 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-2/10f/b/solution

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

For

0 \leq s \leq 1

, independence gives

G_{X + Y} (s) = E [s^{X + Y}] = E [s^{X} s^{Y}] = E [s^{X}] E [s^{Y}] .

Therefore

G_{X + Y} (s) = G_{X} (s) G_{Y} (s) .

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