= Campbell's theorem
{c}
{wiki=Campbell's_theorem_(probability)}
= Campbell theorem
{c}
{synonym}
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 $S_g=\int g\,dN$, the condition $\int|g|\,d\mu<\infty$ gives
$$
\mathbb ES_g=\int g\,d\mu.
$$
If also $\int g^2\,d\mu<\infty$, then
$$
\operatorname{Var}(S_g)=\int g^2\,d\mu.
$$
Expand the square: equal-index terms contribute $\int g^2d\mu$, 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.<Lectures on the Poisson Process by Günter Last and others>[Proposition 2.7 and equation 4.26; ordinary second moments require both L1 and L2 integrability]{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 $\int g^2dx=1$, but the sum is infinite <almost surely>: the ordered points satisfy $Z_i/i\to1$ 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.<Lectures on the Poisson Process by Günter Last and others>[Section 12.1, Lemma 12.2 and Proposition 12.4: compensated-integral isometry and L2 extension]{ref}
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