= Two-sided interarrival test for overlap
{title2=$\mathbb P(\text{overlap})=1-e^{-2\lambda d}$}
Consider a homogeneous <Poisson process> of arrival rate $\lambda$, with each arrival staying for a fixed duration $d>0$. For any fixed arrival index $k\geq2$, its visit overlaps another visit if the preceding interarrival time is below $d$ or the following interarrival time is below $d$. If the nearest preceding arrival has already left then all earlier ones have left; if the next arrival is too late then all later ones are too late. These two gaps are independent <exponential random variables>, so the probability of no overlap is $e^{-\lambda d}e^{-\lambda d}$. Taking its complement gives the formula. This is an unconditional statement for a fixed indexed arrival; conditioning on a fixed calendar closing time would change the problem.
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