I am trying to understand how a joint distribution is formed when two regular conditional distributions are involved that are conditional with respect to different random variables.
Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space, and let there be the three random variables $X:(\Omega, \mathcal{A})\rightarrow (\mathcal{X}, \mathcal{F})$, $Y:(\Omega, \mathcal{A})\rightarrow (\mathcal{Y}, \mathcal{G})$, $Z:(\Omega, \mathcal{A})\rightarrow (\mathcal{Z}, \mathcal{H})$. Let us consider the Markov kernels $\mathbb{P}_{Y|X}$ and $\mathbb{P}_{X|Z}$.
1.) My question now is, if $$\int_{\mathcal{X}}\mathbb{P}_{Y|X=x}(E) \mathbb{P}_{X|Z=z_0}(dx)=\mathbb{P}_{Y, X|Z=z_0}(E)$$ holds (for some fixed $z_0$)?
On the one hand I would say no because intuitively I would assume that we would need a kernel $\mathbb{P}_{Y|X, Z}$ for that. On the other hand, for some fixed $z_0$, $\mathbb{P}_{X|Z=z_0}$ is simply a measure on $\mathcal{X}$ and not a conditional distribution (i.e., a kernel) anymore, so by that logic the above equation would make sense. Can someone shed light on this?
2.) What is the difference between $\mathbb{P}_{Y, X|Z=z_0}(E)$ and $\mathbb{P}_{Y, X}(E)$ anyway, since both are measures on $\mathcal{X}\times\mathcal{Y}$, so how does the conditioning on a fixed $z_0$ even makes a difference? Can somebody tell me how exactly these two quantities differ?
The correct is: $$\int_D d\mathbb{P}_{X/Z}\int_E d\mathbb{P}_{Y/X,Z}=\mathbb{P}_{X,Y/Z}(D\times E)$$ Logically, if $D=\mathcal{X}$: $$\mathbb{P}_{X,Y/Z}(\mathcal{X}\times E)=\mathbb{P}_{Y/Z}(E)$$