# Smallest gap between $n$ Poisson events

Given a Poisson process with parameter $$\lambda,$$ run time forward until the first $$n>1$$ events occur. What is the distribution of the smallest time interval between two events?

This is surely standard (and probably related to Erlang) but I can't find anything, can anyone help?

I'm especially interested in addressing questions of the flavor, "I have $$n$$ events in $$t$$ time and no reason to think anything funny is going on, except that two events are as close as $$y$$. How unlikely would this be in a Poisson process with $$\lambda = t/n$$?". Suggestions, comments, objections, etc. to this methodology would be appreciated (in a comment, unless you're also answering the question).

• I think you may be looking for the exponential distribution: en.m.wikipedia.org/wiki/Exponential_distribution – KirkD_CO Sep 1 at 2:45
• Minimum order statistics of a sample of exponential random variables? – kjetil b halvorsen Sep 1 at 11:27
• Let $T_i$ be the interarrival times, are you interested in finding the distribution of $T=\min (T_1,...,T_n)$ or $T|\sum_{i=1}^n T_i < t$? – gunes Sep 1 at 12:19

As I mention here, the interarrival times $$T_1, T_2, \ldots$$ are distributed as $$\text{Exponential}(\lambda)$$ random variables. You seem to be interested in $$T=\min (T_1,...,T_n)$$ somehow, but as @gunes points out, you aren't clear about whether or not you're assuming that the total amount of time to get $$n$$ events must be less than $$t$$: "I have $$n$$ events in $$t$$ time." You need to tell us whether $$t$$ is random or not.

If it is, ot's more conventional to use uppercase letters, however capital $$T$$ is already taken to mean the minimum of all the interarrival times. The distribution of the minimum can easily be found using the following technique. Here is another link addressing the same issue.

If you are constraining the total amount of time, and conditioning on the fact that it's less than $$t$$, you are looking for the distribution of $$T|\sum_{i=1}^n T_i < t$$ as @gunes points out. The quickest way for me to find the cdf of this distribution is to use the Monte Carlo technique. To do this simulate $$M$$ sets of $$n$$ random variables $$T_i^j$$, and compute

\begin{align*} P\left(T \le s \bigg\rvert \sum_{i=1}^n T_i < t\right) &= P\left(T \le s , \sum_{i=1}^n T_i < t\right)\bigg/ P\left(\sum_{i=1}^n T_i < t\right)\\ &\overset{\text{p}}{\leftarrow} \left(M^{-1}\sum_{j=1}^M 1\left\{T^j \le s , \sum_{i=1}^n T^j_i < t \right\}\right) \\ &\hspace{10mm} \bigg/ \left(M^{-1}\sum_{j=1}^M 1\left\{ \sum_{i=1}^n T^j_i < t .\right\}\right)\end{align*}

Here's some R code:

lambda <- 3
s <- .03
t <- 20
M <- 1000
n <- 20
samps <- matrix(rexp(M*n, lambda), nrow=M)
sum(apply(samps, 1, function(row) (min(row) < s) & (sum(row)< t)  )) / sum(apply(samps, 1, function(row) (sum(row)< t)))


The Erlang distribution arises as the sum of iid exponential random variables. Instead of estimating $$P\left(\sum_{i=1}^n T_i < t\right)$$ with samples, you could swap in the known cdf function here, and it would reduce the variance of the above Monte Carlo estimator.