Fubini's theorem for transition kernels

Question

In probability theorey there is a different form of Fubini's theorem that includes Markov kernels (regular conditional distributions) that does not need independence. Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space and $X:\Omega \rightarrow \mathcal{X}$ and $Y:\Omega \rightarrow \mathcal{Y}$ random variables. Then for a function $f:\mathcal{X}\times\mathcal{Y}\rightarrow \mathbb{R}$ that is either non-negative and measurable or integrable, the theorem is expressed as $$\int_{\mathcal{X}\times\mathcal{Y}}f(x,y)\mathbb{P}(d(x,y)) = \int_{\mathcal{X}}\biggl(\int_{\mathcal{Y}}f(x, y) \mathbb{P}_{Y\mid X}(dy, x)\biggl)\mathbb{P}_X(dx).$$

My question is if, like in the original theorem by Fubini, it is possible to change the order of integration, i.e., to first integrate over $y$ and then over $x$?

More specifically what I mean here is, if it is possible that we first view the conditional distribution as a measurable function on $\mathcal{X}$ for a fixed set $dy$ and then after integrating over $x$ we view it as a measure on $\mathcal{Y}$?

I am not sure how to write this formally, but something like this: $$ \int_{\mathcal{Y}}\biggl(\int_{\mathcal{X}}f(x, y) \mathbb{P}_X(dx)\mathbb{P}_{Y\mid X}(dy, x)\biggl) $$ Is this possible? Maybe yes under certain conditions?

Answer 1

The equality: $$\int_{\mathcal{X}}\biggl(\int_{\mathcal{Y}}f(x, y) d\mathbb{P}_{Y/X=x}(y)\biggl)d\mathbb{P}_X(x) = \int_{\mathcal{Y}}\biggl(\int_{\mathcal{X}}f(x, y) d\mathbb{P}_X(x)\biggl)d\mathbb{P}_{Y/X=x}(y).$$ doesn't make sense, unless $X$ and $Y$ are independent: the right-hand side depends on $x$, the first does not.

Answer 2

It would be more clear when you consider two discrete random variables. Suppose $X,Y$ have p.m.f $p(x,y)$. Then we have $E[f(x,y)]=\sum_{x,y} p(x)p(y|x)f(x,y)$. Summing $y$ first and then $x$, we have $E[f(x,y)]=\sum_x(p(x)\sum_yp(y|x)f(x,y))$, the genral case of which is the one you first mentioned. However you can see it makes no sense to sum $x$ first in $p(x)p(y|x)f(x,y)$. The only way to sum $x$ first is to write $p(x,y)=p(y)p(x|y)f(x,y)$, but it still finds the conditional expectation first.

In conclusion you cannot integrate over the marginal first and then over the conditional distribution.