Since
$$
\mathbb E_{\theta\mid y}(y-\theta)^2
=(y-\widetilde\theta)^2+\sigma_\theta^2,
$$
we obtain
$$
\boxed{
\mathbb E_{\theta\mid y}\log L(\theta)
=-\frac12\log(2\pi\sigma^2)
-\frac{(y-\widetilde\theta)^2+\sigma_\theta^2}{2\sigma^2}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 219 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-219/4/c/solution