<Conditional independence> gives
$$
P(m\mid d_1,\ldots,d_S)
\propto P(m)\prod_{s=1}^SP(d_s\mid m),
\qquad S=N_{\rm sat}.
$$
For each satellite, <Bayes theorem> gives $P(d_s\mid m)\propto P(m\mid d_s)/P(m)$. Therefore
$$
P(m\mid\mathbf d)\propto
P(m)^{1-S}\prod_{s=1}^SP(m\mid d_s).
$$
If $\widehat p_0(m)$ is a <kernel density estimator> for the simulated marginal masses and $\widehat p_s(m)$ estimates the posterior based on satellite $s$, then
$$
\boxed{
\widehat{\bar m}=
\frac{\int m\,\widehat p_0(m)^{1-S}
\prod_{s=1}^S\widehat p_s(m)\,dm}
{\int \widehat p_0(m)^{1-S}
\prod_{s=1}^S\widehat p_s(m)\,dm}.}
$$
The one-dimensional integrals can be evaluated by <numerical integration> on a common mass grid.

Solution

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

 Solution

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