, not on how the player arrived there. Markov chains have wide applications across various fields such as economics, game theory, and genetics, providing a fundamental tool for modeling random processes.
Transition Probability Matrix
A transition probability matrix is a square matrix that describes the probabilities of moving from one state to another in a Markov chain. Each entry in the matrix, located at row i and column j, represents the probability of transitioning from state i to state j. The sum of each row in a transition probability matrix is 1, as the probabilities of moving to all possible next states must encompass all possible outcomes.

For example, in the case of a two-state Markov chain, the transition probability matrix will have two rows and two columns, reflecting the probabilities of staying in the same state or moving to the other state. This matrix is the key to understanding the behavior of the Markov chain over time and can be raised to the power of n to determine the probabilities after n transitions.
Proof Techniques
Proof techniques are the tools and methods used to establish the truth of mathematical statements. One powerful proof technique is mathematical induction, which is used to prove statements about natural numbers. It consists of two steps: the base case and the inductive step.

The base case involves verifying the statement for the initial value, often n=1. Then, the inductive step assumes the statement is true for some arbitrary natural number k and proves that if the statement holds for k, it must also hold for k+1. By demonstrating these two steps, it's concluded that the statement is true for all natural numbers following the base case. This method is pivotal when proving statements about series, sequences, and, as shown in our exercise, results of consecutive matrix multiplications.
Matrix Multiplication
Matrix multiplication is an operation that takes two matrices and produces another matrix. The entry in the resulting matrix at row i, column j, is computed by taking the dot product of the ith row of the first matrix with the jth column of the second matrix. It's important to note that matrix multiplication is not commutative, meaning that the order of multiplication matters.

In our example, matrix multiplication is used to determine \( \mathbf{P}^{(n)} \) by multiplying the transition matrix \( \mathbf{P} \) by itself n times. Each multiplication represents one more step in the Markov process, with the results yielding the probabilities of transitioning from one state to any another after n steps. A proper understanding of matrix multiplication is critical for working with Markov chains and their transition probabilities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

At all times, an urn contains \(N\) balls?-some white balls and some black balls. At each stage, a coin having probability \(p, 0

A DNA nucleotide has any of four values. A standard model for a mutational change of the nucleotide at a specific location is a Markov chain model that supposes that in going from period to period the nucleotide does not change with probability \(1-3 \alpha\), and if it does change then it is equally likely to change to any of the other three values, for some \(0<\alpha<\frac{1}{3}\). (a) Show that \(P_{1,1}^{n}=\frac{1}{4}+\frac{3}{4}(1-4 \alpha)^{n}\). (b) What is the long-run proportion of time the chain is in each state?

Suppose that coin 1 has probability \(0.7\) of coming up heads, and \(\operatorname{coin} 2\) has probability \(0.6\) of coming up heads. If the coin flipped today comes up heads, then we select coin 1 to flip tomorrow, and if it comes up tails, then we select \(\operatorname{coin} 2\) to flip tomorrow. If the coin initially flipped is equally likely to be \(\operatorname{coin} 1\) or \(\operatorname{coin} 2\), then what is the probability that the coin flipped on the third day after the initial flip is coin 1? Suppose that the coin flipped on Monday comes up heads. What is the probability that the coin flipped on Friday of the same week also comes up heads?

Consider a branching process having \(\mu<1\). Show that if \(X_{0}=1\), then the expected number of individuals that ever exist in this population is given by \(1 /(1-\mu)\). What if \(X_{0}=n ?\)

In a good weather year the number of storms is Poisson distributed with mean \(1 ;\) in a bad year it is Poisson distributed with mean 3. Suppose that any year's weather conditions depends on past years only through the previous year's condition. Suppose that a good year is equally likely to be followed by either a good or a bad year, and that a bad year is twice as likely to be followed by a bad year as by a good year. Suppose that last year-call it year 0 -was a good year. (a) Find the expected total number of storms in the next two years (that is, in years 1 and 2 ). (b) Find the probability there are no storms in year 3 . (c) Find the long-run average number of storms per year.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.