= Poisson factorial moment measure
{c}
{title2=$\alpha^{(k)}=\mu^{\otimes k}$}
For a <Poisson point process> of <intensity measure> $\mu$, its $k$th <Poisson factorial moment measure> is $\alpha^{(k)}=\mu^{\otimes k}$. Writing its <counting measure> as $N=\sum_i\delta_{Z_i}$, the identity means that for nonnegative <measurable function> $g$,
$$
\mathbb E\sum_{i_1,\ldots,i_k\ \mathrm{distinct}}g(Z_{i_1},\ldots,Z_{i_k})=\int g\,d\mu^{\otimes k}.
$$
The indices select distinct point occurrences. Their locations need not be different when $\mu$ has an <atom of a measure>. For a <simple point process>, the locations are different as well.<Lectures on the Poisson Process by Günter Last and others>[Section 4.4, Corollary 4.10: Poisson factorial moment measure is the tensor power of its intensity]{ref}
For example, if all intensity is concentrated at one point $a$, with $\mu(\{a\})=\lambda$, then $N(\{a\})$ has <Poisson distribution>. There are $N(\{a\})(N(\{a\})-1)$ ordered pairs of distinct occurrences there, with <expected value> $\lambda^2$, even though both locations equal $a$. Counting only pairs of different locations would give zero and fail the identity.
For $k=2$, separating equal-index terms from distinct-index terms gives the second-moment calculation in <Campbell's theorem>. Dividing the distinct-pair sum by two instead counts unordered pairs, as in the <intersection point process of a Poisson line process>.
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