IAI Actuarial Core Principles · Risk Modelling and Survival Analysis · Markov chains
A Markov chain on {1,2,3} has P(1→2)=1, P(2→3)=1 and P(3→1)=1. Which statement is correct?
The chain is irreducible with period 3. The uniform distribution is stationary, but because of periodicity the n-step probability P^n(1,1) cycles between 1 and 0 and does not converge.
- AThe chain is irreducible and aperiodic, with a limiting distribution
- BThe chain is irreducible with period 3; the uniform distribution is stationary but P^n(1,1) does not convergeCorrect
- CThe chain is reducible with three absorbing states
- DThe chain has no stationary distribution because it is periodic
- States are transient because the chain returns with certainty only after 3 steps
Explanation
All states communicate, so the chain is irreducible, and returns to a state occur only at multiples of 3, so the period is 3. The vector (1/3,1/3,1/3) satisfies πP=π, so a stationary distribution exists. But P^n(1,1) cycles through 1,0,0, so no limit exists.
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