# Is there a stationary point process generated by i.i.d. waiting times of any distribution?

Let $\mathcal{G}$ be the set of non-empty closed subsets of $\mathbb{R}$, equipped with its natural Polish topology, as induced by the metric $$d(A,B) \ = \ d_H\Big(\overline{\mathrm{arc}\tan(A)} \, , \, \overline{\mathrm{arc}\tan(B)}\Big)$$ where $d_H$ is the Hausdorff metric.

For any probability measure $\nu$ on $(0,\infty)$ and any $t \in \mathbb{R}$, define the probability measure $\mathbb{P}_{\nu,t}$ on $\mathcal{G}$ to be the image measure of $\nu^{\otimes \mathbb{N}}$ under the map $$\hspace{40mm} (x_i)_{i \geq 1} \mapsto \left\{ t + \sum_{i=1}^n x_i \, : \, n \geq 1 \right\} \hspace{6mm} \textrm{(defined \nu^{\otimes \mathbb{N}}-almost everywhere).}$$

Is it the case that for every probability measure $\nu$ on $(0,\infty)$, $\,\mathbb{P}_{\nu,t}$ is weakly convergent as $t \to -\infty\,$?

For example, in the case that $\nu=\mathrm{exponential}(\lambda)$, I believe we have that $\mathbb{P}_{\nu,t}$ converges weakly to the law of a two-sided Poisson process of intensity $\lambda$ as $t \to -\infty$. But this is the "easy case", since a one-sided Poisson point process is already stationary (as a one-sided process).

• I realise the topology I've given isn't really the "natural" topology; I think the more natural topology is given by $d(A,B)=d_H(\{-\frac{\pi}{2},\frac{\pi}{2}\} \cup \mathrm{arc}\tan(A), \{-\frac{\pi}{2},\frac{\pi}{2}\} \cup \mathrm{arc}\tan(B))$. In this case, it makes sense to include $\emptyset$ in $\mathcal{G}$ (as in the answer below). Nonetheless, I think that this difference in topology makes no difference to the answers (except regarding convergence to $\delta_\emptyset$), since any limit other than $\delta_\emptyset$ should assign full probability to sets unbounded above and below. – Julian Newman Dec 1 '17 at 18:20

I consulted the specialists, and here is the summary (I denote by $X$ a random variable with distribution $\nu$):

• if $X$ is integrable and non-lattice (does not assume values from $a\mathbb Z$ for some $a$), then $\mathbb{P}_{t,\nu}$ converges to a non-degenerate stationary distribution;
• if $X$ is lattice, then, obviously, $\mathbb{P}_{t,\nu}$ can't converge (see an example below). It does converge if $t$ is taken from the lattice, and $X$ is integrable;
• if $X$ is non-integrable, then $\mathbb{P}_{t,\nu}$ converges to zero (see an example below).

While at the first glance this seems very natural, it is unfortunately false.

A simple immediate counterexample is $\nu = \delta_1$, for which $\mathbb P_{\nu,t}$ is "oscillating between $x+\mathbb Z$, $x\in[0,1)$".

For a continuous distribution, a necessary (and, probably, sufficient) condition is the convergence in distribution, as $t\to \infty$, of the "overshoot" $$\gamma_t = S_{N_t+1} - t,$$ where $$N_t = \max\{n\ge 0: S_n<t \},$$ and $$S_n = X_1 + \dots + X_n$$ is the sum of iid random variables with distribution $\nu$.

If I think a little bit more about this problem, I might come up with a less abstract criterion of convergence, but I am quite confident that this should be well known.

Just for fun, here is a pathological example where all points "escape to infinity". It is discrete, but one can make a continuous example out of it.

Let $(r_n,n\ge1)$ be growing very fast, something like $3^{3^n}$ so that $r_{n+1}\gg r_n^2$, $n\ge 1$. Set $\nu(\{r_n\})= C/r_{n-1}, n\ge 2$.

Let us look at "overshoot" and "undershoot" of the level $r_n$. With positive probability (at least $1-(1-C/r_{n-1})^{r_{n-1}}$), among the variables $X_1,\dots,X_{r_{n-1}}$ there is one not less than $r_n$. Then the overshoot must have happened with adding this variable (the total value of all others is just not enough to overshoot). It follows that both overshoot and undershoot must be huge with positive probability. Appealing to 0-1 law, we have that almost surely, the overshoot and the undershoot converge to infinity, so the limit measure (at least over the subsequence $t= r_n$) is zero!

• Thanks for this wonderful answer! By "zero", I assume you mean the Dirac mass at the empty set. In the third point, are you saying that if $X$ is non-integrable then we definitely have convergence to zero, or just that we may have convergence to zero, as in your example? – Julian Newman Dec 1 '17 at 18:06
• @JulianNewman, by zero I mean the zero measure, viz. $\int f \, d\mu = 0$. AFAIK, there is always convergence to zero. Heuristically, by LLN, the mass of $[0,n]$ divided by $n$ goes to zero, thus we have this effect. – zhoraster Dec 1 '17 at 18:30
• Ah, so I suppose you're working with the wide topology; I was thinking of the narrow topology (intended by my use of the phrase "weakly convergent"), in which convergence of probability measures to zero is impossible. But I expect this makes no difference to the results, except that convergence of $\mathbb{P}_{t,\nu}$ to zero in the wide topology associated to $\mathcal{G}$ is probably equivalent to convergence of $\mathbb{P}_{t,\nu}$ in the narrow topology associated to $\mathcal{G} \cup \{\emptyset\}$ (with the natural topology) to $\delta_\emptyset$. – Julian Newman Dec 8 '17 at 14:09