I'm currently working out of Mathematical Statistics and Data Analysis by John Rice and ran into the following problem from Chapter 2 that I have no idea on how to solve.
Ch. 2 #43
Find the probability density for the distance from an event to its nearest neighbor for a Poisson process in three-dimensional space.
After googling around a bit I found a solution with very little explanation here, but I'm not sure where the use of a sphere comes from. Any help solving this problem or understanding the linked to solution would be appreciated (ie more explanation about the use of the volume of a sphere formula)
Best Answer
For a moment, consider not the 3D case, but the 2D. Let $d$ be the distance between a particular point and the nearest neighbor. Notice that $$\{d >x\}\iff \{\text{No points in the disk of radius $x$}\}$$ The sphere (ball) is the analogous case in 3D.
You are told that you have a Poisson process, and so $N(A)$ is a homogeneous Poisson process of intensity $\lambda >0$, and $$P(N(A) = k) = e^{-\lambda|A|}\frac{(\lambda|A|)^k}{k!},$$ for $k = 0,1,\dotsc$, where $A$ is a subset of $\mathbb R^3$. In this case $A$ is a sphere, so $|A|$ denotes the volume of the sphere.
Because of spatial homogeneity, it doesn't matter where the point is in $\mathbb R^3$, and you can just consider the origin. Thus, if $D$ is the distance of the nearest neighbor in $\mathbb R^3$, then \begin{align*} F_D(x) &= P(D <x)\\ &= 1-P(D> x)\\ &= 1-P(N(A) = 0)\\ &=1- \exp\left\{-\lambda\cdot \frac{4}{3}\pi x^3\right\} \frac{(\lambda(4/3)\pi x^3)^0}{0!}\\ &= 1- \exp\left\{-\lambda\cdot \frac{4}{3}\pi x^3\right\}, \end{align*} where I assume the Poisson process has intensity $\lambda >0$.