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I am not really a professional, but this question has been asked on Math.SE already and in spite of a bounty it was not answered.

That made me decide to give it a try here, and I hope that is acceptable.

It is clear to me that I can build a suitable underlying probability space for a homogeneous Poisson point process.

It is enough to have a probability space $(\Omega,\mathcal A,P)$ with on it iid random variables $X_1,X_2,\dots$ having exponential distribution.

So I could do already with $\mathbb R^{\mathbb N}$ applied with product measure.

Then $N_t$ can be defined as the cardinality of the set $\{n\mid S_n\leq t\}$ where $S_n:=X_1+\dots+X_n$.

If $A$ is a measurable subset of $[0,\infty)$ then I can define random variable $\hat A$ as the (random) cardinality of $\{n\mid S_n\in A\}$ and then $\hat A$ has Poisson-distribution with a (multiple of) $\lambda(A)$ as parameter, where $\lambda$ denotes the Lebesgue measure. In that sense it can be called a Poisson process on base of the Lebesgue measure.

Now my question:

How to build up a probability space allowing me to construct a Poisson process based on an arbitrary chosen measure $\nu$ on $[0,\infty)$ with the property that $\nu([0,t])<\infty$ for each $t$?

So this means that for a measurable $A\subseteq[0,\infty)$ the random cardinality of $\{n\mid S_n\in A\}$ has Poisson-distribution with parameter $\nu(A)$. Further if two such sets $A,B$ are disjoint then $\{n\mid S_n\in A\}$ and $\{n\mid S_n\in B\}$ must be independent (as is also the case described above).

Thank you for your attention and (valuable) time in advance.

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    $\begingroup$ The simplest way is to consider $N_{m(t)}$, where $N_t$ is the usual rate-1 Poisson process and $m(t) = \nu([0, t])$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jun 16, 2018 at 15:12
  • $\begingroup$ @MateuszKwaśnicki That looks indeed promising and simple too! Thank you. I will have a closer look, and if I do not encounter any obstacles then I will ask you to turn your comment into an answer (that will be accepted). $\endgroup$
    – Vera
    Commented Jun 16, 2018 at 15:18
  • $\begingroup$ @MateuszKwaśnicki It works! Please be so kind to turn your comment into an answer, so that I can accept this answer. $\endgroup$
    – Vera
    Commented Jun 19, 2018 at 9:42
  • $\begingroup$ Kostya_I provides a much more general approach, why don't you accept that answer? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jun 19, 2018 at 12:36
  • $\begingroup$ @MateuszKwaśnicki Because I do not completely understand it. See my comment on that question. I don't understand how $N_t$ is defined on base of what is written in that answer. $\endgroup$
    – Vera
    Commented Jun 19, 2018 at 14:09

2 Answers 2

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Another construction, which does not use the structure of $\mathbb{R}$ and works for a sigma-finite measure $\nu$ on arbitrary measurable space $\Omega$, is as follows: let $\Omega=\bigcup_i E_i$ with $\nu (E_i)<\infty$ and $E_i\cap E_j=\emptyset$ for $i\neq j$. For each $i$ independently, sample a Poisson random variable $N_i$ with parameter $\nu(E_i)$, then for each $i$, sample $N_i$ random points in $E_i$ distributed according to $\nu|_{E_i}$, independently of each other and between $i$. Properties of Poisson random variables guarantee that the number of points in two disjoint sets are independent, and the average is given by $\nu$.

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  • $\begingroup$ Then how exactly is $N_t$ (as described in my question) defined here? (Or is $i$ ranging over $[0,\infty)$ maybe?) $\endgroup$
    – Vera
    Commented Jun 17, 2018 at 14:56
  • $\begingroup$ @Vera: $N_t$ is the number of points in $[0, t]$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jun 19, 2018 at 16:50
  • $\begingroup$ @MateuszKwaśnicki I think I understand now. $\endgroup$
    – Vera
    Commented Jun 19, 2018 at 18:29
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How about this: Let $N_t$ be the cardinality of $\{n: S_n\le \nu(0,t)\}.$

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  • $\begingroup$ Yes, that is a nice way of doing it. Thank you! $\endgroup$
    – Vera
    Commented Jan 1, 2022 at 16:12

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