Csdrhrt

Campbell's theorem
gives a method for calculating expectations of sums of measurable functions f(x).

While I was solving my system model considering a Poisson point process, I came across the equation

$operatorname {E}big[big(sum_{xepsilon N} f(x)big)^nbig]$

Where $N$ is a Poisson point process

Is it possible to implement Campbell's theorem to find the expectation of the sums of measurable functions f(x) raised to the power n?

Notice that

$$ mathbf{E}left[ expleft{ -s sum_{x in N} f(x) right} right]
= expleft{ - int left(1 - e^{-sf(x)}right) , Lambda(dx) right}.$$

So the $n$-th moment of the sum $sum_{x in N} f(x)$ can be computed by differentiating both sides by $n$ times and plugging $s = 0$. To this end, we may invoke the following instance of Faa di Bruno's formula,

$$ frac{d^n}{ds^n}e^{g(s)} = e^{g(s)} sum_{lambda vdash n} left( frac{n!}{prod_{i=1}^{n} lambda_i! i^{lambda_i}} right) prod_{i=1}^{n} left( g^{(i)}(s) right)^{lambda_i}, $$

where the sum on the right-hand side runs over all the integer partitions $lambda$ of $n$, i.e., sequences $lambda in mathbb{N}_{0}^{n}$ satisfying $sum_{i=1}^{n} ilambda_i = n$. Plugging $g(s) = - int left(1 - e^{-sf(x)}right) , Lambda(dx)$, The result is that

begin{align*}
mathbf{E}left[ left( sum_{x in N} f(x) right)^n right]
&= sum_{lambda vdash n} left( frac{n!}{prod_{i=1}^{n} lambda_i! i^{lambda_i}} right) prod_{i=1}^{n} left( int f(x)^i , Lambda(dx) right)^{lambda_i}.
end{align*}

Addendum. The right-hand side admits a useful probabilistic expression: Let $S_n$ the symmetric group over the set ${1,cdots,n}$. For each $pi in S_n$ and each $i in {1, cdots, n}$, define $m_i(pi)$ as the number of $i$-cycles in $pi$. If $pi$ is chosen uniformly at random from $S_n$, then

begin{align*}
mathbf{E}left[ left( sum_{x in N} f(x) right)^n right]
&= n! , mathbb{E}left[ prod_{i=1}^{n} left( int f(x)^i , Lambda(dx) right)^{m_i(pi)} right]
end{align*}

Since the cycle structure of random permutation is well-studied, this allows to study the asymptotic behavior of the expectation for large $n$.

You can approximate $sum_{x in N} f(x)$ by $sum_j U_j f(x_j)$ where you divide your domain up into small pieces, $x_j$ is in the $j$'th piece, and $U_j$ is the number of points of your point process in the $j$'th piece. Further assume the pieces are so small that we can neglect values of $U_j$ other than $0$ and $1$.
Then for example
$ left(sum_{xin N} f(x)right)^3$ becomes
$$ sum_{i} U_{i} f(x_i)^3 + 3 sum_{i} sum_{jne i} U_i U_j f(x_i) f(x_j)^2
+ sum_{i} sum_{j ne i} sum_{k ne i,j} U_i U_j U_k f(x_i) f(x_j) f(x_k)$$
so that we get
$$ mathbb Eleft[left(sum_{xin N} f(x)right)^3right]=int f(x)^3; dLambda(x) + 3 iint f(x) f(y)^2 ; dLambda(x); dLambda(y)
+ iiint f(x) f(y) f(z); dLambda(x); dLambda(y); dLambda(z)$$
where $dLambda$ is the intensity of your point process.
Similarly for other powers, where in general for the $n$'th power you need to consider all ways
of partitioning $[1,ldots,n]$ into disjoint nonempty subsets.

Thank you for the answers, it is useful and gave a different perspective. I also found a similar result as Lee suggested, text here, Page 20.