IAI Actuarial Core Principles · Risk Modelling and Survival Analysis · Markov processes
A two-state Markov jump process has rate a from state 0 to 1 and rate b from 1 to 0, with a=0.3 and b=0.5. Starting in state 0, p_00(t) takes which form?
p_00(t) = 0.5/0.8 + (0.3/0.8) e^{-0.8t}. The limit is b/(a+b), the stationary probability of state 0, and the decay rate is a+b. It equals 1 at t=0, as required.
- A0.3/0.8 + (0.5/0.8) e^{-0.8t}
- B0.5/0.8 + (0.3/0.8) e^{-0.8t}Correct
- C0.5/0.8 + (0.3/0.8) e^{-0.2t}
- D0.3/0.8 + (0.5/0.8) e^{-0.2t}
- 0.5 + 0.3 e^{-0.8t}
Explanation
For a two-state chain, p_00(t) = b/(a+b) + a/(a+b) e^{-(a+b)t}. With a+b=0.8 this gives 0.5/0.8 + 0.3/0.8 e^{-0.8t}. At t=0 it equals 1, and as t grows it tends to 0.625. Option 0 swaps the constants and fails the stationary check.
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