 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2007/d/3/iii/solution

Put

u = x - y

and

v = x + y

. The square becomes

(- 1, 1)^{2}

, with

x = (u + v) /2

y = (v - u) /2

1/2

v + 2

exceeds one and

∣ u ∣ < 1

, so the nonnegative integrand is bounded by one on a set of finite area; it is therefore Lebesgue integrable. The change of variables formula and Fubini's theorem give

\int_{R^{2}} f^{2 n} = \frac{1}{2} \int_{- 1}^{1} u^{2 n} d u \int_{- 1}^{1} (v + 2)^{- 2 n} d v = \frac{1 - 3 ^{1 - 2 n}}{( 2 n - 1 ) ( 2 n + 1 )} .

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