For $X=\theta^TZ$,
$$
\mathbb EX=\theta^Tb,\qquad \operatorname{Var}(X)=\theta^TV\theta.
$$
The objective has <gradient> $b-V\theta$ and Hessian $-V$, which is negative definite. Its unique maximizer is therefore
$$
\boxed{\theta_M=V^{-1}b}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 30K / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/30k/a/solution