Poisson Arrival Process and Uniform Distribution I'm brushing up on some basic probability and have this question: 
If we have a Poisson arrival process with arrivals $A_{1}, A_{2}, \dots$, and we know that there is one and only one arrival in a time period, say $[t_{1}, t_{2}]$. Does this mean that the one arrival is distributed uniformly on $[t_{1},t_{2}]$? How would one go about "proving" (it might be trivial, but not sure) such a thing?

Let's call your process $N(t)$. i.e. $N(t)$ is the number of arrivals that have happened up to and including time $t$. And given an interval $(a,b]$, let $N((a,b])$ denote the number of arrivals in the interval $(a,b]$.
The way you would go about proving it is to fix some number $s \in (t_1,t_2]$ and try to calculate
$$
\mathbb{P}\bigl(N(t) < s |\ \text{first arrival is in}\ (t_1,t_2]\bigr).
$$
You are hoping the answer is
$$
\frac{s-t_1}{t_2 - t_1}
$$
The event ''first arrival is in $(t_1,t_2]$'' can be thought of as ''$N(t_1) = 0$ and $N(t_2) \geq 1$''.
