Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 9F / a

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- Solution a

Solution

Because

U

and

V

are bounded, their exponential series may be integrated term by term for every real

t

. Thus their moment-generating functions satisfy

M_{U} (t) = E [e^{t U}] = \sum_{j = 0}^{\infty} \frac{t ^{j}}{j !} E [U^{j}],

and similarly for

V

. Equality of every moment gives equality term by term, so

M_{U} (t) = M_{V} (t) for every t \in R .

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