IAI Actuarial Core Principles · Risk Modelling and Survival Analysis · Markov processes
A two-state process (Alive A, Dead D) has constant force of mortality 0.04 per year, with no other transitions. What is the probability that a life in state A at time 0 is still in state A at time 5, to 4 decimal places?
The probability of survival is exp(-0.2) = 0.8187. With a constant force of 0.04 the holding time in alive is exponential, so survival over 5 years is exp(-0.04 times 5).
- A0.8187Correct
- B0.8000
- C0.9608
- D0.1813
- 0.7788
Explanation
P_AA(5) = exp(-0.04 x 5) = exp(-0.2) = 0.8187. The option 0.1813 is the probability of death, 1 - 0.8187. The option 0.8000 uses the linear approximation 1 - 0.2 which ignores compounding.
Did you get it right without looking?
One question tells you little. A timed set on Markov processes shows your real accuracy, how long you take and where you lose marks.
More Markov processes questions
- A time-homogeneous Markov jump process has states 0 and 1 with generator matrix rows (-3, 3) and (2, -2). What is the expected holding time …
- For a time-homogeneous Markov jump process with transition matrix P(t) and generator matrix A, which statement is correct?
- A Markov chain on states 1, 2, 3 has transition matrix rows: 1: (0.5, 0.5, 0); 2: (0.25, 0.5, 0.25); 3: (0, 0.5, 0.5). What is the stationar…
- Let Y_1, Y_2, ... be independent identically distributed random variables and S_n = Y_1 + ... + Y_n. Which statement about the process {S_n}…
- 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) take…
- In a time-homogeneous continuous-time Markov jump process, the holding time in state i has which distribution, and what is its parameter?