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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?

gtoques
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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 $S=(t_1,t_2]$.

drhab
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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 $$ \mathbb{P}\bigl(N(s) = 1 + N(t_1)| N(t_2) = 1 + N(t_1)\bigr). $$ You are hoping the answer is $$ \frac{s-t_1}{t_2 - t_1} $$

SBK
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  • Are all arrival in $(t_{1}, t_{2}]$ uniformly distributed or is it just the first one? I'm not sure how to prove this. – gtoques Jun 03 '20 at 20:23
  • Sorry, I just edited it to say $N(t_2) \geq 1$ at the end. I don't know if that helps. Well I'm not sure if you just wanted a hint to get started or more of a proof. You just said how to "go about" proving it. You presumably know some formula for certain probabilities associated with $N$? – SBK Jun 03 '20 at 20:32
  • @gtogues Not just the first. When given that the arrival of some event $A_n$ is the only arrival that occurs in interval $(t_1..t_2]$, then the (conditional) probability distribution function for its time of arrival, $T_n$, is evaluated by:$$\mathsf P\big(T_n\leqslant t\mid N_{(t_1..t_2]}=1\cap T_n\in(t_1..t_2]\big)=\dfrac{\mathsf P(N_{(t_1..t]}=1\cap N_{(t..t_2]}=0)}{\mathsf P(N_{(t_1..t_2]}=1)}$$ Where for any time inteval $(u..v]$, the count for arrivals in that interval is Poisson distributed: $N_{(u..v]}\sim\mathcal{Poiss}(\lambda(v-u))$, for an unknown rate $\lambda$. – Graham Kemp Jun 04 '20 at 01:51
  • @gtoques If you know that there are exactly $n$ arrivals after $t_1$ and before or at $t_2$, then the density for any possible ordered $(a_1,a_2,\ldots, a_n)$ is a constant $\frac{n!}{(t_2-t_1)^n}$, but for $n>1$ the marginal density for the first $a_1$ is $\frac{n,(t_2-a_1)^{n-1}}{(t_2-t_1)^n}$ so it is more likely to be near the start than the end, as you might intuitively expect. – Henry Jun 26 '24 at 01:19