$\mathcal{A}\perp_\mathcal{G}\mathcal{B}\wedge\mathcal{H}\subseteq\mathcal{G}\implies\mathcal{A}\perp_\mathcal{H}\mathcal{B}$?

Question

If $\mathcal{A}\perp_\mathcal{G}\mathcal{B}$ and $\mathcal{H}\subseteq\mathcal{G}$, is it the case that $\mathcal{A}\perp_\mathcal{H}\mathcal{B}$? Here $\mathcal{A}$, $\mathcal{B}$, $\mathcal{G}$ and $\mathcal{H}$ are sub-$\sigma$-algebras of some common probability space $\left(\Omega,\mathcal{F},P\right)$ and $\cdot\perp_\square\cdot$ is the conditional independence relation.

Answer 1

Consider a complete probability space for all the following underlying $\sigma$-fields. Then $\mathcal A\cap \mathcal B\subset\mathcal G$ (take $A$ in the intersection, then $P(A\mid\mathcal G)\in\{0,1\}$, and completeness ensures that $A\in\mathcal G$). To get a counter-example, take $\mathcal H:=\mathcal A\cap\mathcal B=\{\emptyset,X\}$ and $\mathcal A$ and $\mathcal B$ are not independent.

Answer 2

I'd like to describe a slight variation on Davide's answer that i find crisper.

Suppose $\mathcal{A}$ and $\mathcal{B}$ are dependent fields that are conditionally independent given $\mathcal{G}$ (see, e.g., the Wikipedia article on Conditional Independence) and let $\mathcal{H}:=\left\{\emptyset,\Omega\right\}$. Then independence conditional on $\mathcal{H}$ is the same as unconditional independence, hence the desired result.