Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 34 / 3 / c / Solution

Assume conditional independence of the Poisson distributions, and write $u_i={\mu }_{i1}$, $S_i={\sum }_k{e}^{{\eta }_{ik}}$. The correct Poisson mass function has $e^{-\mu }$; the positive sign in the PDF's reminder is erroneous. The group likelihood isMultiply by the shape–rate Gamma distribution prior $b^a{u}_i^{a-1}{e}^{-b{u}_i}/\Gamma (a)$ and integrate. The gamma integral givesIndependent baseline priors give the product over groups. For fixed $a,b$, the leading factors do not depend on $\beta$, proving the requested Gamma-integrated baseline Poisson likelihood.