 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-1/30k/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

0 \leq t \leq 1

U

implies

G (t s_{1} + (1 - t) s_{2}) = E [U (t (m + s_{1} Z) + (1 - t) (m + s_{2} Z))] \geq tG (s_{1}) + (1 - t) G (s_{2}),

G

is concave.

Since

E Z = 0

G (s) = E [U (m + s Z)] \leq U (m + s E Z) = U (m) = G (0) .

0 \leq s < t

, concavity and

s = (1 - s / t) 0 + (s / t) t

yield

G (s) \geq (1 - \frac{s}{t}) G (0) + \frac{s}{t} G (t) \geq G (t) .

Hence $G$ is decreasing on $[0, \infty)$ ; this is scaled centered risk under concave utility.

 New to topics? Read the docs here!