# Non-homogeneous Poisson process with intensity function $\lambda\cdot f_{X_1}$

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!

• 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. – Dieter Kadelka Jun 8 '20 at 18:48
• @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 – Math is like Friday Jun 8 '20 at 20:17
• 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 – mike Jun 9 '20 at 6:52

$$\newcommand{\la}{\lambda}$$ Take any $$t_0,\dots,t_n$$ such that $$0=t_0<\dots. For each $$j\in[n]:=\{1,\dots,n\}$$, let $$I_j:=[t_{j-1},t_j)$$ and $$$$\nu_j:=\#\{i\in[Z]\colon X_i\in I_j\}.$$$$ Then, for each nonnegative integer $$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$$, where $$$$p_j:=P(X_1\in I_j).$$$$ So, for any nonnegative integers $$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}$$.
• 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? – Math is like Friday Jun 8 '20 at 21:38
• @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 $t\ge0$. – Iosif Pinelis Jun 8 '20 at 21:48
• 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. – Iosif Pinelis Jun 8 '20 at 21:48
• 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. – Iosif Pinelis Jun 8 '20 at 22:03