Find the Probability density for the distance from an event to its nearest neighbor for a Poisson process in a plane.
What i tried
Assume the plane to be a circle. The points can be represented as $E$ to $F$ with the distance in between be represented by $X$. $E$ is at the centre of the circle with radius $r$ While the area of the circle is $\pi r^{2}$. The distance $X$ have to be constant.Thus there are the possibilities of $P(X<r)$ or $P(X>r)$ For $P(X<r)$, It means both points $F$ lies inside a larger but punctured circle.For $P(X>r)$ it means there is a smaller circle that still have the origin $E$ but lies 'inside of the line $X$. in other words the line $X$ begins at the origin and protrudes out of the circle.And also assume that $\lambda$ represent $1$ unit area of the circle. This also means that the smaller circle lies inside the larger circle and the smaller circle can be cut out from the larger circle to form a hole in the larger circle.Let $U_r$ be the number of events in the punctured circle and $U_r$ is modeled by a Poisson process. Then since the area is $\pi r^{2}$, the Probability density for the distance from an event to its nearest neighbor is represented by the Poisson parameter of $\pi r^{2}\lambda$
Then we want to find $P(X<r)=1-P(X>r)=1-P(U_r=0)$
Could anyone explain more clearly. Thanks
Best Answer
I assume that you have a Poisson point process of constant intensity $\lambda$ in the plane, so a random collection $P$ of points. Then you take some point (say, the origin) and look for a closest point from $P$. Let the distance be $X$. Then $$ P(X>r) = P(\text{there are no points in }B(0,r)) = e^{-\lambda\pi r^2}, $$ where $B(0,r)$ is the ball of radius $r$ centered at origin.