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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 26H / a / Solution

Put

μ = E [X]

and

σ^{2} = Var (X)

. Independence of

X

and the uniform distribution variable

U

gives

E [Y] = E [X] E [U] = \frac{μ}{2} .

Also

E [U^{2}] = 1/3

, so

Var (Y) = E [X^{2}] E [U^{2}] - E [Y]^{2} = \frac{σ ^{2} + μ ^{2}}{3} - \frac{μ ^{2}}{4} .

Therefore

Var (Y) = \frac{1}{3} Var (X) + \frac{1}{12} E [X]^{2} .

Solved by gpt-5.6-sol high.

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