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Chapter 6: Problem 8

Let \(X\) and \(Y\) be Markov chains on the set \(Z\) of integers. Is the sequence \(Z_{n}=X_{n}+Y_{n}\) necessarily a Markov chain?

Short Answer

Expert verified
No, \(Z_n\) is not necessarily a Markov chain.

Step by step solution

01

Understanding the Exercise

We need to determine if the sequence \(Z_n = X_n + Y_n\), where \(X_n\) and \(Y_n\) are two separate Markov chains on the set of integers, is also a Markov chain. A Markov chain has the property that the future is independent of the past given the present.
02

Define the Markov Property

A sequence \(X_n\) is a Markov chain if \(P(X_{n+1} = x_{n+1} \mid X_0 = x_0, X_1 = x_1, \ldots, X_n = x_n) = P(X_{n+1} = x_{n+1} \mid X_n = x_n)\) for any states \(x_0, x_1, \ldots, x_{n+1}\). We need to see if this property holds for \(Z_n\).
03

Analyze Combined Sequence

For \(Z_n = X_n + Y_n\) to be a Markov chain, the conditional probability \(P(Z_{n+1} = z_{n+1} \mid Z_0 = z_0, Z_1 = z_1, \ldots, Z_n = z_n)\) must depend only on \(Z_n\). Since \(Z_n = X_n + Y_n\), dependencies on past values of \(X\) and \(Y\) beyond \(n\) could influence \(Z_{n+1}\), violating the Markov property.
04

Counterexample Construction

Consider simple independent Markov chains \(X_n\) and \(Y_n\) each having possibly different transition rules or even states interacting in \(Z_n\). Changes in \(X_n\) and \(Y_n\) might result in \(Z_{n+1}\) depending on more than \(Z_n\), illustrating the potential for losing the Markov property.
05

Determination of Sequence Type

Conclude that \(Z_n = X_n + Y_n\) is not necessarily a Markov chain because the transitions of \(Z\) could depend on the states contributing \(X_{n+m}\) and \(Y_{n+m}\) where \(m e 0\), violating the necessary Markov memorylessness.

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Key Concepts

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

Conditional Probability in Markov Chains
The concept of conditional probability is central to understanding Markov chains. A Markov chain is characterized by the fact that the probability of moving to the next state depends only on the current state, not on the sequence of events that preceded it. This property is best expressed using conditional probability.
  • In a Markov chain, the conditional probability that describes the process is: \(P(X_{n+1}=x_{n+1} \mid X_0=x_0, \ldots, X_n=x_n) = P(X_{n+1}=x_{n+1} \mid X_n=x_n)\).
  • This equation means that the next state \(X_{n+1}\) only depends on the present state \(X_n\), and this simplifies the complexity involved in predicting the system's evolution over time.
  • The original exercise explores if this conditional dependency still holds when two Markov chains \(X_n\) and \(Y_n\) are combined into a new sequence \(Z_n = X_n + Y_n\).
Transition Rules of Markov Chains
Transition rules define how a state in a Markov chain moves to another state. These rules are depicted by a *transition matrix* that outlines the probabilities of moving from one state to another.
  • In a single Markov chain, the transition from one state to another can be entirely defined by these transition probabilities.
  • For example, if you know the transition matrix of \(X\), you have an understanding of where \(X\) could potentially move next given its current state.
  • In the context of the exercise, combining Markov chains introduces complexity. If \(Z_n = X_n + Y_n \), then how \(X_n\) and \(Y_n\) transition could be independently determined, potentially bringing additional dependencies into \(Z_n\) beyond its current state \(Z_n\).
  • This interaction might cause \(Z_n\) to include dependencies that are not contained in standard Markov transition rules, complicating whether \(Z_n\) can retain the Markov property.
Understanding Memorylessness
Memorylessness is a defining feature of Markov chains; it ensures that only the present state is needed to predict the next state of the chain. This property implies that the process does not "remember" any past states outside of the immediate present.
  • For a sequence to be a Markov chain, it must exhibit memorylessness. This means anything you predict about future steps can be done using only the current step, without needing the full past history.
  • The exercise challenges this property by questioning whether the new sequence \(Z_n=X_n+Y_n\) maintains memorylessness, given that it combines two separate sequences.
  • When combining \(X_n\) and \(Y_n\), any inherent dependency or variation in their states could potentially inform \(Z_{n+1}\) differently from \(Z_n\), thus violating the memorylessness characteristic.
  • This breaking of memorylessness signifies that \(Z_n\) may not behave like a proper Markov chain if it retains information from past state interactions. Therefore, while \(X_n\) and \(Y_n\) are individually memoryless, their sum \(Z_n\) could fail to be so.

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Most popular questions from this chapter

Consider the symmetric random walk in three dimensions on the set of points \((x, y, z): x, y, z=\) \(0, \pm 1, \pm 2, \ldots, 1 ;\) this process is a sequence \(\left[\mathbf{X}_{n}: n \geq 0\right]\) of points such that \(\mathbb{P}\left(\mathbf{X}_{n+1}=\mathbf{X}_{n}+\epsilon\right)=\frac{1}{6}\) for \(\epsilon=(\pm 1,0,0),(0, \pm 1,0),(0,0, \pm 1)\). Suppose that \(\mathbf{X}_{0}=(0,0,0)\). Show that $$ \mathrm{P}\left(\mathbf{X}_{2 n}=(0,0,0)\right)=\left(\frac{1}{6}\right)^{2 n} \sum_{i+j+k=n} \frac{(2 n) !}{(i) j ! k !)^{2}}=\left(\frac{1}{2}\right)^{2 n}\left(\begin{array}{c} 2 n \\ n \end{array}\right) \sum_{i+j+h=n}\left(\frac{n !}{3^{n} i 1 j ! k !}\right)^{2} $$ and deduce by Stirling's formula that the origin is a transient state.

Let \(\left(X_{n}: n \geq 0\right)\) be a persistent irreducible discrete-time Markov chain on the state space \(S\) with transition matrix \(\mathbf{P}\), and let \(x\) be a positive solution of the equation \(x=\mathbf{x P}\) (a) Show that $$ q_{i j}(n)=\frac{x_{j}}{x_{i}} p_{j i}(n), \quad i, j \in S, n \geq 1 $$ defines the \(n\)-step transition probabilities of a persistent irreducible Markov chain on \(S\) whose first-passage probabilities are given by $$ g_{i j}(n)=\frac{x_{j}}{x_{i}} l_{f i}(n), \quad i \neq j, n \geq 1 $$ where \(l_{j i}(n)=\mathrm{P}\left(X_{n}=i, T>n \mid X_{0}=j\right)\) and \(T=\min \left(m>0: X_{m}=j\right\\}\) (b) Show that \(\mathrm{x}\) is unique up to a multiplicative constant. (c) Let \(T_{j}=\min \left(n \geq 1: X_{n}=j\right)\) and define \(h_{i j}=\mathrm{P}\left(T_{j} \leq T_{i} \mid X_{0}=i\right) .\) Show that \(x_{i} h_{i j}=x_{j} h_{j i}\) for all \(i_{+} j \in S\).

Let \(A\) and \(\mathbf{B}=\left(B_{0}, B_{1}, \ldots, B_{n}\right)\) be a discrete random variable and vector, respectively. The conditional entropy of \(A\) with respect to \(\mathrm{B}\) is defined as \(H(A \mid \mathbf{B})=\) \(\mathbb{E}(\mathbb{E}|-\log f(A \mid \mathbf{B})| \mathbf{B} \mid)\) where \(f(a \mid \mathbf{b})=\mathrm{P}(A=a \mid \mathbf{B}=\mathbf{b})\). Let \(X\) be an aperiodic Markov chain on a finite state space. Show that $$ H\left(X_{n+1} \mid X_{0}, X_{1} \ldots, X_{n}\right)=H\left(X_{n+1} \mid X_{n}\right) $$ and that $$ H\left(X_{n+1} \mid X_{n}\right) \rightarrow-\sum_{i} \pi_{i} \sum_{j} p_{i j} \log p_{i j} \quad \text { as } n \rightarrow \infty $$ if \(X\) is aperiodic with a unique stationary distribution \(\pi\).

Pasta property. Let \(X=\mid X(t): t \geq 0\\}\) be a Markov chain having stationary distribution \(\pi\). We may sample \(X\) at the times of a Poisson process: let \(N\) be a Poisson process with intensity \(\lambda\), independent of \(X\), and define \(Y_{n}=X\left(T_{n}+\right)\), the value taken by \(X\) immediately after the epoch \(T_{n}\) of the \(n\)th arrival of \(N\). Show that \(Y=\left\\{Y_{n}: n \geq 0\right\\}\) is a discrete-time Markov chain with the same stationary distribution as \(X\). (This exemplifies the 'Pasta' property: Poisson arrivals see time averages.) (The full assumption of the independence of \(N\) and \(X\) is not necessary for the conclusion. It suffices that \(\mid N(s): s \geq t)\) be independent of \(\\{X(s): s \leq t]\), a property known as 'lack of anticipation'. It is not even necessary that \(X\) be Markov; the Pasta property holds for many suitable ergodic processes.]

Successive offers for my house are independent identically distributed random variables \(X_{1}\). \(X_{2} \ldots\), having density function \(f\) and distribution function \(F\). Let \(Y_{1}=X_{1}\), let \(Y_{2}\) be the first offer exceeding \(Y_{1}\), and generally let \(Y_{n+1}\) be the first offer exceeding \(Y_{a}\) - Show that \(Y_{1}, Y_{2} \ldots .\) are the times of arrivals in a non-homogeneous Poisson process with intensity function \(\lambda(t)=f(t) /(1-F(t))\). The \(Y_{i}\) are called 'record values". Now let \(Z_{1}\) be the first offer received which is the second largest to date, and let \(Z_{2}\) be the second such offer, and so on. Show that the \(Z_{l}\) are the arrival times of a non-homogencous Poisson process with intensity function \(\lambda\).
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