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.
Best Answer
$\newcommand{\D}{\overset{\text{D}}=}\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}$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. $\quad\Box$