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

A Markov chain on states {1, 2, 3} has transition matrix rows: 1: (0, 1, 0); 2: (0.5, 0, 0.5); 3: (0, 1, 0). Which statement about this chain is correct?

The chain is irreducible with period 2 and stationary distribution (0.25, 0.5, 0.25). Every return takes an even number of steps, so n-step transition probabilities oscillate rather than converge, even though the stationary distribution exists.

  1. AIt is aperiodic with stationary distribution (0.25, 0.5, 0.25), and the n-step probabilities converge to it
  2. BIt is reducible because state 2 cannot be reached from state 3
  3. CIt has period 2 with stationary distribution (0.25, 0.5, 0.25), and the n-step probabilities do not convergeCorrect
  4. DIt has period 2 with stationary distribution (1/3, 1/3, 1/3)
  5. It has period 3 with a unique stationary distribution (0.25, 0.5, 0.25)

Explanation

All return paths to a state have even length (1 to 2 to 1, 2 to 1 to 2 etc.), so period is 2. Solving pi=pi P: pi1=0.5pi2, pi3=0.5pi2, pi2=pi1+pi3, giving (0.25,0.5,0.25). Because the chain is periodic, P^n alternates and does not converge, though the stationary distribution exists and is unique.

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