Is the expected distance between events in a spatial Poisson Process the same as the expected value from any point to the nearest event?

442 Views Asked by At

The problem I am trying to solve is modeling the # of coffee shops in a city as a spatial Poisson Process. I found the expected value from any given point to the nearest coffee shop to be $\frac{1}{2\sqrt\lambda}$, but I am also being asked what the expected distance between any store and the nearest other store.

I suspect these will be the same due to either the independent or stationary increments property of a Poisson Process, but am struggling to show why.

2

There are 2 best solutions below

2
On

The results are the same simply because the question is essentially the same. It doesn’t matter whether you look for the nearest coffee shop “from any given point” or from “any store”; in both cases you’re just looking for the nearest coffe shop from some point. The probability density for shops in the vicinity is independent of whether there’s a shop at that point, so the results are the same.

0
On

Came across the answer to my problem - so I'm posting it here in case anyone else is curious in the future.

For the expected distance between an arbitrary point and the nearest store:

Let R be a random variable, denoting the distance to the nearest store, and (x,y) be the arbitrary point. Consider the ball Br centered at (x,y). Finally, N(Br) is the counting process which denotes the number of coffee shops in Br. Then

$P(R > r) = P(N(B_r)=0) = e^{-λ*\pi*r^2}\\\\F_R(r) = P(R \leq r) = 1 - e^{-λ*\pi*r^2}\\\\f_R(r) = \frac{d}{dx}F_R(r)= 2λ\pi*e^{-λ*\pi*r^2}\\\\E[R] = \int_{0}^{\infty}rf_R(r)dr=\frac{1}{\sqrt2*λ}$

For the expected distance between stores, do the same, but define Br as the ball centered at (x,y) with the point (x,y) removed. The area of the ball doesn't change and therefore, because of the stationary increments property of the Poisson process, the calculation doesn't change.