 Solution

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

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

y_{1} \leq y_{2}

, then the utility function is increasing, so for every

X \in X

E [U (X + y_{1})] \leq E [U (X + y_{2})] .

Taking suprema gives

F (y_{1}) \leq F (y_{2})

For

0 \leq t \leq 1

, let

X_{1}

and

X_{2}

attain the suprema at

y_{1}

and

y_{2}

. Because

X

X_{t} = t X_{1} + (1 - t) X_{2} \in X .

U

now gives

F (t y_{1} + (1 - t) y_{2}) \geq E [U (X_{t} + t y_{1} + (1 - t) y_{2})] \geq t E [U (X_{1} + y_{1})] + (1 - t) E [U (X_{2} + y_{2})] = tF (y_{1}) + (1 - t) F (y_{2}) .

Thus $F$ is increasing and concave, as stated by optimized affine shift of concave utility.

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