IAI Actuarial Core Principles · Risk Modelling and Survival Analysis · Markov chains
A chain on {A,B,C} has P(A→B)=1, P(B→A)=0.5, P(B→C)=0.5, P(C→A)=1. What is the stationary probability of state A?
The stationary probability of A is 0.4. Since A moves only to B, πB = πA, and C is reached only from B with probability 0.5, so πC = 0.5πA. Normalising 2.5πA = 1 gives πA = 0.4.
- A1/3
- B0.4
- C0.5Correct
- D0.6
- 2/3
Explanation
Equations: πB=πA, πC=0.5πB, and πA=0.5πB+πC. So πB=πA and πC=0.5πA. Sum: πA(1+1+0.5)=1 gives πA=0.4. Check: 0.5×0.4+0.2=0.4. So 0.4, not 0.5.
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