I have found the solution of this problem to the following paper http://alexanderchernyy.com/pmt.pdf , theorem 2.1, page 4. However I am struggling to convince myself about the statement:
$$ E(X_t Y_t|\mathcal{F}^X_s\vee\mathcal{F}^Y_s)= X_s Y_s $$ implies $$E(X_t Y_t|\mathcal{F}^{XY}_s) = X_s Y_s $$
Where does this implication come from?
On
Note that $\mathcal{F}_s^{XY} \subseteq \mathcal{F}_s^X \vee \mathcal{F}_s^Y$ for all $s \geq 0$; this follows from the fact that the product $X_s Y_s$ is measurable with respect to $\mathcal{F}_s^X \vee \mathcal{F}_s^Y$. Hence, by the tower property of conditional expectation,
$$\mathbb{E}(X_t Y_t \mid \mathcal{F}_s^{XY}) = \mathbb{E} \bigg[ \underbrace{\mathbb{E}(X_t Y_t \mid \mathcal{F}_s^X \vee \mathcal{F}_s^Y)}_{=X_s Y_s} \mid \mathcal{F}_s^{XY} \bigg] = X_s Y_s.$$
I would rather use the $\pi - \lambda$ theorem (and not monotone class theorem). Sets of the type $A\cap B$ with $A \in \mathcal F^{X}_s$ and $B \in \mathcal F^{Y}_s$ form a $\pi$ system and the class defined in that reference is a $\lambda$ system containing this $\pi$ system. Hence it contains the sigma algebra generated, which is $\sigma(\mathcal F^{X}_s \cup \mathcal F^{Y}_s)$. Final step where this last sigma algebra is replaced by $\mathcal F^{XY}_s$ is immediate from definition of conditional expectation.