Is it true that $\int_\mathcal{X}\int_\mathcal{Z} \eta(x,z)d\mathbb{P}_{f(x,Y)}(z)d\mathbb{P}_X(x)=\mathbb{E}_\mathbb{P}(\eta(X,f(X,Y))?$

Question

Suppose $(\Omega,\mathcal{F},\mathbb{P})$ is a probability space, $\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ are separable complete metric spaces, $X:\Omega\to\mathcal{X}$ and $Y:\Omega\to\mathcal{Y}$ are $\mathbb{P}$-independent random variables and $f:\mathcal{X}\times \mathcal{Y}\to\mathcal{Z}$ is measurable. If $\eta :\mathcal{X}\times\mathcal{Z}\to[0,+\infty]$ it seems plausible that $$\int_\mathcal{X}\int_\mathcal{Z} \eta(x,z)\operatorname{d}\mathbb{P}_{f(x,Y)}(z)\operatorname{d}\mathbb{P}_X(x)=\mathbb{E}_\mathbb{P}(\eta(X,f(X,Y)).$$ At least, this is true in the discrete case, since: $$\int_\mathcal{X}\int_\mathcal{Z} \eta(x,z)\operatorname{d}\mathbb{P}_{f(x,Y)}(z)\operatorname{d}\mathbb{P}_X(x)=\sum_{x\in\mathcal{X}}\sum_{z\in\mathcal{Z}}\eta(x,z)\mathbb{P}(f(x,Y)=z)\mathbb{P}(X=x) \\=\sum_{x\in\mathcal{X}}\sum_{z\in\mathcal{Z}}\eta(x,z)\mathbb{P}(f(x,Y)=z\cap X=x) = \sum_{x\in\mathcal{X}}\sum_{z\in\mathcal{Z}}\eta(x,z)\mathbb{P}(f(X,Y)=z\cap X=x)\\ =\sum_{x\in\mathcal{X}}\sum_{z\in\mathcal{Z}}\eta(x,z)\mathbb{P}((X,f(X,Y))=(x,z)) = \mathbb{E}_\mathbb{P}(\eta(X,f(X,Y)),$$ but how to prove it rigorously in the general case? I tried the following:

$$\int_\mathcal{X}\int_\mathcal{Z} \eta(x,z)\operatorname{d}\mathbb{P}_{f(x,Y)}(z)\operatorname{d}\mathbb{P}_X(x)\\ = \int_\mathcal{X}\int_\mathcal{Z} \eta(x,z)\operatorname{d}\mathbb{P}_{f(X,Y)}(z|X=x)\operatorname{d}\mathbb{P}_X(x)\\ = \int_{\mathcal{X}\times \mathcal{Z}} \eta(x,z)\operatorname{d}\mathbb{P}_{(X,f(X,Y))}(x,z)=\mathbb{E}_\mathbb{P}(\eta(X,f(X,Y)).$$ but I don't know how to justify that the first and the second equality holds.

Answer 1

Define an auxiliary mapping

$$(x,y) \mapsto h(x,y):= \eta(x,f(x,y)).$$

By definition, $$h(X,Y) = \eta(X,f(X,Y)).$$ Since the random variables $X$ and $Y$ are independent, we have \begin{align*} \mathbb{E}\eta(X,f(X,Y)) = \mathbb{E}h(X,Y) &= \int h(x,y) \, d\mathbb{P}_{(X,Y)}(x,y) \\&= \int \int h(x,y) \, d\mathbb{P}_Y(y) \, d\mathbb{P}_X(x). \tag{1}\end{align*}

Now note that

$$\int h(x,y) \, d\mathbb{P}_Y(y) = \mathbb{E}\eta(x,f(x,Y)) = \int h(x,z) \, d\mathbb{P}_{f(x,Y)}(dz).$$

Plugging this into $(1)$, gives the assertion.