0

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)

Math_Illiterate
  • 975
  • 1
    $P(D_x>t)=P(N(B(x,t))=0)=e^{-\lambda\cdot volume(B(x,t))}$. where $D_x$ is distance in question (from $x$), $N$ is Poisson count and $B(x,t)$ is ball of radius $t$ centered at $x$ – A.S. Jan 02 '16 at 20:36

1 Answers1

2

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 $\lambda >0$.

Em.
  • 15,981
  • Thanks! I can see now how the sphere is being used – Math_Illiterate Jan 02 '16 at 22:28
  • @Math_Illiterate No, problem. I forgot to mention, you can extend this to $\mathbb R^n$. In $\mathbb R$, $|A|$ is usually the familiar length of time, and in $\mathbb R^2$, $|A|$ is for example the area of the circle. – Em. Jan 02 '16 at 22:37