/*! 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 4 Let \(P=\left[\begin{array}{cc}0... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 4

Let \(P=\left[\begin{array}{cc}0.5 & 0.3 \\ 0.5 & 0.7\end{array}\right]\) be the transition matrix for a Markov chain with two states. Let \(\mathbf{x}_{0}=\left[\begin{array}{l}0.5 \\ 0.5\end{array}\right] b e\) the initial state vector for the population. Find the steady state vector.

Short Answer

Expert verified
The steady state vector is \( \begin{bmatrix} \frac{3}{8} \\ \frac{5}{8} \end{bmatrix} \)."

Step by step solution

01

Understand the Steady State Condition

A steady state vector \( \mathbf{x} \) for a Markov chain's transition matrix \( P \) satisfies the equation \( P \mathbf{x} = \mathbf{x} \). This means that once the system reaches the steady state, it does not change in subsequent transitions.
02

Set Up the Steady State Equation

For the given transition matrix \( P = \begin{bmatrix} 0.5 & 0.3 \ 0.5 & 0.7 \end{bmatrix} \) and steady state vector \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \), the equation becomes: \[ \begin{bmatrix} 0.5 & 0.3 \ 0.5 & 0.7 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}. \] This results in a system of linear equations.
03

Write the System of Equations

Multiply the matrix \( P \) by \( \mathbf{x} \) to get the system:\[ 0.5x_1 + 0.3x_2 = x_1 \]\[ 0.5x_1 + 0.7x_2 = x_2 \].
04

Simplify and Solve the System

Rearrange the equations from Step 3:1. \( 0.5x_1 + 0.3x_2 = x_1 \) simplifies to \( 0.5x_1 = 0.3x_2 \) or \( x_1 = 0.6x_2 \).2. \( 0.5x_1 + 0.7x_2 = x_2 \) simplifies to \( 0.5x_1 = 0.3x_2 \), which is consistent as it is same.Introduce the condition \( x_1 + x_2 = 1 \) for the steady state vector and solve: \( x_1 = 0.6x_2 \) gives \( 0.6x_2 + x_2 = 1 \) resulting in \( 1.6x_2 = 1 \), therefore \( x_2 = \frac{5}{8} \) and \( x_1 = \frac{3}{8} \).
05

Verify the Solution

Ensure that your solution satisfies the requirement of all probabilities summing to 1:\( x_1 + x_2 = \frac{3}{8} + \frac{5}{8} = 1 \). Thus, the steady state vector is \( \begin{bmatrix} \frac{3}{8} \ \frac{5}{8} \end{bmatrix} \).

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.

Transition Matrix
A transition matrix is a square matrix used in Markov chains, where each entry represents the probability of transitioning from one state to another. For the given example:
  • The matrix is given by \[ P = \begin{bmatrix} 0.5 & 0.3 \ 0.5 & 0.7 \end{bmatrix} \].
  • Each column must sum to 1, ensuring that all possible outcomes are accounted for in each state transition.
  • Here, the system consists of two states, and the numbers within the matrix specify the probability of transitioning from one state to another.
Understanding how a transition matrix represents these changes over time is crucial for analyzing Markov processes. Once you comprehend how to interpret the matrix, you can predict future behavior of the Markov chain.
Steady State
A steady state is a condition where the probabilities remain constant over time despite further transitions. In other words:
  • The system no longer changes as transitions occur, implying a sort of "equilibrium" in the chain.
  • To find this steady state, you solve the equation \[ P \mathbf{x} = \mathbf{x} \].
  • In our example, after solving the steady-state equation, we find the vector \[ \mathbf{x} = \begin{bmatrix} \frac{3}{8} \ \frac{5}{8} \end{bmatrix} \].
This vector indicates the long-term stable distribution of states. Understanding the steady state can help in predicting how systems behave in continuous time after many transitions.
Linear Equations
Linear equations form the backbone of solving for the steady state in a Markov chain. The goal is to set up and solve these equations based on the transition matrix. Here's how it works:
  • Consider the matrix multiplication, \[ \begin{bmatrix} 0.5 & 0.3 \ 0.5 & 0.7 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \end{bmatrix} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \], which yields a system of linear equations.
  • These equations are:\[ 0.5x_1 + 0.3x_2 = x_1 \] \[ 0.5x_1 + 0.7x_2 = x_2 \].
  • Simplifying these equations allows us to express one variable in terms of the other, leading to easy substitution and solving.
By manipulating these linear equations, we can derive the steady state vector as a function of the initial probabilities.
Probability Vector
A probability vector is a vector whose components represent probabilities, which sum to 1. This concept is essential when dealing with Markov chains.
  • The vector \( \mathbf{x}_0 \) represents the initial state probabilities as \[ \begin{bmatrix} 0.5 \ 0.5 \end{bmatrix} \].
  • In context, each entry reflects the likelihood of the system being in a particular state.
  • As the Markov chain evolves, this vector changes in accordance with the transition matrix until it converges to a steady state vector.
  • For the steady state, we ensure our derived vector, \[ \begin{bmatrix} \frac{3}{8} \ \frac{5}{8} \end{bmatrix} \], meets the requirement that all elements of a probability vector must add up to 1.
Grasping the concept of a probability vector is simple yet powerful, allowing us to determine the long-term behavior of the system.

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

Determine whether the given consumption matrix is productive. $$\left[\begin{array}{llll} 0.2 & 0.4 & 0.1 & 0.4 \\ 0.3 & 0.2 & 0.2 & 0.1 \\ 0 & 0.4 & 0.5 & 0.3 \\ 0.5 & 0 & 0.2 & 0.2 \end{array}\right]$$

Let \(A\) be an \(n \times n\) matrix, \(A \geq 0 .\) Suppose that \(A \mathbf{x} < \mathbf{x}\) for some \(\mathbf{x}\) in \(\mathbb{R}^{n}, \mathbf{x} \geq 0 .\) Prove that \(\mathbf{x} > \mathbf{0}\).

Find the standard matrix of the composite transformation from \(\mathbb{R}^{2}\) to \(\mathbb{R}^{2}\). Counterclockwise rotation through \(60^{\circ}\), followed by reflection in the line \(y=x\)

Use the Gauss-Jordan method to find the inverse of the given matrix (if it exists). $$\left[\begin{array}{rrrr}0 & -1 & 1 & 0 \\\2 & 1 & 0 & 2 \\\1 & -1 & 3 & 0 \\\0 & 1 & 1 & -1\end{array}\right]$$

If \(A\) is \(m \times n\), prove that every vector in null \((A)\) is orthogonal to every vector in row \((A)\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

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.