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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ia/paper-2/10f/solution

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Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 2 / 10F / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

For a uniform random point in a disk, the event

R < x

occupies a disk of area

π x^{2}

when

0 < x < 1

. Thus

F_{R} (x) = ⎩ ⎨ ⎧ 0, x^{2}, 1, x \leq 0, 0 < x < 1, x \geq 1.

The radial probability density function is

f_{R} (r) = 2 r

on

(0, 1)

. Integrating gives

E R = \int_{0}^{1} 2 r^{2} d r = \frac{2}{3}, E R^{2} = \frac{1}{2}, Var R = \frac{1}{18} .

Choose the polar angle

Θ \in [0, 2 π)

. The Cartesian density is

1/ π

in the disk, and the polar coordinates Jacobian determinant is

r

. Hence

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

The product support and product density prove that $R$ and $Θ$ are independent, with uniform angle.

Reflection symmetry gives

E X = E Y = 0

and

E (X Y) = 0

, so $Cov (X, Y) = 0$ . They are nevertheless not independent. The events

X > 3/4

and

Y > 3/4

each have positive probability, but their intersection is empty inside the disk, because it would require

X^{2} + Y^{2} > 18/16 > 1

. Independence would give a positive product probability for that intersection.

Except at the probability-zero center,

X / R = cos Θ

and

Y / R = sin Θ

. Therefore

E \frac{X}{R} + i E \frac{Y}{R} = E e^{i Θ} = \frac{1}{2 π} \int_{0}^{2 π} e^{i θ} d θ = 0, ξ = Θ (mod 2 π) .

An arbitrary definition at the center does not change these expected values. This example distinguishes uncorrelated Cartesian coordinates from independent polar coordinates.

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