The measurement model has the <Gaussian likelihood>
$$
P(d\mid x)=\frac1{\sqrt{2\pi\sigma^2}}
\exp\!\left[-\frac{(d-x)^2}{2\sigma^2}\right].
$$
Marginalizing the latent angular momentum gives the normalized <posterior distribution>
$$
\boxed{P(m\mid d)=
\frac{\int P(d\mid x)P(x,m)\,dx}
{\iint P(d\mid x)P(x,m)\,dx\,dm}.}
$$
Its <posterior mean> is
$$
\boxed{\bar m=\frac{\iint mP(d\mid x)P(x,m)\,dx\,dm}
{\iint P(d\mid x)P(x,m)\,dx\,dm}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 219 / 2 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-219/2/a/solution