Distribution of interarrival times for a special class of stochastic point processes

Question

I am interested in Poisson-binomial stationary point processes (here on the real line) defined as follows. Let

• $t_k=k/\lambda$, with $k\in\mathbb{Z}$ and $\lambda>0$,
• $F_s(x)$ be a symmetric, continuous cumulative distribution function centered at 0, with $s$ an increasing function of the variance. If $s=0$, the variance is zero, and if $s=\infty$, the variance is infinite.
• $(X_k), k\in \mathbb{Z}$, be a sequence of independent random variables, with $P(X_k<x) = F_s(x-t_k)$.

The points of the process are the $X_k$'s. The parameter $\lambda$ is the intensity of the process, and $E(X_k)=t_k$ by construction. If $B$ is a Borel set and $N(B)$ is the random variable counting the number of points in $B$, then $N(B)$ has a Poisson-binomial distribution of parameters $p_k, k\in\mathbb{Z}$, where $p_k = P(X_k\in B)$. See here for details.

If $s=0$, then $X_k=t_k$, and if $s=\infty$, the process is a stationary point process of intensity $\lambda$ regardless of $F_s$ (this statement still needs to be firmly established, but this is not the purpose of the question, and may be trivial). I am interested in the cases where $F_s$ is uniform on $[-s, s]$ with variance $s^2/3$, or a logistic distribution $F_s(x)=1/(1+\exp(-x/s))$ with variance $\pi^2 s^2/3$.

My question

What is the distribution of the interarrival time $T$, that is, the distance between two successive points of the process? More specifically, I am only interested in the variance of $T$, especially if it can be obtained in closed form, even if only for the uniform or logistic $F_s$. Approximations are OK too. It looks like $E(T)=1/\lambda$ though I did not prove it, and I expect $\mbox{Var}(T)$ to be a function of $s$ only, for a fixed $\lambda$. The final purpose, given a realization of such a process, is to estimate $\lambda$ and $s$. It seems that estimating $\lambda$ is solved already, if $E(T)=1/\lambda$ as guessed. All I know is that $T$ does not have an exponential distribution, unless $s=\infty$.

Suggestions for a solution

There is indeed an exact formula, but it is intractable, of no use, and way too general for my needs. If you look at the order statistics $X_{(k)}, k=1,\dots,n$, the Bapat-Beg theorem (see here) provides the formula for the distribution of order statistics of independent but non identically distributed random variables. Here I am only interested in the distribution of $T_k=X_{(k+1)}-X_{(k)}$, and given the stationarity and the fact that all $X_k$'s have the same distribution except for the location parameter, things should be much simpler, especially if you are only interested in the first two moments. The distribution of $T_k$ should not depend on $k$ when we consider the entire point process with infinitely many points. Even though $k\in \mathbb{Z}$ and restricting ourselves to $k$ between $1$ and $n$ produces side effects, as $n\rightarrow\infty$, these side effects disappear. So we can focus on $\mbox{Var}(T_{\lfloor n/2\rfloor})$ with $n\rightarrow\infty$ and $k\in \{1,\dots,n\}$. This should yield the desired result.

Answer 1

$\newcommand{\D}{\overset{\text{D}}=}\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}$By rescaling, without loss of generality $\lambda=1$, so that $X_k=k+Z_k$, where the $Z_k$'s are iid. As stated in a comment by Mateusz Kwaśnicki, \begin{equation} T\D Y:=U-X_0, \end{equation} where $\D$ means the equality in distribution and $U:=\inf\{X_k\colon X_k>X_0\}$. So, for real $y>0$, \begin{equation} \begin{aligned} P(T>y)&=P(\forall k\in\Z\ X_k\notin(X_0,X_0+y]) \\ &=P(\forall k\in\Z\ k+Z_k\notin(Z_0,Z_0+y]) \\ &=\int_\R P(Z\in dz)\prod_{k\in\Z\setminus\{0\}}(1-P(k+Z\in(z,z+y])), \end{aligned} \tag{1} \end{equation} where $Z:=Z_0$.

The latter integral is apparently the best expression in general for $P(T>y)$.

If $Z$ is uniformly distributed on the interval $[-s,s]$ for some real $s>0$, then (1) yields \begin{equation} \begin{aligned} &P(T>y)=p_s(y) \\ &:=\frac1{2s}\int_{-s}^s dz\,\prod_{k\in\Z\setminus\{0\},\,-2s<k<1+4s}\Big(1-\frac{g(s,z,k,y)}{2s}\Big), \end{aligned} \tag{2} \end{equation} where \begin{equation} g(s,z,k,y):=\max (0,\min (s,-k+y+z)-\max (-s,z-k)). \end{equation} Even in this special case, the integral in (2) can hardly be simplified for general values of $s$. However, for any given particular value of $s>0$, in principle we can get an explicit expression for $P(T>y)=p_s(y)$ and hence for any moments of $T$: $ET^r=\int_0^\infty ry^{r-1}p_s(y)\,dy$ for any real $r>0$.

For instance, $ET=1$ and $Var\,T=16297/29160\approx0.56$ if $Z$ is uniformly distributed on the interval $[-s,s]$ for $s=3/2$; see details of these calculations in the image of a Mathematica notebook below.

Remark: That $ET=1$ in the above example is no coincidence. Indeed, let $(X_{(j)})_{j\in\Z}$ be the sequence of the $X_k$'s rearranged in the increasing order so that, say, $X_{(0)}=X_0$. Suppose, say, that $X_0$ is bounded. Then, letting $T_j:=X_{(j+1)}-X_{(j)}$ and letting $n\to\infty$, we have \begin{equation} \sum_{j=0}^{n-1}T_j= X_{(n)}-X_{(0)}=n+O(1) \end{equation} with a finite nonrandom constant in $O(1)$, whence \begin{equation} ET_0=\frac1n\,E\sum_{j=0}^{n-1}T_j=1+O(1/n)\to1. \end{equation} Thus, $ET_0=1$. The latter equality should similarly hold whenever the tails of the distribution of $X_0$ are light enough. $\quad\Box$

Answer 2

This is not an answer, I already accepted the fist one. But it builds on the previous answer and offers three interesting theorems in the case where $\lambda$ is any positive real number. In theorem C, I prove the convergence to a Poisson process of intensity $\lambda$ when $s\rightarrow\infty$, based on the above answer. Theorem A is easy to prove.

We assume here that $F_s(x)=F(x/s)$. We use the notation $F$ rather than $F_1$, for the case $s=1$.

Theorem A (generalization of Theorem A posted here)

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

Theorem B

Let $T(\lambda,s)$ denotes the interarrival time discussed in my question and the subsequent answer. We have $T(\lambda,s) = T(1,\lambda s)/\lambda$.

Proof

The formula for $P(T>y)$, provided in the above answer, can be restated as follows:

$$P[T(\lambda,s)>y] = \int_{-\infty}^\infty \frac{f(x)}{1-p_0(x,y)} \prod_{k\in \mathbb{Z}} (1-p_k(x,y))dx,$$ with $$p_k(x,y) \equiv P(X_k\in[x, x+y]) = F\Big(\frac{x+y-k/\lambda}{ s}\Big)-F\Big(\frac{ x-k/\lambda}{s}\Big),$$

where $f$ is the density attached to $F$. The expression $F((x+y-k/\lambda)/ s)$ can be rewritten as $F((\lambda\cdot(x+y)-k/\lambda')/ s')$ with $\lambda'=1$ and $s'=\lambda s$. This works too if $y=0$. With the change of variable $\lambda\cdot(x+y)=x'+y$ we have $dx = (dx')/\lambda$ and the expression becomes $F((x'+y-k/\lambda')/ s')$. The variables are $x,x'$, and $y$ is assumed to be fixed. The above integral with respect to $x$, that defines $P(T(\lambda,s)>y)$, must be updated as follows:

• The dummy variable $x$ is replaced by the dummy variable $x'$
• The value of the integral is divided by $\lambda$ because $dx = (dx')/\lambda$
• The bounds are still from $-\infty$ to $\infty$
• $\lambda$ is replaced by $\lambda'=1$ and $s$ by $s'=\lambda s$

That is: $P[T(\lambda,s)>y] = P[T(\lambda',s')/\lambda>y] =P[T(1,\lambda s)/\lambda >y]$, thus $T(\lambda,s)=T(1,\lambda s)/\lambda$.$\blacksquare$

In two dimensions, $x$ is replaced by $(x_1,x_2)$, and $dx$ becomes $dx_1 dx_2$. The product over $k$ becomes a double product over $h,k$. The interarrival times is now the distance between a point of the process, and its nearest neighbor. Also, $F_s(x-k/\lambda)$ is replaced by $F_s(x_1-h/\lambda)F_s(x_2-k/\lambda)$, and $dx_1 = (dx_1')/\lambda, dx_2 = (dx_2')/\lambda$. Finally, $T(\lambda,s)=T(1,\lambda s)/\lambda^2$. In $d$ dimensions, it would become $T(\lambda,s)=T(1,\lambda s)/\lambda^d$. Note that $y$ is still a positive real number, now representing a radius.

Theorem C

If $s\rightarrow\infty$, the process converges to a Poisson process of intensity $\lambda$. It is assumed that $F$ is continuous, $F_s(x)=F(x/s)$ and that $F$ has a density (its derivative), denoted as $f$.

Proof

This is just a sketch. The reader is invited to check if some of my arguments result in some constraints on $F$, for instance about the thickness of its tail.

Part #1

We use the notation $T$ rather than $T(\lambda,s)$, for simplicity. Also, let $$p_k(x,y) \equiv P(X_k\in[x, x+y]) = F\Big(\frac{\lambda (x+y)-k}{\lambda s}\Big)-F\Big(\frac{\lambda x-k}{\lambda s}\Big)=\int_a^b f(u)du,$$ with $b=(\lambda (x+y)-k)\cdot(\lambda s)^{-1}$ and $a=(\lambda x-k)\cdot (\lambda s)^{-1}$. The last integral is denoted as $I_k$. Its interval has length $b-a=y/s$ and midpoint $(a+b)/2=(2x+y - 2k/\lambda)\cdot(2s)^{-1}$. In particular, $$I_k\sim \frac{y}{s}f\Big(\frac{2x+y}{2s}-\frac{k}{\lambda s}\Big) \mbox{ as } s\rightarrow\infty,$$ $$J_n \equiv \sum_{k=-n}^{n} I_k\sim \int_{-n}^{n}I_\nu d\nu= \frac{y}{s}\int_{-n}^n f\Big(\frac{2x+y}{2s}-\frac{\nu}{\lambda s}\Big)d\nu.$$ With the change of variable $\tau=-\nu/(\lambda s)$, we obtain

$$J_n \sim \frac{y}{s}\cdot \Big[\lambda s\int_{-n/(\lambda s)}^{n/(\lambda s)} f\Big(\frac{2x+y}{2s}+\tau\Big)d\tau\Big]=\lambda y\int_{-n/(\lambda s)}^{n/(\lambda s)} f\Big(\frac{2x+y}{2s}+\tau\Big)d\tau.$$

Here $\lambda$ is fixed. When $n\rightarrow\infty$, $s\rightarrow \infty$ and $n/s\rightarrow \infty$ (say $s\sim\sqrt{n}$ or $s\sim n/(\log n)$, we have

$$J_n\rightarrow \lambda y\int_{-\infty}^\infty f\Big(\frac{2x+y}{2s}+\tau\Big)d\tau =\lambda y,$$ because $f$ is a density and thus integrates to one.

Part #2

The formula for $P(T>y)$, provided in the above answer, can be restated as follows:

$$P(T>y) = \int_{-\infty}^\infty \frac{f(x)}{1-p_0(x,y)} \prod_{k\in\mathbb{Z}} (1-p_k(x,y))dx.$$

Note that regardless of $k$, we have $p_k(x,y)=I_k\rightarrow 0$ as $s\rightarrow\infty$. So the denominator $1-p_0(x,y)$ can be ignored in the previous formula (when $s\rightarrow\infty)$, and we also have:

$$\log\Big[\prod_{k\in\mathbb{Z}} (1-p_k(x,y))\Big]=\sum_{k=-\infty}^\infty\log(1-p_k(x,y))\sim -\sum_{k=-\infty}^\infty p_k(x,y) = -J_\infty = -\lambda y.$$ Thus,

$$\prod_{k\in\mathbb{Z}} (1-p_k(x,y))\sim \exp(-\lambda y) \mbox{ as } s\rightarrow \infty.$$

This product does not (at the limit) depend on $x$. Finally, we get

$$P(T>y)\sim \exp(-\lambda y)\int_{-\infty}^\infty \frac{f(x)}{1-p_0(x,y)}dx\sim \int_{-\infty}^\infty f(x) dx = \exp(-\lambda y),$$ as $f$ is a density and thus integrates to one. So, $T$ has an exponential distribution as $s\rightarrow\infty$. This implies that the limiting point process must be Poisson of intensity $\lambda$. $\blacksquare$

The takeaway from the proof of theorem C (see bottom of part #1) is that to simulate a realistic Poisson process as a limit of a Poisson-binomial process (pretty much regardless of $F$), you generate your $2n+1$ points ($k$ between $-n$ and $n$), you choose a large $n$ and a large $s$, but $s$ must be an order of magnitude smaller than $n$, to maker border effects negligeable. For instance, $s=\sqrt{n}$ or $s=n/(\log n)$ will do.