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.


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.


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


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 $k\rightarrow\infty$.


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.


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 $\log\lambda$?

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    $\begingroup$ I wouldn't trust Mathematica in this regard. :-) $\endgroup$ Nov 24, 2021 at 22:29
  • $\begingroup$ Also, some of the \prods in your calculations should be \sums. 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. $\endgroup$ Nov 24, 2021 at 22:33
  • $\begingroup$ @Mateusz: There were many instances where Mathematica was right and I was wrong, but the infinite series for $E[N(B)]$ with the logistic distribution, is a tricking one, made up of a difference of two divergent series that converge when combined together. Though some exact values are known for the finite sums. $\endgroup$ Nov 24, 2021 at 22:35
  • $\begingroup$ I made the change $\mbox{\sum}$ to $\mbox{\prod}$. Will be back in my desk in a few hours. Thank you. $\endgroup$ Nov 24, 2021 at 22:51
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    $\begingroup$ That said, your reasoning of course does show 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. $\endgroup$ Nov 24, 2021 at 23:47

2 Answers 2


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.

  • $\begingroup$ Thank you, I will be back at my desk in a few hours. But it is very counter-intuitive to me that we would always end up with a Poisson process, especially as my goal was to create non-Poisson processes. Wondering if it is Poisson when using the logistic distribution. Like you said, it can't possibly happen with $s=0$, even for the logistic distribution. But what if $s=10^{-50}$? $\endgroup$ Nov 24, 2021 at 22:39
  • $\begingroup$ @VincentGranville : The above proof shows that this cannot be true for all small enough $s$. $\endgroup$ Nov 25, 2021 at 0:59
  • $\begingroup$ Agreed. I did some more research, see my last comment to Mateusz. I will accept your answer, and update my question. $\endgroup$ Nov 25, 2021 at 3:16

This not an answer to my question (I accepted Iosif's answer) but additional information based on my most recent research, that may be helpful to the reader interested in these processes. It is too long for an update of my question.

For simplicity, we use the notation $F(x)=F_0(x)$ and $F(x-k)=F_k(x)$.

Theorem A

Regardless of the distribution $F_k$, if $b-a$ is an integer, then $E[N(B)]=\mu(B)=b-a$. Here $B=[a, b]$ with $a<b$. This is true regardless of the value of the scaling factor $s>0$.


The proof is very easy. We have, for any function $F$, the following trivial equality. Assuming $b=a+1$, $$E_n[B]\equiv \sum_{k=-n}^n \Big(F(b-k)-F(a-k)\Big)=F(b+n)-F(a-n).$$ Since $F$ is a probability distribution, we have $F(b+n)\rightarrow 1$ and $F(a-n)\rightarrow 0$, thus $F(b+n)-F(a-n)\rightarrow 1$ as $n\rightarrow\infty$. Note that $1=b-a$. A similar argument involving $2r$ terms on the right hand side, $r$ of them that tend to $1$, minus $r$ of them that tend to $0$ as $n\rightarrow\infty$, leads to the limit $(r\times 1) - (r \times 0) = r$ if $r=b-a$ is a positive integer. To conclude, note that $E[N(B)]=\lim_{n\rightarrow\infty} E_n[N(B)]$. $\blacksquare$

This result is not true in general if $b-a$ is not an integer. Now we analyze more closely what happens when $F$ is a simple distribution with some closed-form results available.

Case 1: $F$ is a uniform distribution

Let $p_k=F_k(b)-F_k(a)=F(b-k)-F(a-k)$ as in the original question. Here $$F_k(x)=\frac{1}{2}+\frac{x-k}{2s}, \mbox{ with } -s\leq x-k\leq s, \mbox{ and } s>0.$$ We have two cases, each with three sub-cases:

If $b-a\leq 2s$ then

  • If $a-s\leq k\leq b-s$ then $p_k=\frac{1}{2}-\frac{a-k}{2s}$.
  • If $b-s\leq k \leq a+s$ then $p_k=\frac{b-a}{2s}$.
  • If $a+s\leq k \leq b+s$ then $p_k=\frac{1}{2}+\frac{b-k}{2s}$.

If $b-a\geq 2s$ then

  • If $a-s\leq k\leq a+s$ then $p_k=\frac{1}{2}-\frac{a-k}{2s}$.
  • If $a+s\leq k \leq b-s$ then $p_k=1$.
  • If $b-s\leq k \leq b+s$ then $p_k=\frac{1}{2}+\frac{b-k}{2s}$.

If $k\notin [a-s,b+s]$, then $p_k=0$. Let $B=[a,b]$. The above results can be used to compute (in closed form) the quantities $$\mbox{E}[N(B)]=\sum_{k=-\infty}^\infty p_k, \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{Var}[N(B)]=\sum_{k=-\infty}^\infty p_k(1-p_k).$$

In particular, if $a=-s$ and $b=s$, we have:

$$E[N(B)]=-1+2\sum_{0\leq k\leq 2s}\Big(1-\frac{k}{2s}\Big)=-1+\frac{\lfloor 2s\rfloor(1+\lfloor 2s\rfloor)}{2s}.$$

Note that if $s/2$ is an integer, the above result is compatible with Theorem A: $E[N(B)]=2s = b-a$. Also, the guess is that as $s\rightarrow\infty$, the process tends to a stationary Poisson process of intensity $1$. In general though, $E[N(B)]$ is not a function of $\mu(B)=b-a$ unless $b-a$ is an integer.

Case 2: $F$ is a Laplace distribution

Again, let $p_k=F_k(b)-F_k(a)=F(b-k)-F(a-k)$ as in the original question. Here

$$F(x)=\frac{1}{2}+\frac{1}{2}\mbox{sgn}(x)\Big[1-\exp(-|x|/s)\Big] \mbox{ with } s>0$$

where $\mbox{sgn}$ stands for the sign function, with $\mbox{sgn}(0)=0$.

We have three cases:

  • If $k\leq a < b$ then $p_k=\frac{1}{2}\Big[\exp(-(a-k)/s) - \exp(-(b-k)/s)\Big] $
  • If $a\leq k \leq b$ then $p_k=1-\frac{1}{2}\Big[\exp(-(b-k)/s) + \exp((a-k)/s)\Big] $
  • If $a< b\leq k$ then $p_k=\frac{1}{2}\Big[\exp((b-k)/s) - \exp((a-k)/s)\Big] $

The second case is empty if $\lfloor b \rfloor < \lfloor a \rfloor +1$. The values for the expectation and variance of $N(B)$ can be computed in closed form. It is also expected that as $s\rightarrow \infty$, the process tends to a stationary Poisson process of intensity $1$.

Areas of further research

It includes:

  • Simulations (including 2D), finding (via simulations) the point distribution in any interval, and the interarrival times distribution.
  • Estimating $s$ based on data available, and model selection (what is the best fit for a given data set - process with uniform, Gaussian, Laplace, Cauchy, or logistic $F$? Model fitting can be done using $N(B)$.
  • These processes are repulsive processes when $s$ is small: it avoids points clustering. Design an attractive process: instead of replacing $k$ by $X_k$ (a single point), replace $k$ by multiple points close to $k$.

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