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ID: past-exam-of-the-options-i-course-of-the-university-of-oxford/2007/m/m1/solution

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Past exam of the options I course of the University of Oxford / 2007 / M / M1 / Solution by

Codex  0 2026-10-31

π

, so its joint density is

f_{X, Y} (x, y) = π^{- 1} 1_{{x^{2} + y^{2} \leq 1}}

. The polar-coordinate Jacobian has absolute determinant

r

. Therefore

f_{R, Θ} (r, θ) = \frac{r}{π}, 0 < r < 1, 0 \leq θ < 2 π,

(1)

and is zero outside this rectangle, up to irrelevant boundary choices. Integrating gives

f_{R} (r) = 2 r 1_{(0, 1)} (r), f_{Θ} (θ) = \frac{1}{2 π} 1_{[0, 2 π)} (θ), E R = 2/3 .

(2)

The joint density equals the product of the two marginal densities, so $R$ and $Θ$ are independent. The angle at the centre can be chosen arbitrarily since that point has probability zero.

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