Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 31 / 4 / b / Solution

Conditional on $\Theta =\theta$, summing the individual independent Poisson counts gives $Y_j\sim \mathrm{Poisson}(m_j\theta )$. The between-year conditional independence makes the likelihood functionMultiplying by the gamma distribution prior density, proportional to ${\theta }^{\alpha -1}{e}^{-\beta \theta }$, gives the Poisson-gamma conjugacy with unequal exposures:For an action $t$, the posterior squared-error loss decomposes asHence the Bayes estimator under squared error loss is the posterior mean, givingFuture counts are independent of the observed years conditional on $\Theta$, so the law of total expectation then gives the posterior predictive expected countThe Bayesian and credibility estimates coincide exactly. This exact Bühlmann–Straub credibility for Poisson-gamma counts occurs because the posterior mean is already affine in the exposure-weighted experience, and therefore belongs to the class over which the credibility estimate minimizes mean squared error.