 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-2/3f/a/solution

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

G_{X} (s) = E [s^{X}] = \sum_{n = 0}^{\infty} P (X = n) s^{n},

initially for

∣ s ∣ \leq 1

, with

0^{0} = 1

G_{X} (s) = e^{- λ} \sum_{n = 0}^{\infty} \frac{( λ s ) ^{n}}{n !} .

Thus

G_{X} (s) = e^{λ (s - 1)} .

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