 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-32/2/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

In a time-homogeneous continuous-time Markov chain, a state's holding time is exponential with rate equal to its total outgoing transition intensity. Converting the specified times to months gives

a + b = \frac{1}{96}, \frac{a}{a + b} = \frac{1}{2}, c = \frac{1}{36} .

The exit-type probability follows by dividing its transition intensity by the total exit rate. Therefore

a = b = \frac{1}{192}, c = \frac{1}{36}, Q = - 1/96 00 1/192 - 1/36 0 1/192 1/36 0 month^{- 1} .

These are starting values for numerical estimation, rather than further observations or constraints on the final fitted rates.

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