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This question originates from the definition of the Cox point process, but I suspect it might be a more general one.

If we define $$Q(\cdot) = \int_{\mathcal M} P_{\Lambda}(\cdot)Q_{\Psi}(d\Lambda)$$

Then $$\int_{\mathcal N} \mu(B) Q(d\mu) \stackrel{(\ast)}= \int_{\mathcal M} \int_{\mathcal N} \mu(B) P_{\Lambda}(d\mu) Q_{\Psi}(d\Lambda)$$

Where

$\mathcal M$ is a set of locally finite measures

$\mathcal N$ is a set of locally finite integer-valued measures

$P_\Lambda$ is the distribution of a Poisson process with intensity measure $\Lambda$

$\Psi$ is a random (diffusion) measure with distribution $Q_\Psi$

$B$ is a Borel set on the measurable space $X$ on which the measures in $\mathcal N$ are defined.

My question is: How to explain the equality $(\ast)$? Intuitivelly it makes sense. Possibly this could be contrasted with integration w.r.t. $\nu(E) = \int_E f\; d\mu$ which gives $\int_E g \; d\nu = \int_E fg \; d\mu$, if such contrast is helpful in answering the question.

Thank you.

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Ok, here $P$ is a stochastic kernel on $\mathcal N$ given $\mathcal M$ and $Q$ is a measure on $\mathcal N$ that can be shortly written as $Q := Q_\Psi P$. By this definition $$ Q(N) = \int_\mathcal M P_\Lambda(N)Q_\Psi(\mathrm d\Lambda) $$ for any measurable set of measures $N\subseteq \mathcal N$. Now, let $K$ be a stochastic kernel on $X$ (I labeled the underlying point space for $\mathcal N$, and I guess you should have meant there the latter, not $\mathcal M$ as in the OP) given $\mathcal N$, which acts as $K(B|\mu) = \mu(B)$. So the $(*)$ formula is essentially a nested definition of $(Q_\Psi PK)(B)$. For the details you can check any book that talks about stochastic kernels, I think Kallenberg has a chapter that will cover most of your questions. Feel free to ask for clarification here as well.

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  • $\begingroup$ Thank you, I think I will read a bit more on stochastic kernels and re-visit this question and answer later. $\endgroup$ Commented Sep 25, 2016 at 13:59
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Almost exactly two years later, I find myself wondering the same thing, only to find my own question on the topic. Anyway, here's a more rudimentary approach to answering it.

In fact, it's simply an application of the standard measure-theoretic approach (indicator function -> simple function -> non-negative function)

The definition

$$Q(D) = \int_{\mathcal M} P_{\Lambda}(D)Q_{\Psi}(d\Lambda), D \in \mathcal N$$

is the special case for the indicator function. Take $f=1_D$, then we have $$\int_{\mathcal N} f(\mu) Q(d\mu) = \int_{\mathcal M} \int_{\mathcal N} f(\mu) P_\Lambda(d\mu) Q_\Psi(d\Lambda)$$

from which we obtain the same equality for all non-negative $f:\mathcal N \to \mathbb R$ by the standard measure-theoretic argument.

The equality $(\ast)$ is then only an application of that equality for $f(\mu) = \mu(B)$, sometimes called the projection of measure $\mu$.

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