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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-2/12f/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 2 / 12F / b / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

0 < x, y < 1

a = x^{- 1/ n} - 1

and

b = y^{- 1/ n} - 1

. Conditional on

S

,

P (X_{1} \leq x, X_{2} \leq y ∣ S) = e^{- (a + b) S} .

Using part (a),

F_{X_{1}, X_{2}} (x, y) = (x^{- 1/ n} + y^{- 1/ n} - 1)^{- n} .

Differentiation gives

f_{X_{1}, X_{2}} (x, y) = \frac{n + 1}{n} \frac{x ^{- 1 - 1/ n} y ^{- 1 - 1/ n}}{( x ^{- 1/ n} + y ^{- 1/ n} - 1 ) ^{n + 2}} .

Putting

y = 1

yields

F_{X_{1}} (x) = x

, so

X_{1}

is uniform on

(0, 1)

.

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