 Solution

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

n - 1

E [Y^{- 1}] = \frac{θ ^{n}}{( n - 1 )!} \int_{0}^{\infty} y^{n - 2} e^{- θ y} d y = \frac{θ ^{n}}{( n - 1 )!} \frac{( n - 2 )!}{θ ^{n - 1}} = \frac{θ}{n - 1} .

The same argument, valid for

n \geq 3

, gives

E [Y^{- 2}] = θ^{2} / [(n - 1) (n - 2)]

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