 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/6c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

n

and sum. Index shifts give

\overset{μ}{˙} = (B - D) μ .

For the second moment, a birth changes

n^{2}

2 n + 1

and a death changes it by

- 2 n + 1

, so

\frac{d}{d t} E [N^{2}] = 2 (B - D) E [N^{2}] + (B + D) μ .

Since

σ^{2} = E [N^{2}] - μ^{2}

(σ^{2})^{'} = 2 (B - D) σ^{2} + (B + D) μ .

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