 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-1/19h/d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Because

0 \leq U \leq 1

, for fixed

t

and sufficiently large

n

P (n (1 - U) \geq t) = {1, P (U \leq 1 - t / n) = (1 - t / n)^{n}, t < 0, t \geq 0.

Therefore

h (t) = {1, e^{- t}, t < 0, t \geq 0.

Equivalently,

n (1 - θ / θ)

Put

c = lo g (1/ α)

. The limiting probability

P (n (1 - \frac{θ}{θ}) \leq c) ⟶ 1 - e^{- c} = 1 - α

and the certain inequality

θ \leq θ

suggest the approximate confidence interval

[θ, \frac{n θ}{n - lo g ( 1/ α )}] .

Since

E [θ] = n θ / (n + 1)

, its expected length is

(\frac{n θ}{n + 1}) (\frac{lo g ( 1/ α )}{n - lo g ( 1/ α )}),

as required.

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