Chapter 7: Problem 31
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,
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.
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:
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.
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.
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.