= Solution
Let
$$
S_g=\sum_{x\in\Pi}g(x).
$$
The compact support of $g$ ensures that this sum has only finitely many nonzero terms almost surely. The intensity measure of the ideal gas is
$$
\mu(dx)=z\,dx.
$$
Applying the first-moment identity in <Campbell theorem> gives
$$
\boxed{\mathbb E S_g=z\int_{\mathbb R^3}g(x)\,dx}.
$$
For the second moment, separate equal and distinct particles:
$$
\begin{aligned}
\mathbb E S_g^2
&=\mathbb E\sum_{x\in\Pi}g(x)^2
+\mathbb E\sum_{\substack{x,y\in\Pi\\x\ne y}}g(x)g(y)\\
&=z\int_{\mathbb R^3}g(x)^2\,dx
+z^2\left(\int_{\mathbb R^3}g(x)\,dx\right)^2.
\end{aligned}
$$
Subtracting $(\mathbb ES_g)^2$ yields
$$
\boxed{\operatorname{Var}(S_g)
=z\int_{\mathbb R^3}g(x)^2\,dx}.
$$
These are the mean and variance formulas in Campbell's theorem for a homogeneous Poisson point process.
Back to article page
