Characterization of conditional independence Definition: Let $\mathcal{G},\mathcal{K},\mathcal{H}$ be $\sigma $-subalgebras of $\mathcal{F}$, where $\left( {\Omega ,\mathcal{F},\mathbb{P}} \right)$ is a given probability space. We say that $\mathcal{G}$ and $\mathcal{H}$ are conditionally independent given $\mathcal{K}$ if, $\mathbb{P}$-almost surely, $\mathbb{P}\left( {G \cap H|\mathcal{K}} \right) = \mathbb{P}\left( {G|\mathcal{K}} \right)\mathbb{P}\left( {H|\mathcal{K}} \right),\forall G \in \mathcal{G},\forall H \in \mathcal{H}$.
Let $\mathcal{K} \subseteq \mathcal{H}$. Prove that $\mathcal{G}$ and $\mathcal{H}$ are conditionally independent given $\mathcal{K}$ if and only if, $\mathbb{P}$-almost surely, $\mathbb{P}\left( {G|\mathcal{K}} \right) = \mathbb{P}\left( {G|\mathcal{H}} \right),\forall G \in \mathcal{G}$.
My attempt: So, I have to prove $\mathbb{P}\left( {G|\mathcal{K}} \right) = \mathbb{P}\left( {G|\mathcal{H}} \right),\forall G \in \mathcal{G} \Leftrightarrow \mathbb{P}\left( {G \cap H|\mathcal{K}} \right) = \mathbb{P}\left( {G|\mathcal{K}} \right)\mathbb{P}\left( {H|\mathcal{K}} \right),\forall G \in \mathcal{G},\forall H \in \mathcal{H}$.
I wrote down everything:
$\mathbb{P}\left( {G|\mathcal{K}} \right) = \mathbb{E}\left[ {{1_G}|\mathcal{K}} \right],\mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{1_K}} \right] = \mathbb{E}\left[ {{1_G}{1_K}} \right]$
$\mathbb{P}\left( {G|\mathcal{H}} \right) = \mathbb{E}\left[ {{1_G}|\mathcal{H}} \right],\mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}|\mathcal{H}} \right]{1_H}} \right] = \mathbb{E}\left[ {{1_G}{1_H}} \right]$
$\mathbb{P}\left( {G \cap H|\mathcal{K}} \right) = \mathbb{E}\left[ {{1_{G \cap H}}|\mathcal{K}} \right],\mathbb{E}\left[ {\mathbb{E}\left[ {{1_{G \cap H}}|\mathcal{K}} \right]{1_K}} \right] = \mathbb{E}\left[ {{1_{G \cap H}}{1_K}} \right]$
$\mathbb{P}\left( {H|\mathcal{K}} \right) = \mathbb{E}\left[ {{1_H}|\mathcal{K}} \right],\mathbb{E}\left[ {\mathbb{E}\left[ {{1_H}|\mathcal{K}} \right]{1_K}} \right] = \mathbb{E}\left[ {{1_H}{1_K}} \right]$
but so far, I have just been running in circles. How do I show the equivalence?
 A: Suppose $\mathcal{G}$ and $\mathcal{H}$ are conditionally independent given $\mathcal{K}$ and $\mathcal{K} \subseteq \mathcal{H}$. We want to show that, $\mathbb{P}$-almost surely, $\mathbb{P}\left( {G|\mathcal{K}} \right) = \mathbb{P}\left( {G|\mathcal{H}} \right),\forall G \in \mathcal{G}$, i.e. $\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right] = \mathbb{E}\left[ {{1_G}|\mathcal{H}} \right]$.
Since $\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]$ is $\mathcal{K}$-measurable and $\mathcal{K} \subseteq \mathcal{H}$, it follows that $\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]$ is $\mathcal{H}$-measurable, so it suffices to show that for each $H \in \mathcal{H}$ we have $\mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{1_H}} \right] = \mathbb{E}\left[ {{1_G}{1_H}} \right]$.
$\mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{1_H}} \right]\mathop  = \limits_{\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{1_H}{\text{ is }}\mathcal{K}{\text{ - measurable}}}^{\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{1_H} = \mathbb{E}\left[ {\mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{1_H}|\mathcal{K}} \right]} \right]} \mathbb{E}\left[ {\mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{1_H}|\mathcal{K}} \right]} \right] = $
$\mathop  = \limits_{\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{\text{ is bounded and }}\mathcal{K}{\text{ - measurable}}}^{\mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{1_H}|\mathcal{K}} \right] = \mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]\mathbb{E}\left[ {{1_H}|\mathcal{K}} \right]} \mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]\mathbb{E}\left[ {{1_H}|\mathcal{K}} \right]} \right]\mathop  = \limits^{\mathbb{P}\left( {G \cap H|\mathcal{K}} \right) = \mathbb{P}\left( {G|\mathcal{K}} \right)\mathbb{P}\left( {H|\mathcal{K}} \right)} \mathbb{E}\left[ {\mathbb{E}\left[ {{1_{G \cap H}}|\mathcal{K}} \right]} \right]$
$ = \mathbb{E}\left[ {{1_{G \cap H}}} \right] = \mathbb{E}\left[ {{1_G}{1_H}} \right]$
Conversely, suppose that for each $G \in \mathcal{G}$ we have, $\mathbb{P}$-almost surely, $\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right] = \mathbb{E}\left[ {{1_G}|\mathcal{H}} \right]$. Then for any $G \in \mathcal{G}$ and $H \in \mathcal{H}$ we have
$\mathbb{E}\left[ {{1_{G \cap H}}|\mathcal{K}} \right] = \mathbb{E}\left[ {{1_G}{1_H}|\mathcal{K}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {{1_G}{1_H}|\mathcal{H}} \right]|\mathcal{K}} \right] = \mathbb{E}\left[ {{1_H}\mathbb{E}\left[ {{1_G}|\mathcal{H}} \right]|\mathcal{K}} \right] = $
$\mathop  = \limits^{{\text{assumption}}} \mathbb{E}\left[ {{1_H}\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]|\mathcal{K}} \right]\mathop  = \limits^{\mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]{\text{ is bounded and }}\mathcal{K}{\text{ - measurable}}} \mathbb{E}\left[ {{1_G}|\mathcal{K}} \right]\mathbb{E}\left[ {{1_H}|\mathcal{K}} \right]$.
