/*! 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 31 You are given a transition matri... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given a transition matrix \(P .\) Find the steady-state distribution vector: $$ P=\left[\begin{array}{lll} .5 & 0 & .5 \\ 1 & 0 & 0 \\ 0 & .5 & .5 \end{array}\right] $$

Short Answer

Expert verified
The steady-state distribution vector for the given transition matrix P is \(x = \left[\frac{1}{3},\frac{1}{3},\frac{1}{3}\right]\).

Step by step solution

01

Write down the equation

For the given transition matrix P: $$ P=\left[\begin{array}{ccc} .5 & 0 & .5 \\ 1 & 0 & 0 \\ 0 & .5 & .5 \end{array}\right] $$ We want to find the row vector \(x=\left[\begin{array}{ccc} x_{1} & x_{2} & x_{3} \end{array}\right]\) such that \(xP = x\).
02

Set up the system of linear equations

To find the steady-state distribution vector \(x\), we need to set up a system of linear equations using the equation \(xP=x\) and taking into account that the sum of the components in \(x\) should equal 1: $$ \begin{cases} 0.5x_1 + x_2 = x_1 \\ 0.5x_3 = x_2 \\ 0.5x_1 + 0.5x_3 = x_3 \\ x_1 + x_2 + x_3 = 1 \end{cases} $$
03

Solve the system of linear equations

Now we will solve the system of linear equations to find \(x_1\), \(x_2\), and \(x_3\). From the first equation, we have: $$ x_2 = 0.5x_1\enspace(1) $$ From the second equation, we have: $$ x_2 = 0.5x_3\enspace(2) $$ From the third equation, we have: $$ x_1 = x_3\enspace(3) $$ Now we can substitute the third equation into the first equation, and we have: $$ x_2 = 0.5x_3 $$ Comparing with equation (2), we have: $$ 0.5x_1 = 0.5x_3 $$ From equation (3), we can substitute \(x_1\) with \(x_3\), so we have: $$ x_3 = \frac{1}{3}\enspace(4) $$ Now, we can substitute the value of \(x_3\) from equation (4) into equation (3) and (2), and we have: $$ x_1 = \frac{1}{3} \\ x_2 = \frac{1}{3} $$ So, our solution for the steady-state distribution vector is: $$ x = \left[\frac{1}{3},\frac{1}{3},\frac{1}{3}\right] $$
04

Verify the solution

Finally, we will verify our solution by checking if \(xP = x\): $$ \left[\frac{1}{3},\frac{1}{3},\frac{1}{3}\right]\left[\begin{array}{ccc} .5 & 0 & .5 \\ 1 & 0 & 0 \\ 0 & .5 & .5 \end{array}\right] \ = \ \left[\frac{1}{3},\frac{1}{3},\frac{1}{3}\right] $$ Our solution is indeed the steady-state distribution vector for the given transition matrix P.

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Key Concepts

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

Transition Matrix
A transition matrix is a mathematical tool used primarily in probability and statistics, especially in the study of Markov chains. It represents the probabilities of moving from one state to another in a dynamic system. In our case, each element of the matrix specifies the probability of transitioning between states in a single step. The transition matrix is usually denoted by \( P \) and organized such that the sum of probabilities in each row equals one.

This property arises because each row signifies the probability distribution of transitioning from a particular state to all possible future states. For example, in our matrix,
  • Row 1 represents transitions from state 1 to other states.
  • Row 2 represents transitions from state 2.
  • Row 3 is for transitions from state 3.
Each row's elements indicate the probability of moving to the corresponding column's state. Therefore, understanding the transition matrix allows us to predict how the system evolves over time.
System of Linear Equations
Finding the steady-state distribution involves setting up a system of linear equations. This arises because we want to find a distribution that remains unchanged over time when multiplied by the transition matrix.

The steady-state vector \( x \) satisfies \( xP = x \). This leads us to derive a system of equations based on the matrix multiplication of the vector and the matrix. For instance, from the given transition matrix, we create:
  • \( 0.5x_1 + x_2 = x_1 \)
  • \( 0.5x_3 = x_2 \)
  • \( 0.5x_1 + 0.5x_3 = x_3 \)
  • Adding the condition \( x_1 + x_2 + x_3 = 1 \) ensures the vector is a valid probability distribution.
This system must be solved simultaneously to find the values of \( x_1, x_2, \) and \( x_3 \). The insight here is understanding how to reformulate the problem into solvable mathematical equations that can be manipulated to find a steady state.
Stochastic Matrices
Stochastic matrices are used to model stochastic processes, which involve randomness or uncertainty. Specifically, a stochastic matrix, like our transition matrix, is a square matrix used to describe the transitions of a Markov chain. Its key property is that it describes the change of a state by making sure each row sums to one. This ensures that despite the randomness, probabilities are well-defined.

Such matrices help in determining steady-states, which are conditions where the probabilities become stable over time. This occurs when the distribution of probabilities across states does not change as time progresses, leading us to the steady-state distribution. Understanding stochastic matrices aids in revealing the long-term behavior of the modeled system, such as which states will eventually dominate or coexist equitably, as in our resulting steady-state vector \( \left[\frac{1}{3},\frac{1}{3},\frac{1}{3}\right] \). This equilibrium position is especially crucial in fields like economics, biology, and any domain involving random processes.

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