 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ia/paper-2/12f/b/solution

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

Given

X_{n} = k

, the next size is a sum of

k

E [s^{X_{n + 1}} ∣ X_{n} = k] = G (s)^{k}

. Averaging this conditional expectation proves

G_{n + 1} (s) = E [G (s)^{X_{n}}] = G_{n} (G (s)) .

With one initial individual,

G_{0} (s) = s

and

G_{1} (s) = G (s)

, so

G_{n}

is the

n

-fold composition of

G

with itself.

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