 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/3f/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Independence gives

M_{S} (t) = \prod_{j = 1}^{n} M_{X_{j}} (t) = [exp (e^{t} - 1)]^{n} = exp {n (e^{t} - 1)} .

This is the moment generating function of a Poisson distribution with mean

n

. Hence

S \sim Poisson (n) .

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