 Solution

Given $\Theta =\theta$, the sum of the independent yearly counts is $S_n\sim \mathrm{Poisson}(n\theta )$: their probability generating functions multiply to $\exp (n\theta (z-1))$. Averaging this conditional law over the prior gives the mixed Poisson distributionThe likelihood for the full vector depends on $\theta$ only through $s$, so $S_n$ is a sufficient statistic; conditioning on the total or on the full observed vector gives the same posterior for $\Theta$.

For an event of positive probability $p_n(s)>0$, the posterior density is proportional to $e^{-n\theta }{\theta }^s\pi (\theta )$. Put $I_s={\int }_0^{\infty }{e}^{-n\theta }{\theta }^s\pi (\theta )d\theta$. Its mean is $I_{s+1}/I_s$. The future count has conditional mean $\Theta$, soThis is the posterior mean from adjacent mixed Poisson probabilities. Both probabilities refer to the same $n$-year exposure; the numerator is not a probability for $S_{n+1}$. For a proper prior and $n>0$, the integrals defining this posterior mean are finite because ${\theta }^{s+1}{e}^{-n\theta }$ is bounded on $[0,\infty )$.