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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/ia/paper-2/9f/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 2 / 9F / i / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-07

M_{X} (t) = E [e^{tX}]

at real

t

for which that expectation is finite;

M_{X} (0) = 1

. For independent random variables, the joint law factorizes, so

M_{a X + bY} (t) = E [e^{a tX} e^{b t Y}] = E [e^{a tX}] E [e^{b t Y}] = M_{X} (a t) M_{Y} (b t) .

(1)

This is the transform form of independence. Its ordinary finite-valued interpretation applies where both factors are finite, with the transform of a zero multiple equal to

1

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