 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-1/18h/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

π_{1} (θ) = ⎩ ⎨ ⎧ 0, θ + 0.05, 1, θ \leq - 0.05, - 0.05 < θ < 0.95, θ \geq 0.95.

Put

a = 1/ 10

. Test 2 rejects when

U_{1} + U_{2} > 2 - a - 2 θ .

Using the two branches of the triangular distribution gives

π_{2} (θ) = ⎩ ⎨ ⎧ 0, \frac{1}{2} (a + 2 θ)^{2}, 1 - \frac{1}{2} (2 - a - 2 θ)^{2}, 1, θ \leq - a /2, - a /2 < θ < (1 - a) /2, (1 - a) /2 \leq θ < (2 - a) /2, θ \geq (2 - a) /2.

Both functions equal

0.05

at the least favourable null value

θ = 0

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