 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-2/12f/a/solution

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

1 = A \int_{0}^{2 π} (π - ϕ /2) d ϕ = A π^{2}, A = π^{- 2} .

The first two moments are

E Φ = \frac{1}{π ^{2}} \int_{0}^{2 π} ϕ (π - ϕ /2) d ϕ = \frac{2 π}{3}, E Φ^{2} = \frac{1}{π ^{2}} \int_{0}^{2 π} ϕ^{2} (π - ϕ /2) d ϕ = \frac{2 π ^{2}}{3} .

Consequently

E Φ = \frac{2 π}{3}, Var Φ = \frac{2 π ^{2}}{3} - \frac{4 π ^{2}}{9} = \frac{2 π ^{2}}{9} .

F (ϕ) = ϕ / π - ϕ^{2} / (4 π^{2}) = 1 - (1 - ϕ / (2 π))^{2}

[0, 2 π]

. For

U

(0, 1)

Φ = 2 π (1 - 1 - U) .

The inverse lies in the required interval and satisfies

F (Φ) = U

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