The likelihood is
$$
p(d\mid x)=\frac1{\sqrt{2\pi\sigma^2}}
\exp\left[-\frac{(d-x)^2}{2\sigma^2}\right].
$$
Using the conditional independence $d\perp m\mid x$ and the joint prior $p(x,m)$,
$$
p(m\mid d)=
\frac{\int p(d\mid x)p(x,m)\,dx}
{\iint p(d\mid x)p(x,m)\,dx\,dm},
$$
and
$$
\mathbb E[m\mid d]=
\frac{\iint m,p(d\mid x)p(x,m)\,dx\,dm}
{\iint p(d\mid x)p(x,m)\,dx\,dm}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 219 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-219/2/a/solution