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Chapter 11: Problem 23

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 $$ \sum_{h=0}^{n} p_{i j}^{(h)} $$

Short Answer

Expert verified
The expected frequency of state visits converges to its fixed probability \(w_j\) as \(n\) becomes large.

Step by step solution

01

Understanding the Task

We need to establish that the expected value of the number of times a Markov chain is in a specific state, divided by the number of steps, converges to the fixed probability for that state. This needs to be proven for large n, regardless of the starting state of the chain.
02

Expected Value of Visits

Given that if the chain starts in state \(s_i\), the expected value of \(S_{j}^{(n)}\) is \(\sum_{h=0}^{n} p_{ij}^{(h)}\), where \(p_{ij}^{(h)}\) is the probability of transitioning from state \(s_i\) to state \(s_j\) in \(h\) steps.
03

Normalizing the Sum

We seek \(\frac{1}{n}E[S_{j}^{(n)}] = \frac{1}{n} \sum_{h=0}^{n} p_{ij}^{(h)}\). As \(n\) approaches infinity, the effect of the initial state diminishes.
04

Applying Ergodic Theorems

For an ergodic Markov chain (irreducible and aperiodic), the limit \(\lim_{n \to \infty} p_{ij}^{(h)} = w_j\) exists and is equal to the fixed probability \(w_j\). This implies that each state is visited in proportion to its fixed probability.
05

Taking the Limit

As \(n\) approaches infinity, we use the ergodic theorem to show \(\lim_{n \to \infty} \frac{1}{n} \sum_{h=0}^{n} p_{ij}^{(h)} = w_j\). Thus, the normalized expected number of visits approaches \(w_j\).
06

Conclusion

Hence, for large \(n\), \(\frac{1}{n}E[S_{j}^{(n)}]\) tends to \(w_j\), proving that the expected frequency of visiting state \(s_j\) converges to its fixed probability \(w_j\).

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Key Concepts

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

Ergodic Theorems
Ergodic theorems are crucial in the study of Markov chains, particularly for understanding long-term behavior. When we say a Markov chain is ergodic, it means this chain is both irreducible and aperiodic. What does this mean? Let's break it down:
  • **Irreducible:** Every state can be reached from every other state, no state is isolated.
  • **Aperiodic:** There are no cyclic patterns in the state's visits over time.
These properties ensure that the chain doesn't get stuck in a loop or in a particular section of states, leading to a diverse exploration of all states.
What makes ergodic theorems so powerful? They tell us that, as time goes on (or as the number of steps in the chain, \( n \), goes to infinity), each state gets visited proportionally to a fixed probability. This fixed probability ties into the concept of a fixed or steady-state vector that we will discuss further. In essence, ergodic theorems guarantee that the long-term average time spent in any given state aligns with certain predictable probabilities.
State Transition Probabilities
State transition probabilities are the backbone of Markov chains. They describe the likelihood of moving from one state to another in a given number of steps. In mathematical terms, if we represent states as \( s_i \) and \( s_j \), the probability of transitioning from \( s_i \) to \( s_j \) in \( h \) steps is denoted by \( p_{ij}^{(h)} \).
Understanding these probabilities is essential because they enable us to gauge how likely the chain is to be in a specific state over time. In our exercise, the expected value of being in state \( s_j \) over \( n \) steps is essentially the sum of all these probabilities over those steps: \( \sum_{h=0}^{n} p_{ij}^{(h)} \).
This collection of probabilities forms a matrix known as the transition matrix. It helps in calculating the state visits and predicting how the system evolves. As \( n \) becomes very large, particular patterns and stability, as informed by ergodic theorems, begin to emerge.
Fixed Probability Vector
A fixed probability vector, often also referred to as the stationary distribution, is a key concept in the world of Markov chains. This vector, denoted as \( \mathbf{w} \), gives the long-term probabilities of being in each state after many transitions, irrespective of where the chain started.
For a Markov chain to have a fixed probability vector, it needs to be ergodic. The fixed probability vector satisfies the equation \( \mathbf{w}P = \mathbf{w} \), where \( P \) is the transition matrix. This means that when we multiply the vector by the transition matrix, it doesn't change.
In practical terms, when you look at the average number of visits to each state over time in an ergodic Markov chain, it will converge to this vector. This happens because the chain's randomness 'evens out' with time, providing stable, predictable behavior.
In our given exercise, it shows that for a large number of steps \( n \), the proportion of time the chain is expected to spend in each state \( s_j \) aligns perfectly with the fixed probability \( w_j \). This phenomenon plays a large role in fields ranging from queuing theory to economics, where understanding long-term behavior using Markov chains is essential.

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Most popular questions from this chapter

Let \(\mathbf{P}\) be the transition matrix of a regular Markov chain. Assume that there are \(r\) states and let \(N(r)\) be the smallest integer \(n\) such that \(\mathbf{P}\) is regular if and only if \(\mathbf{P}^{N(r)}\) has no zero entries. Find a finite upper bound for \(N(r)\). See if you can determine \(N(3)\) exactly.

Prove that, if a 3 -by-3 transition matrix has the property that its column sums are 1 , then \((1 / 3,1 / 3,1 / 3)\) is a fixed probability vector. State a similar result for \(n\) -by- \(n\) transition matrices. Interpret these results for ergodic chains.

Consider the following process. We have two coins, one of which is fair, and the other of which has heads on both sides. We give these two coins to our friend, who chooses one of them at random (each with probability \(1 / 2\) ). During the rest of the process, she uses only the coin that she chose. She now proceeds to toss the coin many times, reporting the results. We consider this process to consist solely of what she reports to us. (a) Given that she reports a head on the \(n\) th toss, what is the probability that a head is thrown on the \((n+1)\) st toss? (b) Consider this process as having two states, heads and tails. By computing the other three transition probabilities analogous to the one in part (a), write down a "transition matrix" for this process. (c) Now assume that the process is in state "heads" on both the \((n-1)\) st and the \(n\) th toss. Find the probability that a head comes up on the \((n+1)\) st toss. (d) Is this process a Markov chain?

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?

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.
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