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?
 A: For any bounded set $S\subseteq[0,\infty)$ let $N_S$ denote the number of arrivals in set $S$.
If $T$ denotes an arrival in $(t_1,t_2]$ then for $t\in(t_1,t_2]$ we find:
$$\begin{aligned}P\left(T\leq t\mid N\left(t_{1},t_{2}\right)=1\right) & =\frac{P\left(N_{\left(t_{1},t\right]}=1,N_{\left(t,,t_{2}\right]}=0\right)}{P\left(N_{\left(t_{1},t_{2}\right]}=1\right)}\\
 & =\frac{P\left(N_{\left(t_{1},t\right]}=1\right)P\left(N_{\left(t,,t_{2}\right]}=0\right)}{P\left(N_{\left(t_{1},t_{2}\right]}=1\right)}\\
 & =\frac{e^{-\lambda\left(t-t_{1}\right)}\lambda\left(t-t_{1}\right)e^{-\lambda\left(t_{2}-t_{1}\right)}}{e^{-\lambda\left(t_{2}-t_{1}\right)}\lambda\left(t_{2}-t_{1}\right)}\\
 & =\frac{t-t_{1}}{t_{2}-t_{1}}
\end{aligned}
$$
So under the condition that there is only one arrival in $(t_1,t_2]$ (i.e. $T$ is unique as arrival) $T$ will have uniform distribution on $(t_1,t_2]$.

More generally it can be proved for suitable sets $S$ that under the condition that $N_{S}=k$ the set of arrivals has the same distribution as a sample of $k$ iid random variables that have uniform distribution on $S$.
What was handled above is then the special case $k=1$ and $S=(t_1,t_2]$.
A: 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 $N(t_2) \geq 1$''.
