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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-2/4f/solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 4F / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-29

Independence of random variables and the two unit-rate exponential distributions give

f_{X, Y} (x, y) = e^{- x - y} 1_{{x \geq 0, y \geq 0}} .

The linear transformation

(U V) = (6283) (X Y)

has determinant

2

and inverse

x = \frac{3 u - 8 v}{2}, y = - u + 3 v .

The inequalities

x, y \geq 0

become

8 v /3 \leq u \leq 3 v

. By the change-of-variables formula for probability densities,

f_{U, V} (u, v) = \frac{1}{2} e^{- u /2 + v} 1_{{v \geq 0, 8 v /3 \leq u \leq 3 v}} .

The variables

U

and

V

are not independent: the support of their joint density is a wedge rather than a Cartesian product of one-dimensional supports. For example, conditioning on a positive value of

V

constrains

U

to a finite interval depending on that value.

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