Using Campbell's formula to find intensity measure $\Theta$

Question

Problem

Let $X_1, \ldots,X_n \sim K$ i.i.d random variables on $\mathbb{R}^d$, and let $$ \beta_{n,k} = \{ (x_1,\ldots,x_k) \in \{ X_1, \ldots,X_n\}^k \mid i \neq j, x_i \neq x_j \forall i,j \in \{1,\ldots,n\}\} $$ be the set of all $k$-tuples from the $n$ random variables with no repetition in both indices and values. Now for $f_n : \mathbb{R}^{d\times k} \rightarrow \mathbb{R}$ be measurable and define the random measure $$ Z = \frac{1}{k!}\sum_{x \in \beta_{n,k}} \delta_{f_n(x)} $$ and my goal is now to find the intensity measure $\Theta$ of this i.e. compute $\Theta(A):= E[Z(A)]$ for any $A \in \mathcal{B}(\mathbb{R}^d)$.

Attempt

I believe it is possible to show that

$$ E[Z(A)] = \frac{n_{(k)}}{k!} \int_{\mathbb{R}^{d\times k}} 1_A(f_n(x)) K^k(dx) $$ which would also suffice for my purposes. Here $n_{(k)} := n\cdot(n-1)\cdot \ldots\cdot(n-k+1)$. Using Campbell's formula and that $\{ X_1, \ldots,X_n\}^k$ has intensity measure $K^k$ (by independence and identical distribution) , we get that \begin{align*} E[Z(A)] &= \frac{1}{k!}E\bigg[\sum_{x \in \beta_{n,k}} \delta_{f_n(x)}(A)\bigg] \\ & = \frac{1}{k!}E\bigg[\sum_{x \in \{ X_1, \ldots,X_n\}^k} 1_{\beta_{n,k}}(x) 1_A(f_n(x))\bigg] \\ &= \frac{1}{k!} \int_{\mathbb{R}^{d\times k}} \sum_{x \in \{ X_1, \ldots,X_n\}^k} 1_{\beta_{n,k}}(x) 1_A(f_n(x)) K^k(dx), \end{align*} but this is not quite we we wanted. We need to get rid of $1_{\beta_{n,k}}(x)$ and the sum, and somehow replace with this combinatorical factor $n_{(k)}$. However I am not quite sure how this happens.

Can anyone help me?

Answer 1

Instead of considering $\beta_{n,k}$, whose number of elements may vary with $X_1, \dotsc, X_n$, consider a fixed set of $k$-tuples $$ \alpha_{n,k} = \{(x_1, \dotsc, x_k) \in \{X_1, \dotsc, X_n\}^k \, | \, i \neq j \text{ for all } i,j\} , $$ which always has $(n)_k$ elements, and put the "untiedness" condition $x_i \neq x_j$ in each summand: With $$ \gamma = \{(x_1, \dotsc, x_k) \in \{X_1, \dotsc, X_n\}^k \, | \, x_i \neq x_j \text{ for all } i,j\} , $$ we have $$ Z(A) = \frac{1}{k!} \sum_{x \in \alpha_{n,k}} 1_A(f_n(x)) \, 1_\gamma(x) . $$ Linearity of expectation reveals that your answer should actually be $$ \frac{(n)_k}{k!} P(f_n(X) \in A, X \in \gamma) $$ with $X = (X_1, \dotsc, X_k)$, say. The integral you had expected was $P(f_n(X) \in A)$. These would be equal if $K$ had no atom (point with positive probability), but needn't be in general. If $X_1, \dotsc, X_n$ are Bernoulli with probability $1/2$, then $Z$ would be a good name for the random measure when $k > 2$, because $Z$ would be zero.