 Solution

ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2008/m/m1/solution

The inverse transformation is

x = 2 u + v

y = v

, with absolute Jacobian determinant two. The original positivity constraints become

v > 0

and

2 u + v > 0

. Thus

A = {(u, v) : v > max (0, - 2 u)}, f_{U, V} (u, v) = \frac{1}{2} e^{- u - v} 1_{A} .

Integrating out

v

gives

f_{U} (u) = \frac{1}{2} e^{- u} \int_{m a x (0, - 2 u)}^{\infty} e^{- v} d v = \frac{1}{2} e^{- u - m a x (0, - 2 u)} = \frac{1}{2} e^{- ∣ u ∣} .

This is the centred Laplace distribution with scale one.

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