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 $\|x\_1-x\_2\|\_2 \le r$? 
Couldn't find any existing results on this.



Distance between poisson points in two disjoint unit discs
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 $a-1 < r < a+1$ 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 $f\_X(x) = \lambda e^{-\lambda(1-x)}$ and $f\_Y(y) = \lambda e^{-\lambda(y-a)})$ on $[0,1]$ and $[a,a+1]$ respectively.
Thus if $a-1 < r \le a$
$$P(Y - X \le r) = \lambda^2 \int\_{a-r}^1 dx \int\_a^{x+r} dy\; e^{-\lambda(1-x)} e^{-\lambda(y-a)} = {1 + \left( \lambda\,(a- r-1)-1
 \right)\; {{ e}^{\lambda\, \left( a-r-1 \right) }}}
 $$
while if $a \le r < a + 1$
$$P(Y - X \le r) = (1-e^{-\lambda})^2 - \lambda^2 \int\_0^{a+1-r} dx \int\_{x+r}^{a+1}
dy\; e^{-\lambda(1-x)} e^{-\lambda(y-a)} = 1 - 2 e^{-\lambda} + (1 - \lambda(a-r+1)) e^{\lambda(a-r+1)}$$


