 Solution

ID: past-exam-of-the-probability-statistics-and-options-ii-course-of-the-university-of-oxford/2010/m/m4/a/iii/solution

Assume

a, b > 0

, the standard deviations. Substitute the two normal densities into (ii):

f_{V} (v) = \frac{1}{πab} \int_{0}^{\infty} u exp [- \frac{u ^{2}}{2} (\frac{1}{a ^{2}} + \frac{v ^{2}}{b ^{2}})] d u = \frac{ab}{π ( b ^{2} + a ^{2} v ^{2} )} .

The last step uses

\int_{0}^{\infty} u e^{- A u^{2} /2} d u = 1/ A

. With

c = b / a

this is

f_{V} (v) = c / [π (c^{2} + v^{2})]

, the centered Cauchy density of scale

c

The desired event is

∣ V ∣ < 1

, outside the null event

X = 0

. Integrating the density gives

P (∣ Y ∣ < ∣ X ∣) = \frac{2}{π} arctan (1/ c) = \frac{2}{π} arctan (a / b) .

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