This was previously asked on MathSE, but was not answered.

Answering a question, I realized that the nearest point for a planar Poisson point process (with constant intensity $\lambda>0$) is normally distributed.

Indeed, it is easy to see that if $R_x$ measures the distance to nearest neighbor $n_x$ of some point $x$, then $$ P(R_x>r) = P(\text{no points in a ball of radius }r\text{ around }x) = e^{-\lambda\pi r^2}, $$ so $R_x$ has the Rayleigh distribution with parameter $\sigma = 1/(\lambda\pi\sqrt{2})$. And since the direction to nearest neighbor is obviously uniformly distributed, the vector $n_x - x$ is distributed as a pair of independent centered normal variables with variance $\sigma^2$. It is not hard to see that this is also true if we speak about the nearest neighbor of a point taken from the Poisson point process.

Naturally, this property is just a coincidence, it is specific to dimension $2$, and in this sense it is similar to the conformal invariance of Brownian motion. But the latter property implies a lot of interesting facts; you can even prove theorems from complex analysis like little and big Picard, using the planar Brownian motion.

Therefore, the question:

Are there any interesting facts or properties following from the fact that the nearest neighbor distribution for a planar Poisson point process is normal?

  • $\begingroup$ This is not really an answer, but a possibly related reference. Terry Soo and co-authors have some papers where they study "Poisson thinning". That is, they construct for fixed $\mu<\lambda$ a deterministic measurable map which sends an instance of a Poisson process with rate $\lambda$ to an instance of a Poisson process with rate $\mu$. This is all about "extracting randomness" from PPs; maybe the fact that you mention is used somewhere. (NB: of course non-deterministic Poisson thinning is simple) $\endgroup$ Sep 20, 2015 at 6:13
  • $\begingroup$ Normal random variables can be negative, while distances cannot... $\endgroup$
    – Michael
    Sep 20, 2015 at 17:41
  • $\begingroup$ The $e^{-x^2}$ function is associated with the Gaussian density. In your link, the $e^{-x^2}$ is associated with a CDF function. I do not think it is correct to call your distance "normally distributed," even if you overlook the non-negative issue. $\endgroup$
    – Michael
    Sep 20, 2015 at 17:49
  • $\begingroup$ @Michael, maybe I was not very careful writing this. I don't claim that the distance is normally distributed, but rather that the neighbor is. The neighbor is a two dimensional vector, whose length has Rayleigh distribution with cdf $1-e^{-ax^2}$. But the vector does have normal distribution. I copied the paragraph from my MathSE post in order to clarify. $\endgroup$
    – zhoraster
    Sep 20, 2015 at 18:01

1 Answer 1


One "fact" following from the Gaussian form of the nearest-neigbhor distribution is that it is identical to the spherical contact distribution, being the distribution of the radius of a sphere centered at the origin when it first makes contact with a point of the Poisson process (irrespective of whether the origin is itself a point of the Poisson process). This is a useful property if you wish to test whether a given set of points has a Poisson distribution or not. The identity of nearest-neighbor and spherical contact distributions holds in any dimension iff the process is Poisson (Slivnyak’s theorem).

  • $\begingroup$ Doesn't this exist in any dimension? $\endgroup$
    – zhoraster
    Sep 21, 2015 at 14:36
  • $\begingroup$ yes, in any dimension both distributions are the same, the Gaussian form is special for the plane $\endgroup$ Sep 21, 2015 at 14:43

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