 Solution

ID: past-exam-of-the-probability-course-of-the-university-of-oxford/2017/1/a/ii/solution

x = v w, y = (1 - v) w, 0 < v < 1, w > 0.

det (w - w v 1 - v) = w .

Therefore

f_{V, W} (v, w) = w e^{- w} 1_{{0 < v < 1, w > 0}} .

Since

\int_{0}^{\infty} w e^{- w} d w = 1

f_{V} (v) = 1_{(0, 1)} (v)

and

f_{W} (w) = w e^{- w} 1_{(0, \infty)} (w)

. The joint density factorizes as

f_{V} f_{W}

, so $V$ and $W$ are independent random variables, with

V

uniform on

[0, 1]

and

W \sim Gamma (2, 1)

in the shape-rate convention. This is beta-gamma independence with both shape parameters equal to one.

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