Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 207 / 1

The Directed acyclic graph has arrowsThe roots $\lambda$ and ${\rho }_G$ are independent under their prior distributions; $N$ is fixed. Equivalently, the joint distribution factorizes as

The mean condition gives $a/(a+b)=0.1$, hence $b=9a$. The Beta distribution variance is thenSince $0.02985^2=9/10100$, equating variances gives $10a+1=101$. Therefore

For each person, infection occurs with probability $\lambda$ and, conditional on infection, a GP visit occurs with probability ${\rho }_G$. Independent Bernoulli thinning therefore gives visit probability $\lambda {\rho }_G$. More explicitly, the probability generating function isThusand the likelihood is

By Bayes theorem, with respect to Lebesgue measure on $(0,1{)}^2$,The omitted constant includes the binomial coefficient and the normalizing constant of the Beta distribution; the uniform prior contributes only the support indicator.

Use the deterministic evidence-synthesis relationwith supportA uniform prior on this triangle has density $21_{\mathcal{T}}$. Conditional on the prevalences, model the independent surveys bywith the two counts conditionally independent and ${\pi }_2={\pi }_1+\lambda$.

Integrating the constant density $2$ across horizontal slices of the triangular support givesso ${\pi }_1\sim \mathrm{Beta}(1,2)$. For $z={\pi }_2={\pi }_1+\lambda$, the line segment at fixed $z$ has length $z$, and the transformation has unit Jacobian. Henceso

Extend the original Directed acyclic graph withThe existing arrows $\lambda \to I\to Y_G$ and ${\rho }_G\to Y_G$ remain. The survey sample sizes and $N$ are fixed design variables, while ${\pi }_2$ is a deterministic node satisfying ${\pi }_2={\pi }_1+\lambda$.

For parameters $({\rho }_G,{\pi }_1,\lambda )$ on $0<{\rho }_G<1$ and $({\pi }_1,\lambda )\in \mathcal{T}$, conditional independence givesThis is the joint posterior up to its normalizing constant.