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

The measurement model has the Gaussian likelihoodMarginalizing the latent angular momentum gives the normalized posterior distributionIts posterior mean is

Treat the simulated pairs $(x_i,m_i)$ as samples from the prior $P(x,m)$. The numerator and denominator of the posterior mean are then ordinary Monte Carlo estimators, sowhere the normalized importance sampling weights areThe common Gaussian normalizing constant cancels.

For arbitrary nonnegative raw weights $W_i$, let $\bar{W}=K^{-1}{\sum }_i{W}_i$. The empirical squared coefficient of variation, using variance divisor $K$, isSubstitution into the stated definition gives the usual effective sample size of importance samplingFor normalized weights $w_i=W_i/{\sum }_j{W}_j$, this reduces to $\boxed{\widehat{\mathrm{ESS}}=1/\sum _i{w}_i^2}$.

Conditional independence givesFor each satellite, Bayes theorem gives $P(d_s\mid m)\propto P(m\mid d_s)/P(m)$. ThereforeIf ${\hat{p}}_0(m)$ is a kernel density estimator for the simulated marginal masses and ${\hat{p}}_s(m)$ estimates the posterior based on satellite $s$, thenThe one-dimensional integrals can be evaluated by numerical integration on a common mass grid.

Let $p(m)=P(m\mid \mathbf{d})$ and draw independently from an importance density $q$. The unbiased estimatorof $I=\int mp(m)dm$ has one-sample second momentBy the Cauchy-Schwarz inequality,Equality holds precisely when $q(m)\propto |m|p(m)$, giving the optimal importance density for a single integralThis is circular in practice: constructing and normalizing $q^{\ast }$ requires detailed knowledge of the posterior and the expectation of $|m|$. Here log masses are positive, so the unknown normalizer is the posterior mean being estimated. It is also optimal only for this one integral, not for general posterior summaries.