Conditional on $N(t)$, given some $\lambda(t)$ characterizing some Nonhomogenous poisson point process, the distribution of an arrival time $t_i$ is $\lambda(t_i)/\int_{A}\lambda\left(t\right)dt$ where A is the space over which the point process is defined.

Can anyone explain the derivation here?

Thank you

  • $\begingroup$ what's that $\beta$? And what do you mean by distribution? $\endgroup$ – Taylor Jul 13 '16 at 4:51
  • $\begingroup$ I removed beta, it's unnecessary. Re distribution: In a regular (homogenous poisson process), conditional on the number of arrivals $N(t)$ , the distribution of the arrival times themselves are uniformly distruted over $[0,t]$. In a nonhomogenous poisson process, they are distributed as above. Why? $\endgroup$ – wolfsatthedoor Jul 13 '16 at 15:48
  • $\begingroup$ Do you want a heuristic / intuitive or mathematical explanation? $\endgroup$ – jbowman Jul 13 '16 at 18:26
  • $\begingroup$ Both would be ideal. I have vague intuition already $\endgroup$ – wolfsatthedoor Jul 13 '16 at 19:08

Here's the situation for the first arrival time. Let $T_1$ be the first arrival time. Sorry for making this capitalized...I like random variables to be capital letters. Pick any $t_1 < t$, where $t$ is the end time.

Also let $\Lambda(t') = \int_0^{t'}\lambda(s) ds$, $p = \Lambda(t_1) / \Lambda(t)$, and $r = s-1$.

\begin{align*} P(T_1 \le t_1 | N(t) = n) &= \sum_{s=1}^n\frac{P(N(t_1) = s, N(t) =n)}{P(N(t) = n)}\\ &= \sum_{s=1}^n\frac{P(N(t_1) = s)P( N(t)-N(t_1) =n - s)}{P(N(t) = n)}\\ &= \sum_{s=1}^n \frac{e^{-\Lambda(t_1)}\Lambda(t_1)^s e^{-[\Lambda(t)-\Lambda(t_1)] } [\Lambda(t)-\Lambda(t_1)]^{n-s} n! }{s!(n-s)! e^{-\Lambda(t)}\Lambda(t)^n } \\ &= \sum_{s=1}^n {n \choose s} \frac{\Lambda(t_1)^s [\Lambda(t)-\Lambda(t_1)]^{n-s} }{ \Lambda(t)^n }\\ &= \sum_{s=1}^n {n \choose s} p^s (1-p)^{n-s} \\ &= \frac{np}{(r+1)} \sum_{r=0}^{n-1} {n-1 \choose r} p^r (1-p)^{n-1-r} \\ &= \frac{np}{(r+1)} \end{align*}

You're looking for $f_{T_1|N(t)=n}(t_1)$, and that's $\frac{d}{dt_1}\frac{np}{(r+1))}.$ When I take the derivative I get


which is sort of like your answer. In the special case that $n=1$, then $s$ would have to be $1$, and thus $r$ would be $0$. In that case we would get the answer you mentioned in the question.

As far as intuition goes, there's higher density in places where the rate function $\lambda$ is higher. And the denominator is just a normalizing constant.


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