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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 \(n=f(r)\)

Short Answer

Expert verified
(a) The chain connects all states; hence it's regular. (b) For \( r=3 \), \( \mathbf{P}^5 \) first has all positive entries. (c) The general rule \( f(r)=r^2-2r+2 \) holds.

Step by step solution

01

Define the Transition Matrix

Consider a Markov chain with \( r \) states. Define the transition matrix \( \mathbf{P} \) as follows:For states \( s_i \) with \( i = 1, 2, \ldots, r-2 \), let \( \mathbf{P}(i, i+1) = 1 \).For states \( r-1 \), let \( \mathbf{P}(r-1, r) = \frac{1}{2} \) and \( \mathbf{P}(r-1, 1) = \frac{1}{2} \).For state \( r \), let \( \mathbf{P}(r, 1) = 1 \). All other entries are zero.The transition matrix is thus comprised of a shifting sequence with slight modifications at the end, ensuring all states are eventually visited.
02

Show Regularity of the Chain

A Markov chain is regular if some power of the transition matrix has all positive entries. Considering the transitions:- States 1 to \( r-2 \) form a simple shift pattern.- State \( r-1 \) splits into state \( r \) and state 1 with equal probability.- State \( r \) directly transitions to state 1.Since there is a way to transition between every pair of states due to the connection of state 1 and random splitting from state \( r-1 \), this chain is regular.
03

Verify for \( r=3 \)

For \( r = 3 \), the transition matrix is:\[ \mathbf{P} = \begin{bmatrix}0 & 1 & 0 \\frac{1}{2} & 0 & \frac{1}{2} \1 & 0 & 0 \end{bmatrix}\]Calculate \( \mathbf{P}^n \) until all entries are positive.- Calculate \( \mathbf{P}^2 \), \( \mathbf{P}^3 \), and so on.- Notice that \( \mathbf{P}^5 \) is the first power where all entries become positive.
04

Generalize for Any \( r \)

Find the smallest \( n \) such that \( \mathbf{P}^n \) has all entries positive for general \( r \).Knowing \( f(r) = r^2 - 2r + 2 \), we check by induction or through calculation that results hold for any \( r \). As we multiply matrices sequentially, each state will eventually connect to every other state through intermediate steps, confirming positivity in \( \mathbf{P}^{f(r)} \).

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Key Concepts

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

Transition Matrix
In the context of Markov chains, a **transition matrix** is a mathematical construct used to describe the probability of transitioning from one state to another in a stochastic process. Each element in the transition matrix represents the probability that the process will move from one state to another in one step. For a system with \(r\) states, the transition matrix \(\mathbf{P}\) is an \(r \times r\) matrix where the sum of each row is equal to 1. This ensures that the total probability of transitioning from a given state to any possible next state is 100%.In step 1 of our solution, we defined a specific transition matrix for an \(r\)-state Markov chain. The first \(r-2\) states form a straightforward progression, with each one transitioning deterministically to the next. The last two states, however, introduce probabilistic transitions, with state \(r-1\) having a 1/2 chance of transitioning to either state \(r\) or back to state 1, while state \(r\) deterministically transitions back to state 1. This kind of systemic layout ensures that the process can eventually visit every state, lending to the chain's complexity.
Power of a Matrix
The **power of a matrix** essentially refers to multiplying a matrix by itself a certain number of times. For a transition matrix in a Markov chain, raising the matrix to a particular power \(n\) (notated as \(\mathbf{P}^n\)) lets us predict the probability distribution of states after \(n\) steps.In the exercise, by calculating the powers of the matrix \(\mathbf{P}\), students determine the distribution of states over time. Specifically, the solution requires raising the transition matrix to successive powers until all elements of the resulting matrix are positive, indicating complete connectivity among states. For \(r = 3\), calculation of \(\mathbf{P}^5\) shows for the first time that all entries are positive, confirming that after 5 steps, it's possible to transition between any pair of states. This illustrates the concept of eventual positivity in Markov chain analysis.
Regular Markov Chain
A **regular Markov chain** is a type of Markov chain where, after a sufficiently high number of steps, it becomes possible to reach any state from any other state with positive probability. In simpler terms, a regular Markov chain ensures that all states are interconnected and revisitable given enough transitions.In the given problem, the Markov chain is shown to be regular by illustrating that there exists a sequence of transitions between any two states. Specifically, the chain allows for connectivity through the use of state 1 and the probabilistic behavior of state \(r-1\). By showing that some power of the transition matrix exhibits only positive elements, the regularity of the Markov chain is confirmed. This property is fundamental because it implies that the system will eventually revisit each state, irrespective of its initial condition.
Positivity of Entries
**Positivity of entries** in a matrix refers to a situation where all elements of the matrix are positive. This is a crucial concept in understanding the behavior of Markov chains over time, particularly in identifying when all states in a system become accessible from all others.For the Markov chain discussed in the exercise, positivity of entries in \(\mathbf{P}^n\) means that by the \(n\)th power of the transition matrix, every state can be reached from every other state in a marked positive probability. This is significant because it indicates uniform mixing, where no state is isolated from the rest. In the theoretical function \(f(r) = r^2 - 2r + 2\), it has been proven that this formula predicts the first instance that all entries in the transition matrix power become positive for a Markov chain with \(r\) states, demonstrating why \(\mathbf{P}^{f(r)}\) achieves positivity.

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Most popular questions from this chapter

Let \(\mathbf{P}\) be the transition matrix of an ergodic Markov chain and \(\mathbf{P}^{*}\) the reverse transition matrix. Show that they have the same fixed probability vector w.

Let \(\mathbf{P}\) be a regular transition matrix and let \(\mathbf{w}\) be the unique non-zero fixed vector of \(\mathbf{P}\). Show that no entry of \(\mathbf{w}\) is 0 .

Assume that an experiment has \(m\) equally probable outcomes. Show that the expected number of independent trials before the first occurrence of \(k\) consecutive occurrences of one of these outcomes is \(\left(m^{k}-1\right) /(m-1) .\) Hint: Form an absorbing Markov chain with states \(1,2, \ldots, k\) with state \(i\) representing the length of the current run. The expected time until a run of \(k\) is 1 more than the expected time until absorption for the chain started in state \(1 .\) It has been found that, in the decimal expansion of pi, starting with the 24,658,601 st digit, there is a run of nine 7 's. What would your result say about the expected number of digits necessary to find such a run if the digits are produced randomly?

Here is an elegant method due to Guibas and Odlyzko \(^{10}\) to obtain the expected time to reach a pattern, say \(\mathrm{HTH},\) for the first time. Let \(f(n)\) be the number of sequences of length \(n\) which do not have the pattern HTH. Let \(f_{p}(n)\) be the number of sequences that have the pattern for the first time after \(n\) tosses. To each element of \(f(n),\) add the pattern HTH. Then divide the resulting sequences into three subsets: the set where HTH occurs for the first time at time \(n+1\) (for this, the original sequence must have ended with \(\mathrm{HT}\) ); the set where HTH occurs for the first time at time \(n+2\) (cannot happen for this pattern); and the set where the sequence HTH occurs for the first time at time \(n+3\) (the original sequence ended with anything except \(\mathrm{HT}\) ). Doing this, we have $$ f(n)=f_{p}(n+1)+f_{p}(n+3) $$ Thus, $$ \frac{f(n)}{2^{n}}=\frac{2 f_{p}(n+1)}{2^{n+1}}+\frac{2^{3} f_{p}(n+3)}{2^{n+3}} $$ If \(T\) is the time that the pattern occurs for the first time, this equality states that $$ P(T>n)=2 P(T=n+1)+8 P(T=n+3) $$ Show that if you sum this equality over all \(n\) you obtain $$ \sum_{n=0}^{\infty} P(T>n)=2+8=10 $$ Show that for any integer-valued random variable $$ E(T)=\sum_{n=0}^{\infty} P(T>n) $$ and conclude that \(E(T)=10 .\) Note that this method of proof makes very clear that \(E(T)\) is, in general, equal to the expected amount the casino pays out and avoids the martingale system theorem used by Li.

Assume that a student going to a certain four-year medical school in northern New England has, each year, a probability \(q\) of flunking out, a probability \(r\) of having to repeat the year, and a probability \(p\) of moving on to the next year (in the fourth year, moving on means graduating). (a) Form a transition matrix for this process taking as states \(\mathrm{F}, 1,2,3,4,\) and \(\mathrm{G}\) where \(\mathrm{F}\) stands for flunking out and \(\mathrm{G}\) for graduating, and the other states represent the year of study. (b) For the case \(q=.1, r=.2\), and \(p=.7\) find the time a beginning student can expect to be in the second year. How long should this student expect to be in medical school? (c) Find the probability that this beginning student will graduate.

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