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In Exercises \(5-8,\) find the steady-state vector. $$ \left[\begin{array}{ccc}{.7} & {.1} & {.1} \\ {.2} & {.8} & {.2} \\ {.1} & {.1} & {.7}\end{array}\right] $$

Short Answer

Expert verified
The steady-state vector is \( \begin{bmatrix} \frac{1}{4} \\ \frac{1}{2} \\ \frac{1}{4} \end{bmatrix} \).

Step by step solution

01

Define the steady-state vector

The steady-state vector, often represented as \( extbf{v} \), is a probability vector that remains unchanged after applying the transition matrix \( A \). This means \( A \textbf{v} = \textbf{v} \).
02

Set up the equation

If \( extbf{v} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), then \( A\textbf{v} = \textbf{v} \) gives us: \( \begin{bmatrix} 0.7 & 0.1 & 0.1 \ 0.2 & 0.8 & 0.2 \ 0.1 & 0.1 & 0.7 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} x \ y \ z \end{bmatrix} \). This results in the system of equations for the steady-state vector.
03

Create and solve the system of linear equations

From \( A\textbf{v} = \textbf{v} \), rewrite this as \( (A - I)\textbf{v} = \textbf{0} \), where \( I \) is the identity matrix. The resulting system is: 1. \( 0.7x + 0.1y + 0.1z = x \) 2. \( 0.2x + 0.8y + 0.2z = y \) 3. \( 0.1x + 0.1y + 0.7z = z \). Simplify each by rearranging terms.
04

Simplify equations

Subtract each component. 1. \( 0.7x + 0.1y + 0.1z = x \) becomes \( -0.3x + 0.1y + 0.1z = 0 \). 2. \( 0.2x + 0.8y + 0.2z = y \) becomes \( 0.2x - 0.2y + 0.2z = 0 \). 3. \( 0.1x + 0.1y + 0.7z = z \) becomes \( 0.1x + 0.1y - 0.3z = 0 \).
05

Solve the system for one variable

Using the reduced equations, solve for variables. Set \( z = 1 - x - y \) to convert to two variables as these probabilities must add to 1. Solve simultaneously: From equation 2: \( 0.2x + 0.2z = 0.2y \) Substituting \( z = 1 - x - y \), and solving with equation 1 as well: Find \( x \), \( y \), \( z \) subject to \( x + y + z = 1 \).
06

Verify the solutions form a probability vector

Calculate \( x, y, z \) from solved equations, ensure they sum to 1 and each \( x, y, z \) are probabilities (i.e., between 0 and 1). The correct values signify the solution is a valid steady-state probability vector.

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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 to describe changes from one state to another in a system. It is often represented as a square matrix where each entry indicates the probability of transitioning from one state to another. Typically in our context, each column of the matrix adds up to 1, representing total probability distribution among possible transitions.
A transition matrix is essential when dealing with Markov chains, where future states depend only on the current state and not past states. Using this matrix, we can determine long-term behavior of a system by analyzing its effect on probability distributions of states, like finding the steady-state vector, which remains constant as time progresses.
For instance, consider a transition matrix:
  • The element at the first row and second column might indicate the probability of moving from state 1 to state 2.
  • Understanding the transition matrix allows us to predict how the system evolves over time, providing insights into steady-state probabilities.
Probability Vector
A probability vector is a simple yet powerful concept in the realm of probability and linear algebra. It is essentially a row or a column vector where the entries are individual probabilities, each ranging from 0 to 1, and collectively summing up to 1.
This vector represents the distribution of probabilities across different possible states at a specific time or iteration. Within the scope of transition matrices, probability vectors are used to describe the current state distribution of the system. When multiplied by the transition matrix, they provide the next state probabilities.
For example, if we have a probability vector \( \begin{bmatrix} x \ y \ z \end{bmatrix} \), it means:
  • "x" is the probability of being in state 1 at a given time.
  • "y" is the probability of being in state 2.
  • "z" is the probability of being in state 3.
The aim is often to find the steady-state probability vector, where after many transitions, the system stabilizes, and the vector remains unchanged despite further applications of the transition matrix.
System of Linear Equations
In mathematical modeling, a system of linear equations arises naturally when trying to find a steady-state in Markov chains. It allows the conversion of matrix operations into simpler algebraic terms, providing a straightforward computational approach.
When searching for a steady-state vector \( \mathbf{v} \), we solve \( A\mathbf{v} = \mathbf{v} \). This can be recast into a system of equations, \( (A - I)\mathbf{v} = \mathbf{0} \), where \( I \) is the identity matrix and \( \mathbf{0} \) is the zero vector.
Breaking it down step-by-step:
  • Replace the probability vector with its elements. Here \( \mathbf{v} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), and each element corresponds to a state probability.
  • This yields individual linear equations such as \( 0.7x + 0.1y + 0.1z = x \), which rearranges to \( -0.3x + 0.1y + 0.1z = 0 \).
  • By simplifying further, we solve for one or more of these probabilities using techniques like substitution until we meet the condition that all probabilities sum up to 1.
This calculated steady-state vector offers a prediction of long-term trends for each state, assuming the transition probabilities remain constant over time.

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Most popular questions from this chapter

Consider the polynomials \(\mathbf{p}_{1}(t)=1+t, \mathbf{p}_{2}(t)=1-t,\) and \(\mathbf{p}_{3}(t)=2\) (for all \(t ) .\) By inspection, write a linear dependence relation among \(\mathbf{p}_{1}, \mathbf{p}_{2},\) and \(\mathbf{p}_{3} .\) Then find a basis for \(\operatorname{Span}\left\\{\mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3}\right\\}\)

Let \(H\) be an \(n\) -dimensional subspace of an \(n\) -dimensional vector space \(V .\) Show that \(H=V\) .

When a signal is produced from a sequence of measurements made on a process (a chemical reaction, a flow of heat through a tube, a moving robot arm, etc. ), the signal usually contains random noise produced by measurement errors. A standard method of pre processing the data to reduce the noise is to smooth or filter the data. One simple filter is a moving average that replaces each \(y_{k}\) by its average with the two adjacent values: $$ \frac{1}{3} y_{k+1}+\frac{1}{3} y_{k}+\frac{1}{3} y_{k-1}=z_{k} \quad \text { for } k=1,2, \ldots $$ Suppose a signal \(y_{k},\) for \(k=0, \ldots, 14,\) is $$ 9,5,7,3,2,4,6,5,7,6,8,10,9,5,7 $$ Use the filter to compute \(z_{1}, \ldots, z_{13}\) . Make a broken-line graph that superimposes the original signal and the smoothed signal.

Exercises \(27-29\) concern an \(m \times n\) matrix \(A\) and what are often called the fundamental subspaces determined by \(A .\) Justify the following equalities: a. \(\operatorname{dim} \mathrm{Row} A+\operatorname{dim} \mathrm{Nul} A=n\) Number of columns of \(A\) b. \(\operatorname{dim} \operatorname{Col} A+\operatorname{dim} \mathrm{Nul} A^{T}=m\) Number of rows of \(A\)

In Exercises 13 and \(14,\) assume that \(A\) is row equivalent to \(B .\) Find bases for Nul \(A\) and \(\operatorname{Col} A .\) $$ A=\left[\begin{array}{rrrrr}{1} & {2} & {-5} & {11} & {-3} \\ {2} & {4} & {-5} & {15} & {2} \\ {1} & {2} & {0} & {4} & {5} \\ {3} & {6} & {-5} & {19} & {-2}\end{array}\right] $$ $$ B=\left[\begin{array}{rrrrr}{1} & {2} & {0} & {4} & {5} \\ {0} & {0} & {5} & {-7} & {8} \\ {0} & {0} & {0} & {0} & {-9} \\ {0} & {0} & {0} & {0} & {0}\end{array}\right] $$

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