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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

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  • $\begingroup$ The question is not very clear. Let me try to guess. You have a Poisson point process 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 $P$. Is this correct? $\endgroup$
    – zhoraster
    Commented Sep 10, 2015 at 18:00
  • $\begingroup$ Yup should be. Im still trying to edit the question to make it clearer $\endgroup$
    – ys wong
    Commented Sep 10, 2015 at 18:03

1 Answer 1

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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.

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  • $\begingroup$ Yup i meant that. Could u please explain in more detail beacuse i still dont quite understand it. Thanks $\endgroup$
    – ys wong
    Commented Sep 10, 2015 at 18:12
  • $\begingroup$ Do you mean $P(X\lt r)$ instead of $P(X\gt r)$? $\endgroup$
    – Mick A
    Commented Sep 10, 2015 at 18:28
  • $\begingroup$ Sorry, my comment was to @zhoraster. $\endgroup$
    – Mick A
    Commented Sep 10, 2015 at 18:33
  • $\begingroup$ @MickA, no. It is exactly as it written. $\endgroup$
    – zhoraster
    Commented Sep 10, 2015 at 20:18
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    $\begingroup$ @zhoraster I would have thought $P(X>r) = P(\text{there are no points in }B(0,r)) = e^{-\lambda\pi r^2}$. $\endgroup$
    – Mick A
    Commented Sep 11, 2015 at 1:50

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