Geometric count under Bernoulli thinning ($p_{\alpha }=p/(p+(1-p)\alpha )$)

Independently retain each event of a zero-based geometric distribution count with probability $\alpha$. Conditional on the original count, the retained count is binomial, so its probability generating function is $p/[1-(1-p)(1-\alpha +\alpha z)]$. This is again a zero-based geometric distribution, with success parameter $p_{\alpha }$ and mean $\alpha (1-p)/p$. In excess of loss reinsurance, $\alpha$ is the probability that a claim exceeds the retention.

Let $\mathbb{P}(X=n)=p(1-p{)}^n$ for $n\ge 0$ and $Y\mid X=n\sim \mathrm{Binomial}(n,q)$, with $0<p,q<1$. Write $D=p+q-pq$. Summing the joint probability mass function with the differentiated geometric series identity ${\sum }_{n=k}^{\infty }\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^{n-k}=(1-t{)}^{-k-1}$ gives $\mathbb{P}(Y=k)=(p/D)((1-p)q/D{)}^k$. By Bayes' theorem,Thus $X-k$, given $Y=k$, has the failure-count negative binomial distribution for $k+1$ successes and success parameter $D$. Its conditional expectation is $k+(k+1)(1-D)/D$ for $X$ itself. The shift by $k$, the number of retained successes, must be retained when identifying the posterior distribution.