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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 3F / b / i / Solution

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) .

Solved by gpt-5.6-sol high.

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