Chapter 1: Problem 20
Find the steady-state distribution vector for the given transition matrix of a Markov chain. $$\left[\begin{array}{ll}\frac{1}{4} & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{2}\end{array}\right]$$
Short Answer
Expert verified
The steady-state distribution vector is \( \begin{bmatrix} \frac{1}{4} \\ \frac{3}{4} \end{bmatrix} \).
Step by step solution
01
Understand the Transition Matrix
The given transition matrix is \[\begin{bmatrix}\frac{1}{4} & \frac{1}{2} \\frac{1}{4} & \frac{1}{2}\end{bmatrix}\]. Each column sums to 1, indicating a proper stochastic matrix.
02
Define the Steady-State Distribution
The steady-state distribution vector \( \mathbf{v} = \begin{bmatrix} v_1 \ v_2 \end{bmatrix} \) satisfies the equation \( T \mathbf{v} = \mathbf{v} \). This means the result of multiplying the transition matrix by \( \mathbf{v} \) equals \( \mathbf{v} \) itself.
03
Set Up the Equation
To find \( \mathbf{v} \), we solve the equation:\[\begin{bmatrix}\frac{1}{4} & \frac{1}{2} \\frac{1}{4} & \frac{1}{2}\end{bmatrix}\begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} v_1 \ v_2 \end{bmatrix}\].This results in two equations: 1. \( \frac{1}{4}v_1 + \frac{1}{4}v_2 = v_1 \)2. \( \frac{1}{2}v_1 + \frac{1}{2}v_2 = v_2 \).
04
Simplify the Equations
From the first equation, \( \frac{1}{4}v_1 + \frac{1}{4}v_2 = v_1 \), simplify:\( \frac{1}{4}v_1 + \frac{1}{4}v_2 = v_1 \)\( v_1 - \frac{1}{4}v_1 = \frac{1}{4}v_2 \)\( \frac{3}{4}v_1 = \frac{1}{4}v_2 \)\( 3v_1 = v_2 \).
05
Solve the Second Equation
From the second equation, \( \frac{1}{2}v_1 + \frac{1}{2}v_2 = v_2 \), simplify:\( \frac{1}{2}v_1 = \frac{1}{2}v_2 \)\( v_1 = v_2 \).
06
Combine and Solve the Equations
From Steps 4 and 5, we have two simplified equations \( 3v_1 = v_2 \) and \( v_1 = v_2 \). Substitute \( v_2 = v_1 \) into \( 3v_1 = v_2 \) to get:\( 3v_1 = v_1 \).This is only possible if \( v_1 = 0 \), but that wouldn't be a valid probability distribution. Revisiting, ensure \( v_1 + v_2 = 1 \) given probabilities.Assume \( v_1 = x \), then \( v_2 = 1-x \) and replace in equation solutions \( 3x = 1-x \)\( 4x = 1 \)\( x = \frac{1}{4} \), thus \( y = \frac{3}{4} \).
07
Verify the Solution
Verify by inserting the values back:\( v_1 = \frac{1}{4} \), and \( v_2 = \frac{3}{4} \). Check the equation \(\frac{1}{4}(\frac{1}{4}) + \frac{1}{4}(\frac{3}{4}) = \frac{1}{4} \) holds.\( \frac{1}{2}(\frac{1}{4}) + \frac{1}{2}(\frac{3}{4}) = \frac{3}{4} \) holds. Thus,\( \mathbf{v} = \begin{bmatrix} \frac{1}{4} \ \frac{3}{4} \end{bmatrix} \) is correct.
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Steady-State Distribution
In a Markov chain, the steady-state distribution is a probability distribution that remains constant over time, despite the Markov process being applied. Essentially, if the current state has a certain probability distribution, applying the transition matrix repeatedly will eventually bring the process to the steady-state distribution. This is the point where probabilities no longer change as more transitions are applied.
For a Markov chain with a transition matrix represented as \( T \) and a steady-state distribution vector \( \mathbf{v} \), the steady state is achieved when:
For a Markov chain with a transition matrix represented as \( T \) and a steady-state distribution vector \( \mathbf{v} \), the steady state is achieved when:
- \( T \mathbf{v} = \mathbf{v} \)
- The sum of elements in \( \mathbf{v} \) equals 1, maintaining a valid probability distribution
Transition Matrix
The transition matrix is a central component in understanding Markov chains. This matrix, labeled as \( T \) in the context of the problem, defines the probabilities of transitioning from one state to another in a stochastic process. Specifically, in a transition matrix, each element \( T_{ij} \) represents the probability of moving from state \( i \) to state \( j \) in one time step.
Some key characteristics of a transition matrix include:
Some key characteristics of a transition matrix include:
- All elements are non-negative, representing valid probabilities.
- Each column sums to 1, ensuring the sum of probabilities for transitioning to all possible states from a current state equals 100%.
Stochastic Matrix
A transition matrix that adheres to certain probability rules is often referred to as a "stochastic matrix." In a stochastic matrix like the one in our exercise, each column (or sometimes each row, depending on the convention used) represents a complete probability distribution.
Key features of a stochastic matrix include:
Key features of a stochastic matrix include:
- Every column sums to one, making each column itself a probability distribution over the states.
- The entire matrix describes how probabilities are distributed across various states.
Probability Distribution
Probability distribution is a fundamental concept that tells us how probabilities are assigned to different outcomes or states in a process. In the context of Markov chains, each state's probability at any given time is expressed within a probability vector.
Characteristics of a probability distribution include:
Characteristics of a probability distribution include:
- The sum of probabilities is always equal to 1, ensuring all possible outcomes are accounted for.
- Each probability value is between 0 and 1.
- It offers insight into the expected frequency or likelihood of different states being observed.