Laplace functional of a Poisson random measure ($\mathbb{E}{e}^{-M(f)}=\exp (-\int (1-e^{-f})d\mu )$)

For nonnegative measurable $f$, independent Poisson counts first prove this formula for simple functions on disjoint measurable sets. Monotone approximation gives the general formula, allowing infinite integrals with $e^{-\infty }=0$. Conversely, the formula evaluated at $f={\sum }_j{u}_j{1}_{A_j}$ for disjoint finite-intensity sets gives the joint Laplace transform of independent Poisson counts, so characterizes the law of a Poisson random measure.

Let $M$ be a Poisson random measure of intensity $\mu$, let $\mu (A)<\infty$, and let $f\ge 0$ be measurable. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$ and differentiate from the right at $t=0$. On the random side, differentiation is dominated by the integrable random variable $M(A)$; on the intensity side, by $1_A$. This proves the displayed identity when $\int (1-e^{-f})d\mu$ is finite. When this integral is infinite, $e^{-M(f)}=0$ almost surely, and the same identity holds with both sides zero.

For a Poisson random measure on ${\mathbb{R}}^d$ with Lebesgue measure intensity and a nonnegative compactly supported continuous function $f$, put $m_r=\mu (B(0,r))$ and $a_r={\int }_{B(0,r)}{e}^{-f}d\mu$. Continuity gives $a_r/m_r\to e^{-f(0)}$. With $L=\mathbb{E}{e}^{-M(f)}$, the count-weighted Laplace functional of a Poisson random measure and independent inside/outside counts give respectivelyBoth converge to the displayed value. The conditioning is meaningful for every $r>0$; it is not conditioning on a positive-probability point at the origin.