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You are given a transition matrix \(P .\) Find the steady-state distribution vector: $$ P=\left[\begin{array}{ll} .2 & .8 \\ .4 & .6 \end{array}\right] $$

Short Answer

Expert verified
The steady-state distribution vector for the given transition matrix \(P = \begin{bmatrix} 0.2 & 0.8 \\ 0.4 & 0.6 \end{bmatrix}\) is \(\pi = \begin{bmatrix} \dfrac{1}{3} \\ \dfrac{2}{3}\end{bmatrix}\).

Step by step solution

01

Define the Problem

Given the transition matrix P, \[P = \begin{bmatrix} 0.2 & 0.8 \\ 0.4 & 0.6 \end{bmatrix}\] We need to find the steady-state distribution vector that satisfies \[\pi P = \pi\] The vector \(\pi\) must satisfy the following two conditions: 1. \(\pi P = \pi\) 2. The components of \(\pi\) must add up to 1, which means \(\pi_1 + \pi_2 = 1\)
02

Set Up the Equations

To solve for the steady-state distribution vector \(\pi\), we need to set up the equations created from the multiplication of the transition matrix P by the vector \(\pi\), and the additional condition for the components of the vector \(\pi\) to add up to 1. Let \(\pi = \begin{bmatrix} \pi_1 \\ \pi_2\end{bmatrix}\), then the two conditions can be written as: 1. \(\pi P = \pi\) => \(\pi_1(0.2) + \pi_2(0.4) = \pi_1\) and \(\pi_1(0.8) + \pi_2(0.6) = \pi_2\) 2. \(\pi_1 + \pi_2 = 1\)
03

Solve the System of Equations

Now, we'll solve the system of equations: 1. \(0.8\pi_1 = 0.4\pi_2\) 2. \(\pi_1 + \pi_2 = 1\) From equation 1, we can see that: \(\pi_1 = 0.5\pi_2\) Now substitute this expression for \(\pi_1\) in the second equation, \(0.5\pi_2 + \pi_2 = 1\) Solve for \(\pi_2\), \(\pi_2 = \dfrac{1}{1.5} \Rightarrow \pi_2 = \dfrac{2}{3}\) Now, substitute the value of \(\pi_2\) back into the expression for \(\pi_1\), \(\pi_1 = 0.5\pi_2 = 0.5\left(\dfrac{2}{3}\right) \Rightarrow \pi_1 = \dfrac{1}{3}\)
04

Determine the Steady-State Distribution Vector

Now that we have solved for \(\pi_1\) and \(\pi_2\), we can construct the steady-state distribution vector: \[\pi = \begin{bmatrix} \dfrac{1}{3} \\ \dfrac{2}{3}\end{bmatrix}\] So, the steady-state distribution vector for the given transition matrix is: \[\pi = \begin{bmatrix} \dfrac{1}{3} \\ \dfrac{2}{3}\end{bmatrix}\]

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Key Concepts

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

Transition Matrix
In the world of mathematics and particularly in probability theory, a transition matrix represents the probability of moving from one state to another in a stochastic process. This concept is widely used in Markov chains, where the future state depends only on the current state and not on the sequence of events that preceded it.

A transition matrix is composed of rows and columns that denote transition probabilities. Each cell in this matrix indicates the probability of transitioning from the state represented by its row to the state represented by its column. The sum of the probabilities in any given row must equal 1 because the probabilities represent all possible outcomes for the next state.

When it comes to finding the steady-state distribution, you utilize the transition matrix to determine a vector that does not change when multiplied by the transition matrix. This vector is known as the steady-state distribution vector, which is fundamental in understanding long-term behaviors in chain processes.
System of Equations
Solving a system of equations is a common task in algebra that involves finding the set of values that satisfy all equations simultaneously. When we use the transition matrix to find the steady-state distribution vector, we are essentially setting up a system of linear equations where the unknowns are the elements of the distribution vector.

Understanding Linear Combinations

In the context of the steady-state distribution, these equations become a reflection of linear combinations, where each element is represented as a weighted sum of other elements. The condition that the elements of the vector must add up to one adds another equation to the system, which represents the total probability condition. Solving this system efficiently and accurately ensures that we can understand and predict the behavior of dynamic systems over time.
Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra that allows us to compute the product of two matrices. The process involves taking the dot product of the rows of the first matrix with the columns of the second matrix.

When it comes to Markov chains and transition matrices, matrix multiplication helps calculate successive states. For a steady-state distribution vector, the operation is crucial in satisfying the equation \(\pi P = \pi\). This balanced scenario depicts that the system reached equilibrium, where applying the transition matrix does not change the distribution any further.

It's important to note that matrix multiplication is not commutative; the order in which you multiply matrices matters. The rules governing this process must be strictly followed to ensure the correct calculation of the steady-state distribution vector in Markov processes.

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Most popular questions from this chapter

M a r k e t ~ S h a r e : ~ L i g h t ~ V e h i c l e s ~ I n ~ \(2003,25 \%\) of all light vehi- cles sold (SUVs, pickups, passenger cars, and minivans) in the United States were SUVs and \(15 \%\) were pickups. Moreover, a randomly chosen vehicle sold that year was five times as likely to be a passenger car as a minivan. \(^{29}\) Find the associated probability distribution.

Suppose two dice (one red, one green) are rolled. Consider the following events. A: the red die shows \(1 ; B:\) the numbers add to \(4 ; C:\) at least one of the numbers is \(1 ;\) and \(D:\) the numbers do not add to 11. Express the given event in symbols and say how many elements it contains. HINT [See Example 5.] The red die shows 1 and the numbers add to \(4 .\)

Use counting arguments from the preceding chapter. Horse Races The seven contenders in the fifth horse race at Aqueduct on February 18,2002, were: Pipe Bomb, Expect a Ship, All That Magic, Electoral College, Celera, Cliff Glider, and Inca Halo. \({ }^{5}\) You are interested in the first three places (winner, second place, and third place) for the race. a. Find the cardinality \(n(S)\) of the sample space \(S\) of all possible finishes of the race. (A finish for the race consists of a first, second, and third place winner.) b. Let \(E\) be the event that Electoral College is in second or third place, and let \(F\) be the event that Celera is the winner. Express the event \(E \cap F\) in words, and find its cardinality.

Determine whether the information shown is consistent with a probability distribution. If not, say why. \(P(A)=.2 ; P(B)=.4 ; P(A \cap B)=.3\)

(Refer to the preceding exercise.) How can the modeled probability of winning the lotto be nonzero if you have never won it, despite playing 600 times?

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