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 1


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.

  • $\begingroup$ Hi thanks for your answer! Could you also answer the other 2 questions? $\endgroup$
    – guest1
    Dec 19, 2023 at 8:36
  • $\begingroup$ Also my third question was about whether i could interchange the order of Integration when dealing with the densities $\endgroup$
    – guest1
    Dec 19, 2023 at 9:03
  • $\begingroup$ 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 $(x,y)$. $\endgroup$
    – Speltzu
    Dec 19, 2023 at 17:06

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