Let's say I find that in an interval of 5 ns, with event rate 50000/s, that the probability of an event occurring twice in that interval is $$P_2 = {e^{-\lambda} \lambda^{2} \over 2!} = 3.124 \cdot 10^{-8}$$

I believe it is true that if I perform $2 \cdot 10^8 $ separate 5 ns trials, I should expect $P_2 *2 \cdot 10^8 = 6.25$ trials on average to have two events. But instead if I run a trial for 1 second, how many times on average should I get 2 events within 5 ns?

When I run a simulation, I get almost exactly twice the above at 12.5 $\pm$ 0.1. What is the correct method of determining the number of times an event occurs with seperation dt or less in time interval T?


1 Answer 1


The gaps between events are exponentially distributed with rate $50000/s$, so the probability a particular gap is less than $5ns$ is $1-e^{-50000\times 5\times 10^{-9}}\approx 0.00024997$

With about $50000$ gaps in a second, you would then expect about $50000 \times 0.00024997 \approx 12.498$ of them to be shorter than $5ns$

This approach gives a good approximation when the number of gaps expected is large, but it is not so good when the expected number of gaps is small. One reason is that in the large number of gaps case, you get to consider the gap after the second event; another is that in the small number of gaps case seeing a big gap can reduce the total number of other gaps


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