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You are given a transition matrix \(P .\) Find the steady-state distribution vector: $$ P=\left[\begin{array}{ll} .1 & .9 \\ .6 & .4 \end{array}\right] $$

Short Answer

Expert verified
The steady-state distribution vector \(v\) for the given transition matrix \(P\) is: \[v = \begin{bmatrix} 0.4 \\ 0.6 \end{bmatrix}\]

Step by step solution

01

Define the steady-state distribution vector

Let's call the steady-state distribution vector \(v\), which has components \(v_1\) and \(v_2\). Our task is to find the values of \(v_1\) and \(v_2\) that satisfy the equation \(vP = v\), where P is the given transition matrix.
02

Write out the equations

Translate the equation \(vP = v\) into the system of linear equations using the given matrix and the components of the steady-state vector: \[ \begin{cases} 0.1v_1 + 0.6v_2 = v_1 \\ 0.9v_1 + 0.4v_2 = v_2 \end{cases} \]
03

Rearrange the equations

Rearrange the system of linear equations to express one variable in terms of the other: \[ \begin{cases} 0.9v_1 = 0.6v_2 \\ 0.9v_1 = 0.6v_2 \end{cases} \] Notice that both equations are now identical, which means we have a degenerate system. Since we are looking for a probability vector, we can use the fact that \(v_1 + v_2 = 1\) to express one variable as a function of the other. For example, express \(v_2\) in terms of \(v_1\): \[v_2 = 1 - v_1\]
04

Substitute and solve for one variable

Substitute the expression for \(v_2\) found in the previous step into the first equation and solve for \(v_1\): \[0.9v_1 = 0.6(1 - v_1)\] \[0.9v_1 + 0.6v_1 = 0.6\] \[1.5v_1 = 0.6\] \[v_1 = 0.4\]
05

Solve for the second variable

Now that we have found the value of \(v_1\), we can plug it into the expression for \(v_2\) and find the value of \(v_2\): \[v_2 = 1 - v_1 = 1 - 0.4 = 0.6\]
06

Present the steady-state distribution vector

With the values of \(v_1\) and \(v_2\) found, the steady-state distribution vector \(v\) can be written as: \[v = \begin{bmatrix} 0.4 \\ 0.6 \end{bmatrix}\] This is 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
In the world of mathematics, a transition matrix is a square matrix used to describe the transitions of a Markov process. This matrix has a special property: the sum of the entries in each of its rows must be equal to 1, which reflects the notion that the total probabilities of transitioning from any state to all possible states must be 100%.

In our exercise, the transition matrix is written as:
\[ P=\left[\begin{array}{ll} .1 & .9 \ .6 & .4 \end{array}\right]\]
Here, the matrix sizes up the probabilities of moving from one state to another. For example, the probability of transitioning from the first state back to itself is 0.1, and from the first state to the second state is 0.9. Understanding the transition matrix allows one to predict the long-term behavior of the system under study.
Probability Vector
A probability vector, often utilized in stochastic processes and Markov chains, is an array of probabilities that describe the state of a system at a given time. Each component of the vector represents the probability of the system being in a particular state. The sum of all probabilities in a probability vector must equal 1, ensuring that the probabilities are normalized.

In our transition matrix problem, the aim is to find the steady-state distribution vector, which is essentially a probability vector that does not change after applying the transition matrix. This vector represents the long-term state distribution of the Markov process.
Linear Algebra
Linear algebra is a fundamental branch of mathematics focused on vector spaces and the linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces. Linear algebra is vital in many fields of science and engineering and provides the language in which the laws of physics and the rules of statistics and probability are expressed.

In our context, linear algebra concepts are applied to solve the system of linear equations derived from the steady-state conditions of the Markov process. Understanding and manipulating these equations are crucial steps towards finding the steady-state distribution vector.
System of Linear Equations
A system of linear equations is a collection of one or more linear equations involving the same set of variables. The objective is to find the values of these variables that satisfy all of the equations simultaneously. When we're dealing with Markov processes, we create such a system to find the steady-state distribution vector that remains unchanged by the application of the transition matrix.

In the exercise, once we translate the equation \(vP = v\) into the system of linear equations, we uncover a relationship between the values of \(v_1\) and \(v_2\). It is crucial to identify when a system is degenerate, which implies a dependency among the equations that can either simplify the problem, like in this case, or indicate that there's no unique solution.

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Most popular questions from this chapter

You are given a transition matrix \(P\) and initial distribution vector \(v\). Find (a) the two-step transition matrix and (b) the distribution vectors after one, two, and three steps. $$ P=\left[\begin{array}{ccc} .1 & .1 & .8 \\ .5 & 0 & .5 \\ .5 & 0 & .5 \end{array}\right], v=\left[\begin{array}{lll} 0 & 1 & 0 \end{array}\right] $$

Public Health A study shows that \(75 \%\) of the population has been vaccinated against the Martian ague, but \(4 \%\) of this group gets this disease anyway. If \(10 \%\) of the total population gets this disease, what is the probability that a randomly selected person has been neither vaccinated nor has contracted Martian ague?

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