 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-40/5/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

r \in [0, 1]

, an English auction makes no sale if both valuations are below

r

. If exactly one exceeds it, the price is

r

; if both exceed it, the price is their minimum. Thus

R_{c} (r) = r [2 r (1 - r)] + \int_{r}^{1} 2 t (1 - t) d t = \frac{1}{3} + r^{2} - \frac{4}{3} r^{3} .

The first term accounts for one qualifying buyer, and the integral uses the density

2 (1 - t)

of the smaller valuation. Differentiation gives

R_{c}^{'} (r) = 2 r (1 - 2 r)

. It increases up to

r = 1/2

and decreases afterwards, while the endpoints give

1/3

and zero. Therefore

r_{*} = \frac{1}{2}, R_{c}^{*} = \frac{5}{12} .

The strict sale condition at the reserve changes only a probability-zero event.

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