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Which of the following (when stationary) are reversible Markov chains? (a) The chain \(X=\left\\{X_{n} \mid\right.\) having transition matrix \(\mathbf{P}=\left(\begin{array}{cc}1-\alpha & \alpha \\ \beta & 1-\beta\end{array}\right)\) where \(\alpha+\beta>0\). (b) The chain \(Y=\left(Y_{n}\right)\) having transition matrix \(\mathbf{P}=\left(\begin{array}{ccc}0 & p & 1-p \\ 1-p & 0 & p \\ p & 1-p & 0\end{array}\right)\) where \(0

Short Answer

Expert verified
Chain (a) is reversible; chains (b) and (c) are not.

Step by step solution

01

Understand the Reversibility Criterion

A stationary Markov chain is reversible if, for its stationary distribution \( \pi \), it holds that \( \pi(i)P(i,j) = \pi(j)P(j,i) \) for all states \( i \) and \( j \). We will check if each of the chains is reversible by verifying this condition.
02

Check Reversibility for Chain (a)

For the chain \( X \), assume a stationary distribution \( (\pi_0, \pi_1) \). The reversibility condition is \( \pi_0(1-\alpha) = \pi_1\beta \) and \( \pi_0\alpha = \pi_1(1-\beta) \). Solving these equations with \( \pi_0 + \pi_1 = 1 \) gives a valid stationary distribution, verifying that it is indeed reversible.
03

Check Reversibility for Chain (b)

For chain \( Y \), assume a stationary distribution \( (\pi_0, \pi_1, \pi_2) \). Checking the condition for reversibility, \( \pi_0 p = \pi_1 p \), \( \pi_1 p = \pi_2 p \), and \( \pi_2 (1-p) = \pi_0 (1-p) \), solving these equations leads to contradictions except for trivial cases where \( p = 1 \). Hence, chain \( Y \) is not reversible for \( 0 < p < 1 \).
04

Check Reversibility for Chain (c)

Chain \( Z_n = (X_n, Y_n) \) is the combination of the independent chains \( X \) and \( Y \). Because \( X \) and \( Y \) are independent and the product of their stationary distributions is a stationary distribution for \( Z \), we check whether the product chain \( Z \) is reversible. To be reversible, both components \( X \) and \( Y \) need to be reversible, but since \( Y \) is not reversible, \( Z \) is not reversible.

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Key Concepts

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

Stationary Distribution
The stationary distribution in a Markov chain refers to a probability distribution that remains unchanged as the system progresses. Think of it as balance—once the chain has reached this distribution, the overall probabilities of being in each state remain constant over time. Here's how it works:- **Existence:** Stationary distributions might not exist for every Markov chain, but when they do, they provide a stable long-term prediction of the system's states.- **Condition:** For a Markov chain with transition matrix \( P \), a stationary distribution \( \pi \) satisfies \( \pi P = \pi \). This means multiplying the stationary distribution with the transition matrix results in the same distribution.In our exercise, finding the stationary distribution involves solving equations derived from \( \pi P = \pi \) along with the constraint \( \pi_0 + \pi_1 = 1 \) of probabilities summing up to one. Understanding the stationary distribution is essential as it helps determine characteristics like reversibility in Markov chains.
Transition Matrix
The transition matrix is the core element in a Markov chain that defines how the system transitions from one state to another. - **Structure:** It is typically represented as a square matrix where each element \( P(i,j) \) describes the probability of moving from state \( i \) to state \( j \).- **Properties:** Each row of the transition matrix sums to one, reflecting that the total probability of transitioning from one state to any of the possible states (including itself) is 1.For example, in Exercise (a), the transition matrix is:\[P = \begin{pmatrix}1-\alpha & \alpha \\beta & 1-\beta\end{pmatrix}\]This matrix tells us about how the process moves between two states depending on the parameters \( \alpha \) and \( \beta \).- **Application:** Using these matrices, we can construct equations to find stationary distributions and verify properties like reversibility by directly examining transition probabilities.
Reversibility Criterion
Reversibility is a feature of some Markov chains that says you can 'run the chain backward' without altering the behavior in terms of the stationary distribution. A Markov chain is reversible if it satisfies the following condition for its stationary distribution \( \pi \):- **Condition:** \( \pi(i)P(i,j) = \pi(j)P(j,i) \) for all states \( i \) and \( j \). This equation means the flow from state \( i \) to state \( j \) in the equilibrium matches the flow back from \( j \) to \( i \).In our exercise:- For chain \( X \) (Exercise a), by solving the reversibility equations, we observe it meets this criterion.- For chain \( Y \) (Exercise b), assumptions lead to contradictions, so reversibility isn't satisfied.- For chain \( Z \) (Exercise c), since one component \( Y \) isn't reversible, the whole combination \( Z \) isn't reversible either.Understanding the reversibility criterion helps in determining the symmetry of process flows in Markov chains, making it fundamental for certain applications like physics and queueing theory.

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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 \(\lambda=\mu ?\)

(a) Show that for cach pair \(i, j\) of states of an irreducible aperiodic chain, there exists \(N=N(i, j)\) such that \(p_{i j}(n)>0\) for all \(n \geq N\). (b) Let \(X\) and \(Y\) be independent irreducible aperiodic chains with the same state space \(S\) and transition matrix P. Show that the bivariate chain \(Z_{n}=\left(X_{n}, Y_{n}\right), n \geq 0\), is irreducible and aperiodic. (c) Show that the bivariate chain \(Z\) may be reducible if \(X\) and \(Y\) are periodic.

Let \(\left\\{X_{r}: r \geq 1\right\\}\) be independent exponential random variables with parameter \(\lambda\), and set \(S_{n}=\) \(\sum_{r=1}^{n} X_{r .}\) Show that: (a) \(Y_{k}=S_{k} / S_{n}, 1 \leq k \leq n-1\), have the same distribution as the order statistics of independent variables \(\left|U_{k}: 1 \leq k \leq n-1\right|\) which are uniformly distributed on \((0,1)\), (b) \(Z_{k}=X_{k} / S_{n}, 1 \leq k \leq n\), have the same joint distribution as the coordinates of a point \(\left(U_{1}, \ldots, U_{n}\right)\) chosen uniformly at random on the simplex \(\sum_{r=1}^{n} u_{r}=1, u_{r} \geq 0\) for all \(r\).

If the intensity function \(\lambda\) of a non-homogeneous Poisson process \(N\) is itself a random process, then \(N\) is called a doubly stochastic Poisson process (or Cox process), Consider the case when \(\lambda(t)=\Lambda\) for all \(t\), and \(A\) is a random variable taking cither of two values \(\lambda_{1}\) or \(\lambda_{2}\), each being picked with equal probability \(\frac{1}{2}\). Find the probability generating function of \(N(t)\), and deduce its mean and variance.

A particle has velocity \(V(t)\) at time \(t\), where \(V(t)\) is assumed to take values in \(\left(n+\frac{1}{2} ; n \geq 0\right]\). Transitions during \((t, t+h)\) are possible as follows: $$ P(V(t+h)=w \mid V(t)=v)= \begin{cases}\left(v+\frac{1}{2}\right) h+o(h) & \text { if } w=v+1 \\ 1-2 v h+o(h) & \text { if } w=v \\\ \left(v-\frac{1}{2}\right) h+o(h) & \text { if } w=v-1\end{cases} $$ Initially \(V(0)=\frac{1}{2}\), Let $$ G(s, t)=\sum_{n=0}^{\infty} s^{n} \mathrm{P}\left(V(t)=n+\frac{1}{2}\right) $$ (a) Show that $$ \frac{\partial G}{\partial t}=(1-s)^{2} \frac{\partial G}{\partial s}-(1-s) G $$ and deduce that \(G(s, t)=(1+(1-s) t\\}^{-1}\). (b) Show that the expected length \(m_{n}(T)\) of time for which \(V=n+\frac{1}{2}\) during the time interval \([0, T]\). is given by $$ m_{n}(T)=\int_{0}^{T} \mathrm{P}\left(V(t)=n+\frac{1}{2}\right) d t $$ and that, for fixed \(k, m_{k}(T)-\log T \rightarrow-\sum_{i=1}^{k} i^{-1}\) as \(T \rightarrow \infty\). (c) What is the expected velocity of the particle at time \(t\) ?

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