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Chapter 11: Problem 7

Find the matrices \(\mathbf{P}^{2}, \mathbf{P}^{3}, \mathbf{P}^{4},\) and \(\mathbf{P}^{n}\) for the Markov chain determined by the transition matrix \(\mathbf{P}=\left(\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right) .\) Do the same for the transition matrix \(\mathbf{P}=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right) .\) Interpret what happens in each of these processes.

Short Answer

Expert verified
The identity matrix remains the same, while the swap matrix alternates states with each odd power.

Step by step solution

01

Understanding Transition Matrices

A transition matrix describes the probabilities of moving from one state to another in a Markov chain. In this exercise, we have two transition matrices: First matrix: \( \mathbf{P} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \) is the identity matrix, meaning it keeps the system in the same state with probability 1.Second matrix: \( \mathbf{P} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \) swaps the states with probability 1.
02

Calculating Powers of the Identity Matrix

Let's compute \( \mathbf{P}^n \) where \( \mathbf{P} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \) for any integer \( n \geq 1 \). This is the identity matrix, so each power of \( \mathbf{P} \) remains unchanged:- \( \mathbf{P}^2 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \)- \( \mathbf{P}^3 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \)- \( \mathbf{P}^4 = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \)Thus, \( \mathbf{P}^n = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \) for any \( n \geq 1 \). This maintains the original state persistence over time.
03

Calculating Powers of the Swap Matrix

Now, consider \( \mathbf{P} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \). Let's compute successive powers:- \( \mathbf{P}^2 = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \times \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \), which is the identity matrix.- \( \mathbf{P}^3 = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \times \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \)- \( \mathbf{P}^4 = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \times \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \)Thus, for even \( n \), \( \mathbf{P}^n = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \). For odd \( n \), \( \mathbf{P}^n = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \). This means states swap with each odd power.
04

Interpretation of Processes

For the first matrix, the identity matrix \( \mathbf{P} \), each state is self-absorbing, and no transitions occur over time.For the second matrix \( \mathbf{P} \, = \, \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix} \), each application of the matrix represents a state swap. Thus, with every odd power, states are swapped, while every even power returns them to their original configuration.

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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 central element in a Markov chain, capturing the probabilities of moving from one state to another. Each entry of the matrix represents the probability of moving from a given initial state to a subsequent state.
In this exercise, two distinct transition matrices were considered:
  • The identity matrix: \[\mathbf{P} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\]In this matrix, every state transitions to itself with probability 1, which means no change occurs over time.
  • The swap matrix:\[\mathbf{P} = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}\]In this case, each state shifts to the other with probability 1, representing a swap between states.
Understanding these matrices is crucial for analyzing how systems behave and evolve over time under defined probabilistic rules.
Identity Matrix
The identity matrix is a special type of transition matrix where no state transitions occur. Each entry along the diagonal is 1, while all other entries are 0. Mathematically, the identity matrix for a Markov chain with two states is written as:\[\mathbf{I} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\]Computing its powers results in the matrix remaining unchanged. More formally, for any integer \( n \), \[\mathbf{I}^n = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix}\]This means that all states remain persistent in their original configuration with no transition from one to another over time. The identity matrix thus can be seen as a matrix that upholds state persistence indefinitely.
State Persistence
Consider a Markov chain with an identity matrix as its transition matrix. The concept of state persistence arises naturally. Here, persistence refers to the fact that as you take successive powers of the identity matrix, the system remains unchanged.
Analyzing our example:
  • The identity matrix indicates that each state persists since no probabilities allow for state changes.
  • This stabilization can be enormously useful for scenarios where state constancy is desired, such as simulations or models requiring fixed conditions.
Examining the role of state persistence in Markov processes provides insight into scenarios where dynamic change is not a factor, allowing for simplified computations and analyses.
Powers of Matrices
Understanding the powers of matrices is essential for evaluating long-term behavior in Markov chains. When you raise a matrix to a power, you are essentially applying the transition rules repeatedly.
Starting with the identity matrix:
  • The powers of the identity matrix do not alter the matrix鈥攊t stays the same no matter how many times it is multiplied by itself.
With the swap matrix:
  • The pattern changes as powers of the swap matrix alternate between the swap matrix and the identity matrix.
  • For odd powers, the swap matrix remains and for even powers, the result is the identity matrix.
Utilizing this knowledge aids in determining the stability or periodic nature of states within systems affected by Markov chains. Exploring and understanding these patterns allows for better predictions and interpretations regarding the systematic behavior of the Markov processes.

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Most popular questions from this chapter

In the Leontief economic model, \(^{8}\) there are \(n\) industries \(1,2, \ldots, n\). The ith industry requires an amount \(0 \leq q_{i j} \leq 1\) of goods (in dollar value) from company \(j\) to produce 1 dollar's worth of goods. The outside demand on the industries, in dollar value, is given by the vector \(\mathbf{d}=\left(d_{1}, d_{2}, \ldots, d_{n}\right) .\) Let \(\mathbf{Q}\) be the matrix with entries \(q_{i j}\). (a) Show that if the industries produce total amounts given by the vector \(\mathbf{x}=\left(x_{1}, x_{2}, \ldots, x_{n}\right)\) then the amounts of goods of each type that the industries will need just to meet their internal demands is given by the vector \(\mathbf{x} \mathbf{Q}\) (b) Show that in order to meet the outside demand \(\mathbf{d}\) and the internal demands the industries must produce total amounts given by a vector \(\mathbf{x}=\left(x_{1}, x_{2}, \ldots, x_{n}\right)\) which satisfies the equation \(\mathbf{x}=\mathbf{x} \mathbf{Q}+\mathbf{d}\) (c) Show that if \(\mathbf{Q}\) is the \(\mathbf{Q}\) -matrix for an absorbing Markov chain, then it is possible to meet any outside demand \(\mathbf{d}\). (d) Assume that the row sums of \(\mathbf{Q}\) are less than or equal to 1 . Give an economic interpretation of this condition. Form a Markov chain by taking the states to be the industries and the transition probabilites to be the \(q_{i j}\). Add one absorbing state 0 . Define $$ q_{i 0}=1-\sum_{j} q_{i j} $$ Show that this chain will be absorbing if every company is either making a profit or ultimately depends upon a profit-making company. (e) Define \(\mathbf{x} \mathbf{c}\) to be the gross national product. Find an expression for the gross national product in terms of the demand vector \(\mathbf{d}\) and the vector t giving the expected time to absorption.

A certain calculating machine uses only the digits 0 and \(1 .\) It is supposed to transmit one of these digits through several stages. However, at every stage, there is a probability \(p\) that the digit that enters this stage will be changed when it leaves and a probability \(q=1-p\) that it won't. Form a Markov chain to represent the process of transmission by taking as states the digits 0 and 1 . What is the matrix of transition probabilities?

Assume that an ergodic Markov chain has states \(s_{1}, s_{2}, \ldots, s_{k} .\) Let \(S_{j}^{(n)}\) denote the number of times that the chain is in state \(s_{j}\) in the first \(n\) steps. Let \(\mathbf{w}\) denote the fixed probability row vector for this chain. Show that, regardless of the starting state, the expected value of \(S_{j}^{(n)}\), divided by \(n\), tends to \(w_{j}\) as \(n \rightarrow \infty .\) Hint : If the chain starts in state \(s_{i},\) then the expected value of \(S_{j}^{(n)}\) is given by the expression $$ \sum_{h=0}^{n} p_{i j}^{(h)} $$

Consider the Markov chain with transition matrix $$ \mathbf{P}=\left(\begin{array}{ccc} 1 / 2 & 1 / 3 & 1 / 6 \\ 3 / 4 & 0 & 1 / 4 \\ 0 & 1 & 0 \end{array}\right) $$ (a) Show that this is a regular Markov chain. (b) The process is started in state \(1 ;\) find the probability that it is in state 3 after two steps. (c) Find the limiting probability vector w.

Define \(f(r)\) to be the smallest integer \(n\) such that for all regular Markov chains with \(r\) states, the \(n\) th power of the transition matrix has all entries positive. It has been shown, \(^{14}\) that \(f(r)=r^{2}-2 r+2\). (a) Define the transition matrix of an \(r\) -state Markov chain as follows: For states \(s_{i},\) with \(i=1,2, \ldots, r-2, \mathbf{P}(i, i+1)=1, \mathbf{P}(r-1, r)=\mathbf{P}(r-1,1)=\) \(1 / 2,\) and \(\mathbf{P}(r, 1)=1\). Show that this is a regular Markov chain. (b) For \(r=3\), verify that the fifth power is the first power that has no zeros. (c) Show that, for general \(r,\) the smallest \(n\) such that \(\mathbf{P}^{n}\) has all entries positive is \(n=f(r)\)
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