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In an M-M-1 queue with Bernoulli feedback, external arrivals form a Poisson process of rate $\lambda$, service times are independent exponential random variables of rate $\mu$, and each completed service independently returns its customer with probability $p\in [0,1)$. A returned customer immediately rejoins the same queue and requires a fresh service time. The queue population does not change at a feedback completion. Only completions that leave permanently produce downward jumps, at rate $\mu (1-p)$ while the queue is nonempty.

For a continuously busy server, independent Bernoulli thinning of its Poisson completion clock gives an exponential time of rate $\mu (1-p)$ until the next permanent departure. Thus the queue population has exactly the same birth-death process as an M/M/1 queue with service rate $\mu (1-p)$. It has a Stationary distribution precisely when $\lambda <\mu (1-p)$; its geometric parameter is then $\rho =\lambda /[\mu (1-p)]$. This reduction describes the population and external departures, while individual customers can undergo several visits. For example, $p=1/2$ doubles the mean number of services needed per external customer from one to two.

For excess of loss reinsurance with retention $M$, the direct insurer caps its payment at $M$ and the reinsurer pays the excess. ThusLet $I_i=1_{\{X_i>M\}}$ and $\alpha =\mathbb{P}(X_i>M)=1-F_{X_1}(M)\in (0,1)$. These are independent Bernoulli random variables, independent of $N$. Since $N_R={\sum }_{i=1}^N{I}_i$, its conditional law given $N=n$ is a binomial distribution. Conditioning on $N$ therefore gives the Bernoulli thinning identityHere the probability generating function may be evaluated on $0\le z\le 1$, and the identity extends wherever the transforms converge.

The positive reinsurer marks have the conditional law $W\stackrel{d}{=}{X}_1-M\mid X_1>M$. HenceThe independent-mark representation is justified as follows. Conditional on any fixed count and pattern of exceedance indicators, the positive excesses are independent with this same conditional law; that law depends neither on the number of exceedances nor on their locations. Averaging over the indicator pattern thus leaves the positive mark sequence independent of $N_R$. We may extend it by additional independent copies if needed, giving $S_R={\sum }_{j=1}^{N_R}{W}_j$ with the stated independence.

Now write $q=1-p$. For exponential distribution claim sizes of mean $\mu$,Thus $W$ is exponential with the original mean $\mu$, by memorylessness of the exponential distribution. The original zero-based geometric distribution count has $G_N(z)=p/(1-qz)$. ConsequentlyThis is the geometric count under Bernoulli thinning, withFinally use the Laplace transform of a nonnegative random variable, keeping the mass at zero. For $t\ge 0$,The last expression is the transform of a mixture distribution: mass $p_R$ at zero and, with probability $q_R$, an exponential distribution of rate $p_R/\mu$. Uniqueness of the Laplace transform identifies this zero-based geometric sum of exponential variables:Equivalently, $\mathbb{P}(S_R=0)=p_R$ and its positive density is $q_R(p_R/\mu )e^{-p_{R}s/\mu }$. Therefore the requested exceedance probability isThe prefactor is essential: the aggregate is zero whenever no claim crosses the retention.

For each person, infection occurs with probability $\lambda$ and, conditional on infection, a GP visit occurs with probability ${\rho }_G$. Independent Bernoulli thinning therefore gives visit probability $\lambda {\rho }_G$. More explicitly, the probability generating function isThusand the likelihood is