Writing Radon Nikodym derivative of joint distributions in terms of marginals and conditionals Given two random variables $X, Y$ and two joint probability distributions  $P_{X, Y}$ and $Q_{X, Y}$ on the same product space $(\mathcal{X}\times\mathcal{Y}, \mathcal{A}\times\mathcal{B})$ that are given by ($A\in\mathcal{A}, B\in\mathcal{B}$)
$$ P_{X, Y}(A\times B) = \int_{A\times B} P_{X, Y}(dx, dy)=\int_{A}P_X(dx)P_{Y|X}(B, x).$$
Likewise for Q:
$$ Q_{X, Y}(A\times B) = \int_{A\times B} Q_{X, Y}(dx, dy)=\int_{A}Q_X(dx)Q_{Y|X}(B, x).$$
Then if $P_{X, Y}$ is absolutely continuous w.r.t. $Q_{X, Y}$ their Radon Nikodym derivative exists:
$$P_{X, Y}(A\times B) = \int_{A\times B} \frac{dP_{X, Y}}{dQ_{X, Y}}(x,y)Q_{X, Y}(dx, dy). $$
I would now like to somehow express the derivative $\frac{dP_{X, Y}}{dQ_{X, Y}} $ now in terms of $P_X, P_{Y|X}, Q_X, Q_{Y|X}$.
Is that possible and if so why and under which conditions?
So I could obviously write it as $\frac{d(\int_{A}P_X(dx)P_{Y|X}(B, x))}{d(\int_{A}Q_X(dx)Q_{Y|X}(B, x))}$ but can that be further simplified? For instance when the conditionals are absolutely continuous w.r.t. to each other or something like that?
 A: I will try to use your notation though I am not quite familiar with it:
\begin{align*}
P_{X, Y}(A\times B)
&=
\int_{A}P_X(dx) \int_{B} P_{Y|X}(dy, x)
\\
&=
\int_{A}\frac{dP_{X}}{dQ_{X}}(x) \, Q_X(dx) \int_{B} \frac{dP_{Y|X}}{dQ_{Y|X}}(y, x)\, Q_{Y|X}(dy, x)
\\
&=
\int_{A \times B}\frac{dP_{X}}{dQ_{X}}(x) \frac{dP_{Y|X}}{dQ_{Y|X}}(y, x) \, Q_{X,Y}(dx, dy),
\end{align*}
so
$$
\frac{dQ_{X,Y}}{dP_{X,Y}}(x,y) = \frac{dP_{X}}{dQ_{X}}(x) \frac{dP_{Y|X}}{dQ_{Y|X}}(y, x).
$$
Note that the derivation only holds if the corresponding Radon-Nikodym derivatives exist!
Is this what you were looking for? Hope it helps..
UPDATE: What happened in each step:
First step: wrote out the probability as an integral and using conditional distributions (almost identical to your first equation)
Second step: Used Radon-Nikodym derivatives (under the assumption that they exist!) of $P_{X}$ w.r.t $Q_{X}$ and of $P_{Y|X}$ w.r.t $Q_{Y|X}$ in order to rewrite the integrals with respect to the measures $Q_{X}$ and $Q_{Y|X}$, respectively.
Third step: "Pulled out" the integral over $B$ and replaced $Q_X(dx)Q_{Y|X}(dy, x)$ by $Q_{X,Y}(dx, dy)$ (the same conditioning argument that you used).
Final implication: Comparing my derivation with your last equation yields the result.
