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) 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 $\log\lambda$?

Mathematicain this regard. :-) $\endgroup$`\prod`

s in your calculations should be`\sum`

s. The meaning of $o(1)$ is not clear: the diagonal terms $p_i^2/(1-p_i)^2$ clearly have a non-vanishing impact on the sum. $\endgroup$doesshow something: if $s \to \infty$, then indeed the contribution of the diagonal terms is $o(1)$ (under minimal assumptions on the distribution) and you recover a Poisson point process. But this is fairly standard, I think, and it looks like you are more interested in small $s$, in which case Iosif's answer clearly tells you that you get a trivial limit. $\endgroup$3more comments