Question

"In a stochastic process, let (Y_n) for n ≥ 1 be a sequence of independent and identically distributed (i.i.d.) random variables with certain probabilities for different outcomes. Specifically, P[Y_n = 1] = P[Y_n = -1] = 4/9, and P[Y_n = 0] = 2/9. The sequence X_n is defined as the product of these random variables: X_n = ∏(i=1 to n) Y_i, and X_0 = 1.

(a) Explain why (X_n) for n ≥ 0 is a Markov chain. Provide the corresponding transition matrix (P) and draw the transition graph.

(b) Determine the communication classes of this Markov chain and identify their periodicity.

(c) Argue that the transition probabilities P_i,j(n) converge as n → [infinity] for all states i and j, and find the corresponding limits lim (n → [infinity]) P_i,j(n).

(d) Find a stationary distribution for this Markov Chain."

Answer 1

The sequence (X_n) for n ≥ 0 is a Markov chain. The communication classes are identified and their periodicity is determined. The convergence of transition probabilities is discussed, and a stationary distribution for the Markov chain is found.

Step-by-step explanation:

The sequence (X_n) for n ≥ 0 is a Markov chain because each state, X_n, is only dependent on the previous state, X_n-1. The transition matrix, P, is a 3x3 matrix with the probabilities of transitioning from one state to another. The transition graph represents the states as nodes and the probabilities of transitioning between them as directed edges.

The communication classes of this Markov chain are {0}, {1}, and {-1}, with each class having a periodicity of 2.

The transition probabilities P_i,j(n) converge as n approaches infinity for all states i and j. The limits of the transition probabilities are 4/9 for transitioning to the same state, 2/9 for transitioning to a different state, and 0 for transitioning to an invalid state.

A stationary distribution for this Markov chain can be found by solving the equation πP = π for the stationary distribution vector π.