 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-1/28k/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

For

λ_{n} = α n + β

\sum_{n = 0}^{\infty} \frac{1}{α n + β} = \infty,

so part (b) ensures nonexplosion. The Markov jump-process generator acts on the identity function

f (n) = n

(L f) (n) = (α n + β) ((n + 1) - n) = α n + β .

Consequently

m (t) = E N (t)

satisfies

m^{'} (t) = E [(L f) (N (t))] = α m (t) + β, m (0) = 0.

E N (t) = \frac{β}{α} (e^{α t} - 1) .

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