Boolean model

Remarks

Under the integrability assumption, only finitely many grains $x_i + K_i$ meet each bounded set almost surely, so $\Xi$ is a random closed set. Its law is invariant under translations of $\R^d$.

Example: let $d = 2$ and let every grain be the closed disc of radius $r > 0$ centred at $0$. Then $V_2(K + B) = \pi (r + 1)^2$ is finite, so the assumption holds. A point $y$ is uncovered exactly when $\Phi$ has no point in the disc of radius $r$ around $y$, which has probability $e^{-\lambda \pi r^2}$. So the covered fraction of the plane is $p = 1 - e^{-\lambda \pi r^2}$. In general, $P(y \in \Xi) = 1 - e^{-\lambda E[V_d(K)]}$ for every $y \in \R^d$.

Non-example: put the same discs at the points of the lattice $\Z^2$ instead of a Poisson process. The union is a union of discs, but the germs do not form a Poisson point process, so it is not a Boolean model.

The Boolean model is also called the Poisson germ-grain model.