# Exponentially distributed events in time-partitions

The exponential distribution $p(t) = \lambda e^{-\lambda t}$ is a probability distribution that describes the time between events in a Poisson process. If $X$ is an exponential random variable, then write $X \sim \exp(\lambda)$.

Suppose I want to generate a random sample of exponentially-distributed events on an interval, say $[0,100]$. One approach is to initialise the starting time $t_0 = 0$, and create events at times \begin{align} t_1 &= t_0 + X_1 \\ t_2 &= t_1 + X_2 \\ t_3 &= t_2 + X_3 \\ \end{align} $$\vdots$$ where each $X_i \sim \exp(\lambda)$, until I reach an $n$ such that $t_n < 100$ and $t_{n+1} > 100$. Then my sample comprises the set $\lbrace t_1,\ldots,t_n\rbrace$.

However, what if I took another approach: I partition the interval into $p$ partitions, so that e.g. for $p = 10$, I would have the partitions \begin{align} \lbrace [0,10], (10,20], (20,30],\ldots,(90, 100]\rbrace. \end{align} If I performed the above algorithm on each partition separately, then took the union of all events, would my final sample come from the same distribution as the sample $\lbrace t_1,\ldots,t_n\rbrace$ obtained with the other method?

Intuitively I feel like it should, but I don't know how to prove it. The main issue that arises is that as intervals become shorter, more 'overshooting' happens, so that more events get rejected, and many intervals have few or no events. I don't know if this changes the final sample.

• The number of arrivals in the interval $[0,100]$ has a Poisson distribution with expected value $100/\lambda$. Given the number of arrivals, the conditional distribution of the times of arrival is exactly the same as the distribution of the order statistics of a random sample whose observations are independent and uniformly distributed in that interval. So one way to simulate is (1) generate one observation from the Poisson distribution, then (2) generate a sample of that size from the uniform distribution on the interval, then (3) sort them into increasing order. $\qquad$ – Michael Hardy Oct 11 '16 at 17:39