 Solution

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

a, b

, substitute the two normal densities:

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

1/ (a^{- 2} + v^{2} b^{- 2})

, by substituting the exponent's quadratic argument. Thus, with

c = b / a

f_{V} (v) = \frac{ab}{π ( b ^{2} + a ^{2} v ^{2} )} = \frac{c}{π ( c ^{2} + v ^{2} )} .

This is a centered Cauchy distribution of scale

c

. Since

P (X = 0) = 0

, the desired probability is

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

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