# Questions about the conditional Radon-Nikodym derivative

Let $$(\Omega, \mathcal{A}, \mathbb{P})$$ be a probability space and $$X:(\Omega, \mathcal{A})\rightarrow (\mathcal{X}, \mathcal{F})$$ and $$Y:(\Omega, \mathcal{A})\rightarrow (\mathcal{Y}, \mathcal{G})$$ be random variables. Assume there exists a regular conditional distribution $$\mathbb{P}_{Y|X}$$

My first question revolves around the Radon Nikodym derivative of the regular conditional distribution $$\mathbb{P}_{Y|X}$$ w.r.t. some dominating measure $$\lambda$$ on $$\mathcal{Y}$$. So as I understand it, $$\frac{d\mathbb{P}_{Y|X=x}}{d\lambda}$$ is for each fixed $$x$$ a function in $$y$$. I.e., for each $$x$$ we get a different function. Each of these functions is measurable in $$y$$. Is this correct so far?

1.) If we fix $$y$$, is it a function in $$x$$ then? If so, is it measurable in $$x$$? (if it is not measurable what would be the conditions for it to be measurabe?) How would I write this notation-wise? I am wondering about the notation, since then I wouldnt condition on $$X=x$$, right? Does the factorization lemma come into play that gives the relation between $$\mathbb{P}_{Y|X=x}$$ and $$\mathbb{P}_{Y|X}$$? Also is $$\frac{d\mathbb{P}_{Y|X=x}}{d\lambda}$$ only depending on $$y$$ but for instance $$\frac{d\mathbb{P}_{Y|X}}{d\lambda}$$ would be a function of both $$x$$ and $$y$$?

2.) would it also make sense to form a Radon-Nikodym derivative when the dominating measure was some other conditional distribution $$Y|X$$. I.e., assume there is another probability measure $$R$$ on $$\Omega$$. Then let $$R_{Y|X}$$ be a regular conditions distribution such that $$R_{Y|X=x}$$ dominates $$\mathbb{P}_{Y|X=x}$$ for each $$x$$. So would it make sense/be well defined to build the Radon nikodym derivative of these two conditional distributions? What would be the interpretation and would it depend only on $$y$$ or both on $$x$$ and $$y$$? Could we also form the Radon Nikodym derivative of $$\mathbb{P}_{Y|X}$$ and $$R_{Y|X}$$? Would it also make sense if the dominating measure was a distribution defined on the product space $$\mathcal{X}\times\mathcal{Y}$$? Would that change on what arguments the radon nikodym derivative would depend on then?

3.) What I am ultimately interested in is the following: So as far as I understand it, in Fubinis theorem for transition kernels $$\int_{\mathcal{X}\times{\mathcal{Y}}}f(x, y)\mathbb{P}_{X, Y}(d(x,y))=\int_{\mathcal{X}}\int_\mathcal{Y}f(x,y)\mathbb{P}_{Y|X=x}(dy)\mathbb{P}_X(dx),$$ we cannot exchange the integral. However, my question is, if this is possible if I use the above mentioned Radon-Nikodym derivative: $$\int_{\mathcal{X}}\int_\mathcal{Y}f(x,y)\mathbb{P}_{Y|X=x}(dy)\mathbb{P}_X(dx)=\int_{\mathcal{X}}\int_\mathcal{Y}f(x,y)\frac{d\mathbb{P}_{Y|X=x}}{d\lambda}\lambda(dy)\mathbb{P}_X(dx)=\int_{\mathcal{Y}}\int_\mathcal{X}f(x,y)\frac{d\mathbb{P}_{Y|X=x}}{d\lambda}\mathbb{P}_X(dx)\lambda(dy)$$ Would this be allowed?

## 1 Answer

Fubini's theorem only applies in a product space $$Z=X\times Y$$ when $$X$$ and $$Y$$ are independent, that is, if $$P_Z(A\times B)=P_X(A)\times P_Y (B)$$.
In case of non-independence: $$\int f(x,y)dP_Z=\int \int f(x,y)dP_XdP_Y$$ is no longer valid, but this other one: $$\int f(x,y)dP_Z=\int dP_X\int f(x,y)dP_{Y/X}$$ If pdfs existed for $$P_X(p(x)$$) and for $$P_{Y/X}(q(x,y))$$ (which is not always the case), we could write: $$\int f(x,y)dP_Z=\int p(x)dx\int f(x,y)q(x,y)dy=\int\int p(x)q(x,y)f (x,y)d(x,y)$$ Logically, in this last integral we can apply Fubini's theorem.

• Hi thanks for your answer! Could you also answer the other 2 questions? Dec 19, 2023 at 8:36
• Also my third question was about whether i could interchange the order of Integration when dealing with the densities Dec 19, 2023 at 9:03
• If it exists, $\frac{dP_{Y/X}}{d\Lambda}$ would be a $(P_X\otimes\Lambda)$-measurable function : $q(x,y)$ (defined a.e.). The measures $P_X$ and $\Lambda$ are independent, so the last equation in part (3) is perfectly valid. Regarding section (2), the Radon-Nikodim derivative will exist as long as the necessary conditions are met ($\sigma$-finitude, absolute continuity, etc.), whether conditional measures or not. For example, $\frac{dP_{Y/X}}{dQ_{Y/X}}$ might make sense, which would obviously be a function of $(x,y)$. Dec 19, 2023 at 17:06