Here is a problem I couldn't solve.
Given a probability space $ (\Omega, \mathcal{A},\mathbb{P}) $, and $ \mathcal{F}, \mathcal{G}, \mathcal{B} $ sub-$\sigma$-algebras of $\mathcal{A}$.
Is it true that coniditional independence of $\mathcal{B}$ and $\mathcal{F}$ given $\mathcal{G}$ which we can note $\mathcal{F}\;\amalg_{\mathcal{G}} \mathcal{B}$ implies the following property :
$\mathbb{P}(F\cap B \mid G) = \mathbb{P}(F \mid G)\mathbb{P}(B \mid G),$ almost surely $ \;\; \forall \;(F,G,B)\in \mathcal{F} \times \mathcal{G}\times \mathcal{B}$
Where I note :
$\mathbb{P}(A \mid G) =\mathbb{E}[1_A|\sigma(G)], \forall A\in \mathcal{A}$ or more generally given any sub sigma algebra $\mathcal{C}$ of $\mathcal{A}$ :
$\mathbb{P}(A \mid \mathcal{C}) =\mathbb{E}[1_A|\mathcal{C}], \forall A\in \mathcal{A}$
And where conditional independence $\mathcal{F}\;\amalg_{\mathcal{G}} \mathcal{B}$ is defined by :
$\mathbb{P}(F\cap B \mid \mathcal{G}) = \mathbb{P}(F \mid \mathcal{G})\mathbb{P}(B \mid \mathcal{G}),$ almost surely $ \;\; \forall \;(F,B)\in \mathcal{F} \times \mathcal{B}$.
I think this implication is not true but I can't find a counter-example.
Moreover I think the reverse implication holds true.(check edit)
Best Regards
Edit : I thought I had a proof for the reverse implication but as gnometorule suggested it wasn't true, I rechecked it and found a mistake, so I take that last comment back, as I don't know if it is true or not.