My construction is as follows: Let $X\_k$ be a real-valued continuous random variable centered at $k$ (an integer), having distribution $F\_k(x,s)$ where $k$ is the location parameter and $s$, a strictly positive real number, is the scale (fixed, not depending on $k$). Thus $s$ is typically a monotonic increasing function of the variance of $X\_k$. We assume that the support domain of $F\_k$ is the set of all real numbers $x$, and that the $X\_k$'s are independently distributed, with $k \in \mathbb{Z}$ . The points of my point process are the $X\_k$'s, and this is how my point process is defined.


Obviously if the support domain of $F\_k$ is compact, say equal to $[k-s, k+s]$, I end up with a non-Poisson point process. But let's focus on processes meeting the criteria mentioned above. In addition, add these constraints: if $s=0$ then $X\_k=k$ (the variance is zero). If $s\rightarrow\infty$ then the variance of $X\_k$ tends to infinity.


What is weird, and counter-intuitive to me, is that the resulting point process, however small or large $s$ is, is always stationary Poisson with intensity equal to $1$. I will illustrate this in the case when $F\_k$ is the logistic distribution as computations are somewhat simple, but if you look at my computations in the example section, it seems that it would also be true whether $F\_k$ has a Cauchy, Gaussian, Laplace, or other symmetric standard continuous distribution on the full real line.


**Question**:


Can you confirm (by checking my computations in the example below) whether this is true or not? Are there any exceptions? What happens if $F\_k$ is not a symmetric distribution centered at $k$? Or is there something wrong in my computations?


Generalization to two dimensions, with $k$ replaced by $(k,l)\in \mathbb{Z}^2$ and $F\_{k,l}$ being the bivariate distribution attached to $X\_{k,l}$, would be nice to discuss, but this is not part of my question. In particular, what if the joint distribution $F\_{k,l}(x,y)$ is not equal to the product of its marginals, and the parameter $s$ is replaced by a covariance matrix? Also, by increasing the granularity of the underlying lattice by a factor $\lambda$, the resulting process is expected to be Poisson with intensity $\lambda$, but this is not part of my question either.


**Example**:


Let $B=[a, b]$ with $a<b$ be an interval on the real line, and $N(B)$ be the random variable that denotes the number of points from the point process defined above, that are in $B$. Let $F\_k(x)$ be the logistic distribution (CDF) defined by


$$F\_k(x) = \frac{1}{2}+\frac{1}{2}\tanh \Big(\frac{x-k}{2s}\Big).$$


Let $p\_k=p\_k(B)$ be the probability that $X\_k \in B$, with $B=[a,b]$. That is,


$$p\_k=\frac{1}{2}\Big[\tanh \Big(\frac{b-k}{2s}\Big)-\tanh \Big(\frac{a-k}{2s}\Big)\Big].$$


The expected number of points in $B$ is
$$E[N(B)]=\sum\_{k=-\infty}^{\infty} p\_k = b-a.$$


I used Mathematica to compute the value of the above infinite series, and surprisingly, it does not depend on $s$. Of course it is supposed, based on intuition, to be proportional to $b-a$ but the proportionality factor is always $1$. Also, one would guess that the number of points in non-overlapping Borel sets, to be independent. Then I decided to compute the distribution of $N(B)$, with $B=[a, b]$. Let $q\_n=P(N(B)=n)$, for $n=0,1,2$ and so on.


$$q\_0 = \prod\_{k=-\infty}^\infty (1-p\_k)$$


$$q\_1 = q\_0 \prod\_{k=-\infty}^\infty \frac{p\_k}{1-p\_k},$$


*Note: the remaining of this section is wrong. I decided to leave it as is so that readers can relate to the comments / answer provided by other authors. The fixed version can be found in the last section called update, added recently at the bottom of my question.*


$$\begin{align}
q\_2 = &\frac{q\_0}{2!} \prod\_{i\neq j} \frac{p\_i p\_j}{(1-p\_i)(1-p\_j)} \\
 = &
\frac{q\_0}{2!}\Big[\Big(\prod\_i \frac{p\_i}{1-p\_i}\Big)\Big(\prod\_j\frac{p\_j}{1-p\_j}\Big) (1+o(1))\Big]\\
= & \frac{q\_0}{2!}\cdot\Big(\frac{q\_1}{q\_0}\Big)^2,
\end{align}$$


$$\begin{align}
q\_3 = & \frac{q\_0}{3!} \prod\_{i\neq j\neq l} \frac{p\_i p\_j p\_l}{(1-p\_i)(1-p\_j)(1-p\_l)}\\
 = & 
\frac{q\_0}{3!}\Big[\Big(\prod\_i \frac{p\_i}{1-p\_i}\Big)\Big(\prod\_j\frac{p\_j}{1-p\_j}\Big) \Big(\prod\_l\frac{p\_l}{1-p\_l}\Big)(1+o(1))\Big]\\ 
 = & \frac{q\_0}{3!}\cdot\Big(\frac{q\_1}{q\_0}\Big)^3,
\end{align}$$
and continuing iteratively, one finds that


$$q\_n=\frac{q\_0}{n!}\cdot\Big(\frac{q\_1}{q\_0}\Big)^n.$$


So we are dealing with a Poisson distribution, and we showed that its expectation is $b-a = \mu(B)$. With the assumed independence assumptions between disjoint Borel sets, we meet all the criteria to conclude that we are dealing with a Poisson process of intensity $1$. I used the notation $o(1)$ assuming the indices in the products for $q\_2$ and $q\_3$ were varying from $-k$ to $+k$, with $k\rightarrow\infty$.


**Background**:


Initially, I wanted to study the behavior of series such as $\zeta(z)=\sum\_{k=1}^\infty k^{-z}$ when replacing the index $k$ by $X\_k$, where $(X\_k)$ are the points of a stochastic process as defined in the introduction. The idea was to have a distribution $F\_k$ attached to $X\_k$, centered at $k$, and such that when $s\rightarrow 0$, $X\_k \rightarrow k$, as for the logistic distribution. In other words, replacing $\zeta(z)$ by a random function, with a limiting case being $\zeta(z)$ itself, and studying the properties, especially as $s\rightarrow 0$.


For this to work, I needed a point Process that is non-Poisson regardless of $s$ of course. So far, I haven't found a solution yet since I ended up with pure Poisson processes. While I failed, I've found it interesting enough to further explore these original constructions of Poisson processes, discovered by accident.


**Update**


Regardless of the distribution $F\_k$, the distribution of $N(B)$ is not Poisson as claimed, but rather Poisson-Binomial of parameters $p\_k$, with $k\in \mathbb{Z}$ and $p\_k=P(X\_k\in B)$. The only difference with a standard Poisson-binomial distribution (see [here](https://en.wikipedia.org/wiki/Poisson_binomial_distribution)) is that in my case, the number of parameters is infinite. In particular, my values for $E[N(B)], q\_0, q\_1$ are correct but those for $q\_n$ with $n>1$ are wrong. Also,


$$\mbox{Var}[N(B)]= \sum\_{k=-\infty}^\infty (1-p\_k)p\_k.$$


I'll check if I can get a closed form when $F\_k$ is the logistic distribution.


An example when Poisson-binomial becomes Poisson at the limit, is as follows. Pick up randomly an integer between $0$ and $n-1$, another one between $0$ and $n$, then another one between $0$ and $n+1$, and so on, and stop after picking a last one between $0$ and $\lambda n$ (here $\lambda > 1$). This problem arises in a Sieve-like algorithm, where the numbers you pick up are residues of a number massively larger than $\lambda n$, modulo $n, n+1, n+2$ and so on. When $n\rightarrow\infty$, the probability that exactly $k$ of the numbers selected are zero, is $(\log \lambda)^k \cdot \lambda^{-1}/k!$. Does this mean that the chance that a number much larger than $\lambda n$ has exactly $k$ divisors between $n$ and $\lambda n$ is $(\log \lambda)^k \cdot \lambda^{-1}/k!$, as $n\rightarrow\infty$, with the average number of divisors in that interval being $\log\lambda$?



Does my construction always result in a stationary Poisson point process of intensity $1$? How so?
This is not true. Indeed, suppose that $X\_k=X\_{s;k}=k+sZ\_k$, where $s\downarrow0$ and the $Z\_k$'s are **any** iid random variables (r.v.'s).


To obtain a contradiction, suppose that, for the random Borel measure $\mu\_s$ over $\mathbb R$ defined by $\mu\_s(B):=\sum\_{k\in\mathbb Z}1(X\_{s;k}\in B)$, the distribution of the random variable (r.v.) $\mu\_s(B)$ is Poisson with parameter $\lambda(s)|B|$ for some $\lambda(s)>0$ and all Borel $B$, where $|B|$ is the Lebesgue measure of $B$.


Note that
\begin{equation}
 \mu\_s((-1/2,1/2))\to1 \tag{1}
\end{equation}
in probability (see details on (1) below). Therefore and because the r.v. $\mu\_s((-1/2,1/2))$ has the Poisson distribution with parameter $\lambda(s)$,
necessarily $\lambda(s)\to0$ and hence $\mu\_s((-1/2,3/2))\to1$ in probability. However, similarly to (1) we have $\mu\_s((-1/2,3/2))\to2$ in probability, a contradiction.


So, the random measure $\mu\_s$ cannot be Poisson for all $s>0$.




---


*Proof of (1):* Note that $\mu\_s(B)=\sum\_{k\in\mathbb Z}1(Z\_k\in\frac{B-k}s)$ and hence
\begin{equation}
 1-\mu\_s((-1/2,1/2))=s\_1-s\_2,
\end{equation}
where
\begin{equation}
 s\_1:=1-1\Big(Z\_0\in\Big(\frac{-1/2}s,\frac{1/2}s\Big)\Big),
\end{equation}
and
\begin{equation}
 s\_2:=\sum\_{k\in\mathbb Z\setminus\{0\}}1\Big(Z\_k\in\Big(\frac{-1/2-k}s,\frac{1/2-k}s\Big)\Big). 
\end{equation}  

Next,
\begin{equation}
 Es\_1=1-P\Big(Z\_0\in\Big(\frac{-1/2}s,\frac{1/2}s\Big)\Big)\to0
\end{equation}
and
\begin{equation}
 Es\_2=\sum\_{k\in\mathbb Z\setminus\{0\}}P\Big(Z\_0\in\Big(\frac{-1/2-k}s,\frac{1/2-k}s\Big)\Big)\le Es\_1. 
\end{equation}
Therefore and because $s\_1,s\_2\ge0$, we have  

\begin{equation}
 E|\mu\_s((-1/2,1/2))-1|\le Es\_1+Es\_2\to0. 
\end{equation}
So, by Markov's inequality, (1) follows.


