Let $F$ and $G$ be two sigma algebras where $F$ is contained in $G$. Normally we will have the following:
$E[E[X|G]|F] = E[X|F]$, so I wonder if $E[E[X|F]|G] = E[X|G]$.
I am asking the above question because I have the following :
$E[X|F]*E[X|G]$, since $F$ is contained in G, and $E[X|F]$ is $F$ measurable, then $E[X|F]$ is $G$ measurable , so $E[X|F]*E[X|G] = E[E[X|F]*X|G]$.
Of course not. Try $F=\{\varnothing,\Omega\}$ and $G=\sigma(X)$, then $E(X\mid F)=E(X)$ hence $E(E(X\mid F)\mid G)=E(X)$ while $E(X\mid G)=X$.