Campbell's theorem

Let $N$ be a Poisson point process with intensity measure $\mu$, and let $g$ be a real-valued measurable function. For the ordinary Poisson sum $S_g=\int gdN$, the condition $\int |g|d\mu <\infty$ givesIf also $\int g^{2}d\mu <\infty$, thenExpand the square: equal-index terms contribute $\int g^{2}d\mu$, while the Poisson factorial moment measure makes distinct-index terms contribute $(\int gd\mu {)}^2$. Subtracting the squared expected value proves the variance formula. The Campbell first-moment formula also holds for general point processes; this variance identity uses the Poisson assumption.^[ref]

Square integrability alone is insufficient for the ordinary sum on an infinite-intensity space. Take intensity $dx$ on $[1,\infty )$ and $g(x)=1/x$. Then $\int g^{2}dx=1$, but the sum is infinite almost surely: the ordered points satisfy $Z_i/i\to 1$ by the strong law for Poisson arrival times, so comparison with the harmonic series proves divergence.

For this same $g$, the compensated Poisson integral exists as a mean-zero limit under convergence in Lp with $p=2$ and has variance $1$. Its definition is the L2 construction of compensated Poisson integrals, rather than subtraction of two infinite ordinary integrals. When the total intensity is finite, Cauchy-Schwarz inequality makes L2 imply L1, explaining why the shorter variance statement works in that special case.^[ref]

For a Poisson random measure $\Pi$ with intensity $\mu$ and measurable nonnegative $h$, $\mathbb{E}\int hd\Pi =\int hd\mu$, allowing both sides to be infinite. For signed $h$, the same identity holds if $\int |h|d\mu <\infty$. This first-moment statement does not require square integrability. Applying it to a nonnegative volume mark gives the expected total volume in a marked Poisson model.