Two disjoint unit discs $D_1$ and $D_2$. Inside them there are random poisson points with intensity $\lambda$. For a given real $r>0$, what is the probability that there exist a poisson point $x_1\in D_1$ and a poisson point $x_2\in D_2$ such that $\x_1x_2\_2 \le r$? Couldn't find any existing results on this.

$\begingroup$ How about 1D case, i.e., $D_1$ and $D_2$ are line segments? $\endgroup$– user32828Apr 19 '14 at 0:51
If $M$ is the $4$dimensional Lebesgue measure of $\{(x,y) \in D_1 \times D_2: \x  y\_2 \le r\}$, then the number of such pairs of points is a Poisson random variable with parameter $M \lambda^2$, so the probability that there is at least one is $1  \exp(M \lambda^2)$.
EDIT: Oops, this is wrong. It is not Poisson.
EDIT: Consider the case where $D_1$ and $D_2$ are line segments, say $[0,1]$ and $[a,a+1]$ where $a > 1$. Of course we need $a1 < r < a+1$ to make the question nontrivial. Let $X$ be the maximum of the Poisson points in $D_1$ ( $\infty$ if there are none) and $Y$ the minimum of the Poisson points in $D_2$ ($+\infty$ if there are none). $X$ and $Y$ are independent, with densities $f_X(x) = \lambda e^{\lambda(1x)}$ and $f_Y(y) = \lambda e^{\lambda(ya)})$ on $[0,1]$ and $[a,a+1]$ respectively. Thus if $a1 < r \le a$ $$P(Y  X \le r) = \lambda^2 \int_{ar}^1 dx \int_a^{x+r} dy\; e^{\lambda(1x)} e^{\lambda(ya)} = {1 + \left( \lambda\,(a r1)1 \right)\; {{ e}^{\lambda\, \left( ar1 \right) }}} $$ while if $a \le r < a + 1$ $$P(Y  X \le r) = (1e^{\lambda})^2  \lambda^2 \int_0^{a+1r} dx \int_{x+r}^{a+1} dy\; e^{\lambda(1x)} e^{\lambda(ya)} = 1  2 e^{\lambda} + (1  \lambda(ar+1)) e^{\lambda(ar+1)}$$


$\begingroup$ Integrate over $x \in D_1$ the area of the intersection of $D_2$ with the disk of radius $r$ about $x$. I doubt that you'll end up with a closedform expression. $\endgroup$ Apr 8 '13 at 18:41

$\begingroup$ Some anon user wished to comment via an edit and an answer that "The number of pairs (x,y) is not a Poisson random variable." and that this creates problems of the answer. I leave this purely as a service as the technicalities of the site prevent to record it otherwise. $\endgroup$– user9072Apr 17 '14 at 15:54

$\begingroup$ Oops, it looks like the anon user is correct. The product of two independent Poisson processes is not a Poisson process. $\endgroup$ Apr 18 '14 at 3:46

$\begingroup$ However the idea here could be used to find the expected number of pairs of points within a distance r. With that, the Aldous Poisson Clumping Heuristic could be used to get a decent approximation to the probability. I doubt that its been done for this particular problem though. $\endgroup$– guestApr 20 '14 at 11:21