 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2008/m/1/solution

x = 2 u + v

y = v

, with absolute Jacobian determinant two. Positivity of

x, y

gives

A = {(u, v) : v > 0, 2 u + v > 0} .

f_{U, V} (u, v) = \frac{1}{4} e^{- (2 u + 2 v) /2} \cdot 2 = \frac{1}{2} e^{- u - v} ((u, v) \in A) .

For fixed

u

, integrate over

v > max (0, - 2 u)

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}, \frac{1}{2} e^{u}, u \geq 0, u < 0.

Thus

f_{U} (u) = \frac{1}{2} e^{- ∣ u ∣}

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