 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iii/7/a/i/solution

θ

u = θ x

gives

E [X] = \frac{1}{θ} \int_{0}^{\infty} u e^{- u} d u = \frac{1}{θ}, E [X^{2}] = \frac{1}{θ ^{2}} \int_{0}^{\infty} u^{2} e^{- u} d u = \frac{2}{θ ^{2}} .

The integrals are one and two by integration by parts. Hence

E [X] = 1/ θ, Var (X) = 1/ θ^{2}

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