 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/9f/iv/solution

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

Conditioning on

N

and using the moment-generating function of a unit exponential variable,

E (e^{θ Y}) = n = 1 \sum \infty p (1 - p)^{n - 1} (\frac{1}{1 - θ})^{n} = \frac{p}{1 - θ} m = 0 \sum \infty (\frac{1 - p}{1 - θ})^{m} = \frac{p}{p - θ},

where the geometric series converges precisely when

θ < p

. This is the moment-generating function of an exponential variable of rate

p

Y \sim Exp (p) .

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