 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ii/paper-2/26j/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

For

i \neq = j

the generator entry is

q_{ij} = r_{j}

. Its diagonal is the negative total departure rate,

q_{ii} = - (R - r_{i})

. Thus

Q = 1 r^{T} - R I, r = (r_{1}, \dots, r_{s})^{T} .

Every row sums to zero, as required for a continuous-time Markov-chain generator.

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