Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / D / D4 / iii

Table of contents
- Solution iii

Solution

h

is measurable and nonnegative. In first-quadrant polar coordinates, including the Jacobian, its integral is

(\int_{0}^{\infty} e^{- r^{2}} r^{3} d r) (\int_{0}^{π /2} (cos θ)^{3/2} (sin θ)^{1/2} d θ) .

The radial factor is

1/2

by the supplied fact and the angular integrand is at most one, giving finiteness. The nonnegative change-of-variables formula is valid before finiteness is known, and this bound proves

h \in L^{1}

.

Fubini may now factor the Cartesian integral. Substituting

u = x^{2}

,

v = y^{2}

in the respective factors gives

\iint h d x d y = \frac{1}{4} (\int_{0}^{\infty} e^{- u} u^{1/4} d u) (\int_{0}^{\infty} e^{- v} v^{- 1/4} d v) .

Both are finite: their powers at zero exceed

- 1

, and the exponential controls infinity. Comparing with the polar expression proves

\int_{0}^{π /2} (cos θ)^{3/2} (sin θ)^{1/2} d θ = \frac{1}{2} Γ (5/4) Γ (3/4),

which is exactly the product of integrals requested.

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