= Solution
Write $z=(x,y)$ and $\mu(dz)=\lambda(z)\,dz$. Treat the nonnegative plant weights as <independent> marks conditional on the locations, with a <measurable> mark law $K_z(dw)$ depending only on $z$. Its first and second moments are
$$
\int w\,K_z(dw)=m(z),\qquad\int w^2\,K_z(dw)=m(z)^2+v(z).
$$
This is the independent-marking interpretation appropriate for location-dependent weights.
The general results needed are the following. The <Independent marking theorem for Poisson point processes> says that such marks turn a <Poisson point process> of intensity $\mu$ into a <Poisson point process> on location-mark space with intensity $\nu(dz,dw)=\mu(dz)K_z(dw)$. The <Campbell first-moment formula> says that for any nonnegative <measurable> $g$, $\mathbb E\sum_{u\in\Pi}g(u)=\int g\,d\nu$, allowing infinite values. The second <Poisson factorial moment measure> identity says that for nonnegative <measurable> $h$,
$$
\mathbb E\sum_{u,u'\in\Pi,\ u\ne u'}h(u,u')=\iint h(u,u')\,\nu(du)\nu(du').
$$
These statements apply to sigma-finite intensities and therefore do not require finitely many plants in the entire field. We use these general results without proving them.
Define the total as the nonnegative sum $W=\sum_{(z,w)\in\Pi}w$, or equivalently as the increasing limit of its finite-intensity truncations. the <Campbell theorem> gives
$$
\mathbb EW=\int_S\int_0^\infty w\,K_z(dw)\,\mu(dz)=\int_Sm(z)\lambda(z)\,dz=:M.
$$
Since $M<\infty$, the nonnegative <random variable> $W$ cannot be infinite on an event of positive <probability>. Thus \b[$W$ is finite <almost surely>], even if the number of plants is infinite.
Separate the square into its diagonal and ordered distinct-pair terms. Nonnegativity allows the use of <Tonelli's theorem> and increasing limits, giving
$$
\begin{aligned}
\mathbb EW^2
&=\mathbb E\sum_{(z,w)\in\Pi}w^2+\mathbb E\sum_{(z,w)\ne(z',w')}ww'\\
&=\int_S\bigl(m(z)^2+v(z)\bigr)\lambda(z)\,dz+M^2.
\end{aligned}
$$
The assumed second <integral> is finite, so $W$ is <square-integrable>. Subtracting $(\mathbb EW)^2$ yields
$$
\boxed{\mathbb EW=\int_Sm\lambda\,dz,\qquad\operatorname{Var}W=\int_S(v+m^2)\lambda\,dz.}
$$
The term $v$ accounts for random weights at fixed locations; the term $m^2$ accounts for the <Poisson> fluctuations in the locations and count. Nonnegativity is natural for weights. For signed marks, absolute first-moment integrability, rather than just a conditionally convergent <integral> of the signed <mean>, would be needed for the ordinary sum.
Back to article page
