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Stop new arrivals at time $t$ in an initially empty general-service infinite-server queue. The number remaining $v$ time units later is Poisson with mean $\lambda {\int }_v^{t+v}\mathbb{P}(S>u)du$. Thus the probability that clearance has not occurred is one minus the exponential of the negative of that mean. Letting $t\to \infty$ gives the displayed equilibrium formula, since ${\int }_v^{\infty }\mathbb{P}(S>u)du=\mathbb{E}(S-v{)}_{+}$. The clearance time is zero when no customers remain at closing.

The model is a general-service infinite-server queue, usually denoted $M/G/\infty$: arrivals form a Poisson process, each wandering time is a service time with distribution that of $S$, and trapping ends service. Infinitely many servers mean that mosquitoes wander concurrently without waiting or delaying one another. General service times are not necessarily exponential, so the count alone need not be a Markov chain.