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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 $[0,\infty )$.

For a mixed Poisson distribution with known exposure $n>0$, multiply the unnormalized posterior density ${\theta }^s{e}^{-n\theta }\pi (\theta )$ by $\theta$ and compare its integral with the next probability mass. The factors $n^s/s!$ give the displayed identity whenever $p_n(s)>0$. Both mass probabilities have the same exposure $n$. For a future conditionally independent unit-exposure Poisson distribution count, this is also its posterior predictive expected value. It holds for arbitrary proper mixing distributions, with integrals taken against the prior measure when there is no density.