I want to show that there is a non-homogeneous Poisson process with a certain intensity function, but I have some problems while showing that this Poisson process satisfies the axioms(?). I am using the axioms as follows:

  • $N(0) = 0$
  • if $s\leq t$, then $N(s)\leq N(t)$
  • etc

as usual. Now the problem is given by:

Let $X_1,X_2,\dots$ be i.i.d. continuous random variables with common probability density function $f_{X_1}$ and let $Z \xrightarrow{d}Po(\lambda)$ be independent of $X_1,X_2,\dots$. Now define the random point set $\mathcal{P} = \{X_1,\dots,X_Z\}$ ($\mathcal{P}$ = $\emptyset$ if $Z = 0$).

Show that $\mathcal{P}$ is an inhomogeneous Poisson point process with intensity function $\lambda\cdot f_{X_1}$.

Sorry if my question is not clear enough, and thank you!

  • $\begingroup$ Indeed, to me the problem is not clear. What do you want: a Poisson point process or a Poisson process? Please try to clarify the assumptions. $\endgroup$ – Dieter Kadelka Jun 8 '20 at 18:48
  • $\begingroup$ @DieterKadelka If I am right, either Poisson point process or Poisson process means the same. If not, I was just meaning a normal Poisson process which is defined by the set of random points $\endgroup$ – Math is like Friday Jun 8 '20 at 20:17
  • $\begingroup$ I see you are done with this, but I would like to point out that in case the $X_i$ are uniform on (0,1) , this is the well know connection between the times of the poisson process and the order statistics of uniforms $\endgroup$ – mike Jun 9 '20 at 6:52

$\newcommand{\la}{\lambda}$ Take any $t_0,\dots,t_n$ such that $0=t_0<\dots<t_n=\infty$. For each $j\in[n]:=\{1,\dots,n\}$, let $I_j:=[t_{j-1},t_j)$ and \begin{equation} \nu_j:=\#\{i\in[Z]\colon X_i\in I_j\}. \end{equation} Then, for each nonnegative integer $z$, the joint conditional distribution of $(\nu_1,\dots,\nu_n)$ given $Z=z$ is the multinomial distribution with parameters $z,p_1,\dots,p_n$, where \begin{equation} p_j:=P(X_1\in I_j). \end{equation} So, for any nonnegative integers $k_1,\dots,k_n$ \begin{align} P(\nu_1=k_1,\dots,\nu_n=k_n) &=\sum_{z=0}^\infty P(Z=z)P(\nu_1=k_1,\dots,\nu_n=k_n|Z=z) \\ &=\sum_{z=0}^\infty \frac{\la^z e^{-\la}}{z!} \frac{z!}{k_1!\cdots k_n!}\, p_1^{k_1}\cdots p_n^{k_n}\,1\{k_1+\dots+k_n=z\} \\ &=\prod_{j\in[n]} \frac{(\la p_j)^{k_j}e^{-\la p_j}}{k_j!}. \end{align} So, $\nu_1,\dots,\nu_n$ are independent random variables and $\nu_j\sim Poisson(\la p_j)$ for each $j\in[n]$.

This means that indeed $\mathcal{P}$ is an inhomogeneous Poisson point process with intensity function $\la f_{X_1}$.

  • $\begingroup$ I am curious because there was a comment that pointed out the difference between Poisson process and Poisson point process and you said that $\mathcal{P}$ is a Poisson point process. Are they actually different or is it just a more precise name of it? $\endgroup$ – Math is like Friday Jun 8 '20 at 21:38
  • 1
    $\begingroup$ @MathislikeFriday : The Poisson point process (say on the real line) is a random integer-valued measure, which may be identified with the random set of points, whose realizations are the support sets of the corresponding realizations of the random measure; this is what your random set $\mathcal P$ is. Also, the random integer-valued measure, say $\mu$, that is the Poisson point process may be identified with its counting function $N_\cdot$ defined by $N_t:=\mu([0,t])$ for $t\ge0$. $\endgroup$ – Iosif Pinelis Jun 8 '20 at 21:48
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    $\begingroup$ Previous comment continued: The counting function $N_\cdot$ may also be considered as the stochastic process $(N_t)$, which is the Poisson process corresponding the Poisson point process. $\endgroup$ – Iosif Pinelis Jun 8 '20 at 21:48
  • $\begingroup$ Thank you for very detailed answer! It's clear now $\endgroup$ – Math is like Friday Jun 8 '20 at 21:51
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    $\begingroup$ Given $Z=z$, we have $z$ independent trials, each trial with $n$ possible outcomes. The probabilities of the outcomes in each trial are $p_1=P(X_1\in I_1),\dots,p_n=P(X_1\in I_n)$. Here instead of $X_1$ you can write $X_i$ with any natural $i$, since the $X_i$'s are iid. $\endgroup$ – Iosif Pinelis Jun 8 '20 at 22:03

Basically a Poisson point process can be defined on the real line by considering the number of points of the process in the interval (a.b], which is \lambda*(F(b)-F(a)) (\X_i s are iid), then use the definition of the intensity…


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