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Chapter 6: Problem 6

Let \(i\) and \(j\) be two states of a discrete-time Markov chain. Show that if \(i\) communicates with \(j_{3}\) then there is positive probability of reaching \(j\) from \(i\) without revisiting \(i\) in the meantime. Deduce, that, if the chain is irreducible and persistent, then the probability \(f_{i j}\) of ever reaching \(j\) from \(i\) equals 1 for all \(i\) and \(j\).

Short Answer

Expert verified
If the chain is irreducible and persistent, then \(f_{ij} = 1\) for all states \(i\) and \(j\).

Step by step solution

01

Understanding Communication in Markov Chains

In a Markov chain, a state \(i\) 'communicates' with a state \(j\) if there is a non-zero probability of reaching \(j\) from \(i\) and vice versa. Since \(i\) communicates with \(j_3\), there exists a path that allows the transition from \(i\) to \(j_3\). This implies that there are multiple paths that could potentially reach state \(j\) if \(j_3\) is, or can eventually lead to, state \(j\).
02

Constructing Path from i to j

Since we know \(i\) communicates with \(j_3\), and \(j\) can be reached from \(j_3\), there exists a sequence of transitions connecting \(i\) to \(j\): \(i \rightarrow j_1 \rightarrow j_2 \rightarrow \ldots \rightarrow j_3 \rightarrow j\). As \(j_3\) is part of this sequence, we can ascertain a direct path from \(i\) to \(j\) across these transitions, ensuring no immediate return to \(i\) through the definition of communication.
03

Considering Positive Probability of Non-Revisiting Path

Given the above transition from \(i\) to \(j\) via intermediate states, the probability of reaching \(j\) from \(i\) without revisiting \(i\) depends on the product of the non-zero transition probabilities along this path. Therefore, there is a positive probability for this non-revisiting path if each transition in the sequence has a positive probability.
04

Chain Properties: Irreducibility and Persistence

An irreducible Markov chain implies that every state is accessible from every other state, i.e., all states communicate. Persistence implies that recurrent visits to a state happen almost surely. When the chain is irreducible and persistent, \(i\) communicating with \(j\) means that any state \(j\) can eventually be reached from any starting state \(i\).
05

Concluding Probability of Ever Reaching j

Since \(i\) communicates with \(j\) and the Markov chain is both irreducible and persistent, the probability of eventually reaching state \(j\) from state \(i\) is 1. This implies all states are recurrent in the long run, and \(f_{ij} = 1\) for all pairs of states \(i, j\).

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Key Concepts

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

Irreducibility in Markov Chains
In the context of Markov chains, irreducibility is a fundamental property where every state can be reached from any other state. This means there is no subset of states that the chain gets trapped in, unable to reach states outside this subset. Irreducibility ensures that there is communication across all states, promoting the idea of a connected network.
This property is essential because it guarantees that, given enough time, the process can go from any state to any other state eventually. Irreducibility is a hallmark of fully connected systems in which every piece is accessible from every other piece, like interconnected cities on a railway network where you can travel from any city to any other city without eventually getting stuck permanently in one region.
State Communication in Markov Chains
State communication between two states, say state \(i\) and state \(j\), means there is a positive probability of moving from \(i\) to \(j\), and vice versa, over potentially several steps. This bilateral accessibility implies that these states are effectively within the same 'territory' of the chain.
The concept of communication forms a backbone in understanding the flow and transition between states in a Markov chain. It ensures that states can transition between each other, given enough steps, allowing for a dynamic movement within the chain. This dynamic is captured in sequences of states where each state could potentially 'talk' to another, denoting possible routes for progress or connections like going from city to city with viable road pathways.
Transition Probability in Markov Chains
Transition probability is crucial in Markov chains as it quantifies the likelihood of moving from one state to another in a single time step. Transition probabilities are typically represented in a matrix where each entry \(P_{ij}\) specifies the probability of transitioning from state \(i\) to state \(j\).
These probabilities must sum up to 1 for any given state \(i\), ensuring that the entire probability space is covered. High transition probabilities between states suggest a frequent shift, while lower probabilities indicate rarer transitions. Understanding these probabilities allows one to predict the movement within the chain over time, akin to weather forecasting based on current conditions and probable changes.
Persistence or Recurrence in Markov Chains
Persistence, also known as recurrence, describes a state that is revisited infinitely often with probability 1. In a persistent state, if you leave it, you'll almost surely return at some point. This persistence ensures stability within a Markov chain.
When all states in a chain are persistent, it reinforces the idea of a balanced system where states aren't just temporary stops but destinations that will be revisited. It's akin to repeatedly returning to city hubs on a road trip, ensuring that once you've been there, you will circle back eventually. Persistent states are key in ensuring long-term stability and recurrent behavior in Markov chains, serving as assurance that the chain doesn't lose connectivity over time.

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Most popular questions from this chapter

Let \(X\) be a birth-death process with \(\lambda_{n}=n \lambda\) and \(\mu_{n}=n \mu\), and suppose \(X(0)=1\). Show that the time \(T\) at which \(X(t)\) first takes the value 0 satisfies $$ \mathrm{E}(T \mid T<\infty)= \begin{cases}\frac{1}{\lambda} \log \left(\frac{\mu}{\mu-\lambda}\right) & \text { if } \lambda<\mu \\\ \frac{1}{\mu} \log \left(\frac{\lambda}{\lambda-\mu}\right) & \text { if } \lambda>\mu\end{cases} $$ What happens when \(\lambda=\mu ?\)

Let \(\mathrm{P}\) be a stochastic matrix on the finite set \(\Theta\) with stationary distribution \(\pi\). Define the inner product \((\mathbf{x}, \mathbf{y})=\sum_{k \in \Theta} x_{k} y_{k} \pi_{k}\), and let \(I^{2}(\pi)=\left\\{\mathbf{x} \in \mathrm{R}^{\Theta}:(\mathbf{x}, \mathbf{x})<\infty\right\\} .\) Show, in the obvious notation, that \(\mathbf{P}\) is reversible with respect to \(\pi\) if and only if \((x, P y)=\langle\mathbf{P} \mathbf{x}, \mathbf{y}\rangle\) for all \(\mathbf{x}, \mathbf{y} \in l^{2}(\pi)\),

Insects land in the soup in the manner of a Poisson process with intensity \(\lambda\), and each such insect is green with probability \(p\), independently of the colours of all other insects. Show that the arrivals of green insects form a Poisson process with intensity \(\lambda p\).

Customers entering a shop are served in the order of their arrival by the single server. They arrive in the manner of a Poisson process with intensity \(\lambda\), and their service times are independent exponentially distributed random variables with parameter \(\mu\). By considering the jump chain, show that the expected duration of a busy period \(B\) of the server is \((\mu-\lambda)^{-1}\) when \(\lambda<\mu\). (The busy period nuns from the moment a customer arrives to find the server free until the earliest subsequent time when the server is again free.)

A transition matrix is called doubly stochastic if all its column sums equal 1, that is, if \(\sum_{i} p_{i j}=1\) for all \(j \in S\) (a) Show that if a finite chain has a doubly stochastic transition matrix, then all its states are non-null persistent, and that if it is, in addition, irreducible and aperiodic then \(P_{i j}(n) \rightarrow N^{-1}\) as \(n \rightarrow \infty\), where \(N\) is the number of states. (b) Show that, if an infinite irreducible chain has a doubly stochastic transition matrix, then its states are cither all null persistent or all transicnt.
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