 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/29k/c/solution

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

Differentiating

f (μ, σ) = E [U (μ + σ Z)]

twice gives the Hessian matrix

Hess f = (E [U^{''} (X)] E [Z U^{''} (X)] E [Z U^{''} (X)] E [Z^{2} U^{''} (X)]) = E [U^{''} (X) (1 Z) (1 Z)] .

For every vector

(a, b)

(a b) (Hess f) (a b) = E [U^{''} (X) (a + b Z)^{2}] \leq 0.

f is jointly concave in (μ, σ) .

Equivalently, with

W = - U^{''} (X) \geq 0

E W E [Z^{2} W] - (E [Z W])^{2} \geq 0

by the Cauchy-Schwarz inequality, while both diagonal entries are nonpositive.

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