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[M] Examine powers of a regular stochastic matrix. a. Compute \(P^{k}\) for \(k=2,3,4,5,\) when $$ P=\left[\begin{array}{cccc}{.3355} & {.3682} & {.3067} & {.0389} \\ {.2663} & {.2723} & {.3277} & {.5451} \\ {.1935} & {.1502} & {.1589} & {.2395} \\\ {.2047} & {.2093} & {.2067} & {.1765}\end{array}\right] $$ Display calculations to four decimal places. What happens to the columns of \(P^{k}\) as \(k\) increases? Compute the steady-state vector for \(P\) . b. Compute \(Q^{k}\) for \(k=10,20, \ldots, 80,\) when $$ Q=\left[\begin{array}{ccc}{.97} & {.05} & {.10} \\ {0} & {.90} & {.05} \\\ {.03} & {.05} & {.85}\end{array}\right] $$ (Stability for \(Q^{k}\) to four decimal places may require \(k=116\) or more.) Compute the steady-state vector for \(Q .\) Conjecture what might be true for any regular stochastic matrix. c. Use Theorem 18 to explain what you found in parts (a) and \((b) .\)

Short Answer

Expert verified
Powers of a regular stochastic matrix approach a matrix with identical steady-state columns. The steady-state vector is a fixed vector obtained from the matrix eigenstructure.

Step by step solution

01

Calculate Power Matrices of P

Calculate \( P^2, P^3, P^4, \) and \( P^5 \) by multiplying the matrix \( P \) by itself the necessary number of times. Matrix multiplication involves multiplying rows by columns. Ensure calculations are precise to four decimal places.
02

Analyze Columns of P^k

As \( k \) increases, observe the values in the columns of \( P^k \). Regular stochastic matrices have the property that as the power \( k \) of a regular stochastic matrix increases, the columns of \( P^k \) tend to approach a fixed vector, known as the steady-state vector.
03

Calculate Steady-State Vector for P

Find the steady-state vector \( \mathbf{v} \) for the matrix \( P \) by solving the equation \( P \mathbf{v} = \mathbf{v} \) under the constraint that the sum of the elements of \( \mathbf{v} \) is 1. This generally involves solving a system of linear equations derived from \( (P - I) \mathbf{v} = 0 \).
04

Calculate Power Matrices of Q for Multiple Powers

Compute \( Q^{10}, Q^{20}, ..., Q^{80} \) similarly as in Step 1. Observe patterns in the power matrices as \( k \) increases. For precision, consider going further to \( Q^{116} \) or more if necessary, as stability may require higher \( k \).
05

Calculate Steady-State Vector for Q

Solve \( Q \mathbf{w} = \mathbf{w} \) to find the steady-state vector \( \mathbf{w} \), again ensuring that the vector components sum to 1. This involves the same method as in Step 3.
06

Conjecture for Regular Stochastic Matrices

Based on observations from Steps 2 and 4, postulate that for any regular stochastic matrix, its powers approach a matrix whose columns are comprised of its steady-state vector as \( k \) approaches infinity.
07

Explanation Using Theorem 18

Theorem 18 states that for a regular stochastic matrix \( A \), \( A^k \) will approach a matrix with equal columns all equal to the steady-state vector. This explains the findings in both parts (a) and (b), as the powers \( P^k \) and \( Q^k \) stabilize towards their respective steady-state vectors.

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Key Concepts

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

steady-state vector
The concept of a steady-state vector is crucial when examining stochastic matrices. Imagine a situation where you want to see how a system behaves in the long run. A steady-state vector provides a glimpse into that future. It is a vector that doesn't change when multiplied by the matrix. This means, in mathematical terms, for a matrix \( P \), the steady-state vector \( \mathbf{v} \) satisfies the equation \( P \mathbf{v} = \mathbf{v} \).

This vector represents the equilibrium point of the system, where probabilities of various states remain constant over time. To find it, one typically solves the linear system derived from \( (P - I) \mathbf{v} = 0 \) with the constraint that the sum of the vector's components is 1. This constraint ensures the vector is a true probability distribution, as all probabilities sum to unity.
matrix multiplication
Matrix multiplication is a fundamental operation in linear algebra, especially when working with stochastic matrices. Here, matrices are multiplied by performing the dot product of rows from the first matrix with columns of the second matrix. For instance, given two matrices \( A \) and \( B \), the element in the resulting matrix at position \( (i, j) \) is computed as the sum of the products of corresponding elements of row \( i \) of \( A \) and column \( j \) of \( B \).

The process can be simplified as:
  • Align the rows of the first matrix with the columns of the second.
  • Multiply each element of the row by the corresponding element of the column.
  • Add the results to get the final value for the element \( (i, j) \).
For stochastic matrices like in this exercise, multiplying the matrix by itself \( k \) times (i.e., raising it to the power of \( k \)) is crucial to discovering the behavior of its steady-state vector.
linear equations
Solving systems of linear equations is often necessary to find the steady-state vector of a stochastic matrix. These equations represent the constraints under which the steady-state exists. Given a stochastic matrix \( P \), we solve \( (P - I) \mathbf{v} = 0 \) to identify its steady-state vector \( \mathbf{v} \).

Key steps to solve these equations include:
  • Subtracting the identity matrix \( I \) from \( P \).
  • Setting up the resulting homogeneous system of equations.
  • Applying methods like Gaussian elimination or matrix inversion to solve for \( \mathbf{v} \).
Ensuring the sum of the vector's elements equals 1 is crucial to maintain the probabilistic nature of the solution. This step may involve balancing the system with an additional equation \( \sum v_i = 1 \).
Theorem 18
Theorem 18 plays a pivotal role in understanding the behavior of powers of a regular stochastic matrix. This theorem states that for a regular stochastic matrix \( A \), as the power \( k \) increases, the matrix \( A^k \) converges to a matrix where every column is the steady-state vector.

This convergence is significant because it implies stability in stochastic systems. Regardless of the initial state, as you apply the matrix successively (i.e., raise it to a power), the system's state distribution tends to stabilize. This stabilizing behavior is why stochastic matrices are often used in modeling long-term processes, such as Markov chains.

In simpler terms, no matter your start point, repeatedly applying the matrix leads all paths to the steady-state vector, predicting the behavior of the system in the long term.

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Most popular questions from this chapter

Show that every \(2 \times 2\) stochastic matrix has at least one steady-state vector. Any such matrix can be written in the form \(P=\left[\begin{array}{cc}{1-\alpha} & {\beta} \\ {\alpha} & {1-\beta}\end{array}\right],\) where \(\alpha\) and \(\beta\) are constants between 0 and \(1 .\) (There are two linearly independent steady-state vectors if \(\alpha=\beta=0 .\) Otherwise, there is only one.)

Exercises 31 and 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let \(V\) and \(W\) be vector spaces, let \(T : V \rightarrow W\) be a linear transformation, and let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) be a subset of \(V .\) Suppose that \(T\) is a one-to-one transformation, so that an equation \(T(\mathbf{u})=T(\mathbf{v})\) always implies \(\mathbf{u}=\mathbf{v} .\) Show that if the set of images \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{p}\right)\right\\}\) is linearly dependent, then \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set (because in this case the set of images cannot be linearly dependent).

Let \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\\}\) and \(\mathcal{C}=\left\\{\mathbf{c}_{1}, \mathbf{c}_{2}\right\\}\) be bases for \(\mathbb{R}^{2} .\) In each exercise, find the change-of-coordinates matrix from \(\mathcal{B}\) to \(\mathcal{C}\) and the change-of-coordinates matrix from \(\mathcal{C}\) to \(\mathcal{B} .\) \(\mathbf{b}_{1}=\left[\begin{array}{l}{-6} \\ {-1}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{l}{2} \\ {0}\end{array}\right], \mathbf{c}_{1}=\left[\begin{array}{r}{2} \\ {-1}\end{array}\right], \mathbf{c}_{2}=\left[\begin{array}{r}{6} \\ {-2}\end{array}\right]\)

At time \(k=0,\) an initial investment of \(\$ 1000\) is made into a savings account that pays 6\(\%\) interest per year compounded monthly. (The interest rate per month is \(005 . )\) Each month after the initial investment, an additional \(\$ 200\) is added to the account. For \(k=0,1,2, \ldots,\) let \(y_{k}\) be the amount in the account at time \(k,\) just after a deposit has been made. a. Write a difference equation satisfied by \(\left\\{y_{k}\right\\}\) b. \([\mathbf{M}]\) Create a table showing \(k\) and the total amount in the savings account at month \(k,\) for \(k=0\) through \(60 .\) List your program or the keystrokes you used to create the table. c. \([\mathbf{M}]\) How much will be in the account after two years (that is, 24 months), four years, and five years? How much of the five-year total is interest?

Let \(M_{2 \times 2}\) be the vector space of all \(2 \times 2\) matrices, and define \(T : M_{2 \times 2} \rightarrow M_{2 \times 2}\) by \(T(A)=A+A^{T},\) where \(A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]\) a. Show that \(T\) is a linear transformation. b. Let \(B\) be any element of \(M_{2 \times 2}\) such that \(B^{T}=B .\) Find an \(A\) in \(M_{2 \times 2}\) such that \(T(A)=B\) . c. Show that the range of \(T\) is the set of \(B\) in \(M_{2 \times 2}\) with the property that \(B^{T}=B\) . d. Describe the kernel of \(T\)

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