Campbell Theorem for $n$-th moment

Question

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

$\operatorname {E}\big[\big(\sum_{x\epsilon N} f(x)\big)^n\big]$

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?

Answer 1

Notice that

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

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 $\sum_{i=1}^{n} i\lambda_i = n$. 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 $n$.

Answer 2

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_{x\in N} f(x)\right)^3$ becomes $$ \sum_{i} U_{i} f(x_i)^3 + 3 \sum_{i} \sum_{j\ne i} U_i U_j f(x_i) f(x_j)^2 + \sum_{i} \sum_{j \ne i} \sum_{k \ne i,j} U_i U_j U_k f(x_i) f(x_j) f(x_k)$$ so that we get $$ \mathbb E\left[\left(\sum_{x\in N} f(x)\right)^3\right]=\int f(x)^3\; d\Lambda(x) + 3 \iint f(x) f(y)^2 \; d\Lambda(x)\; d\Lambda(y) + \iiint f(x) f(y) f(z)\; d\Lambda(x)\; d\Lambda(y)\; d\Lambda(z)$$ where $d\Lambda$ 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.

Answer 3

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.