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.