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

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- Solution b

Solution

The map

(x, u) \mapsto (x, y) = (x, xu)

has inverse

u = y / x

and Jacobian magnitude

1/ x

. Hence the density of

(X, Y)

is

f_{X, Y} (x, y) = \frac{g ( x )}{x} 1_{{x > 0, 0 \leq y \leq x}} .

For

(y, z) = (xu, x (1 - u))

, the inverse is

x = y + z, u = \frac{y}{y + z},

and the Jacobian magnitude of

(y, z) \mapsto (x, u)

is

1/ (y + z)

. Thus the uniform split of a random total has joint density

f_{Y, Z} (y, z) = \frac{g ( y + z )}{y + z} 1_{{y, z \geq 0}} .

Integrating out the other piece yields

f_{Y} (y) = \int_{0}^{\infty} \frac{g ( y + z )}{y + z} d z = h (y), y \geq 0.

The transformation

U \mapsto 1 - U

preserves the uniform law, so

Z

has the same density:

f_{Z} (z) = h (z), z \geq 0.

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