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IAI Actuarial Core Principles · Risk Modelling and Survival Analysis · Stochastic processes

A Markov chain on states {A, B} has transition matrix with rows A: (0, 1) and B: (1, 0). Which statement about its long-term behaviour is correct?

The chain has the unique stationary distribution (0.5, 0.5), but it is periodic with period 2, so the n-step transition probabilities alternate and do not converge. Stationarity exists, while limiting convergence requires aperiodicity.

  1. AIt has no stationary distribution because the chain is reducible
  2. BIt has a unique stationary distribution (0.5, 0.5), but the n-step probabilities do not converge to it because the chain is periodicCorrect
  3. CIt has a unique stationary distribution (0.5, 0.5) and P^n converges to it as n tends to infinity
  4. DIt has infinitely many stationary distributions because it is irreducible
  5. It has stationary distribution (1, 0) because A is the starting state

Explanation

The chain is irreducible with period 2, so it has a unique stationary distribution (0.5, 0.5), which satisfies πP=π. However, P^n alternates between the identity and P, so there is no convergence. Convergence needs aperiodicity.

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