 Geometric count under Bernoulli thinning

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.