Is the maximum nearest neighbor distance between points of the process, over all the infinitely many points of a stationary Poisson point process of intensity $\lambda$ in $\mathbb{R}^d$, almost surely finite? What is its distribution?

I am interested in the case $d=1$ and $d=2$. Also, if it is any easier, you can replace between points of the process by between an arbitrary location in $\mathbb{R}^d$ and the closest point of the process.


1 Answer 1


$\newcommand\la\lambda\newcommand\de\delta\newcommand\R{\mathbb R}$This maximum (or, rather, supremum) distance, say $M$, is $\infty$ almost surely (a.s.).

Indeed, recall that a (simple) Poisson point process of intensity $\la\in(0,\infty)$ on $\R^d$ is a random Borel measure $m$ over $\R^d$ such that, for any pairwise disjoint bounded Borel subsets $A_1,\dots,A_k$ of $\R^d$, the random variables (r.v.'s) $m(A_1),\dots,m(A_k)$ are independent Poisson r.v.'s with respective parameters $\la|A_1|,\dots,\la|A_k|$, where $|\cdot|$ is the Lebesgue measure. It is not hard to show (see e.g. Proposition 9.1.III (ii, iii), p. 4) that $m=\sum_{i=1}^\infty\de_{X_i}$, where $\de_x$ is the Dirac delta measure supported on $\{x\}$, for $x\in\R^d$, and the $X_i$'s are random points in $\R^d$ that are a.s. pairwise distinct. So, $$M=\sup_i\min\{\|X_k-X_i\|\colon k\ne i\}$$ a.s., where $\|\cdot\|$ is the Euclidean norm.

Now take any real $a>0$ and any natural $n$. Take the hypercube $C_{a,n}:=[0,3na)^d$ and partition it naturally into $n^d$ congruent smaller hypercubes each with edgelengths $3a$. In each of these $n^d$ hypercubes $C_j$ ($j=1,\dots,n^d$) each with edgelengths $3a$, take the central sub-hypercube, say $B_j$, with edgelengths $a$. Note that
$$p:=P\big(m(B_j)=1,m(C_j\setminus B_j)=0\big) \\ =P\big(m(B_1)=1\big)\,P\big(m(C_1\setminus B_1)=0\big)>0$$ for each $j=1,\dots,n^d$.

Then the probability $P(M\ge a)$ will be no less than the probability that at least one of the $n^d$ congruent smaller hypercubes $C_j$ contains exactly one point of the random points $X_i$ and that one point is in $B_j$. The latter probability is $$1-(1-p)^{n^d}\to1$$ as $n\to\infty$. So, $P(M\ge a)\ge1$ for all real $a>0$, and hence $P(M=\infty)=1$.

  • $\begingroup$ Thank you! The reason I asked the question is because for Poisson-binomial point processes discussed in my previous questions, if the $F$ distribution is uniform, the maximum nearest neighbor distance is always finite regardless of the scaling factor $s$, and when $s\rightarrow\infty$, the Poisson-binomial process is a Poisson process. However, the maximum may not be bounded as $s\rightarrow\infty$, leaving open the possibility that for a Poisson process, $M=\infty$ (which you proved). $\endgroup$ Commented Dec 31, 2021 at 20:19

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