Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 43 / 4 / i / Solution

Let $s={\sum }_{j=1}^n{x}_j$. The conditional joint likelihood function of the observed counts isThe prior is a gamma distribution with shape two and rate $\lambda$. Multiplication by its density gives the unnormalized posterior density ${\theta }^{s+1}{e}^{-(n+\lambda )\theta }$. Normalizing with the gamma function yieldsThus Poisson-gamma conjugacy gives $\Theta \mid \mathbf{x}\sim \mathrm{Gamma}(s+2,n+\lambda )$. Conditional on $\Theta$ and the past data, the next count has mean $\Theta$, so the law of total expectation givesWith $\overline{x}=s/n$ and prior mean $m_0=2/\lambda$, this is exactlyHere $n\ge 1$, $\lambda >0$, and the credibility factor increases to one as the number of observed years grows. The denominator uses the prior rate $\lambda$ and total exposure $n$; it does not add two to the exposure when adding the prior shape.