Consider $X$ and $Y$ tow random variables $\mathcal F_2$-measurable where $\mathcal F_1$ and $\mathcal F_2$ are two $\sigma$-algebras such that $\mathcal F_1 \subseteq \mathcal F_2 $.
Can we always say that $\mathbb E [X ~Y| \mathcal F_1] = \mathbb E[X~\mathbb E [Y | \mathcal F_2] | \mathcal F_1]$ ? Shouldn't $X$ and $Y$ be independent ?
Many thanks,
You only need $X$ is $\mathcal{F}_2$-measurable, which gives: $$ E[XY|\mathcal{F_2}]=X E[Y|\mathcal{F}_2]. $$ Now, using $\mathcal{F}_1\subset\mathcal{F}_2$ and iterated conditioning for the first equality below, we have $$ E[XY|\mathcal{F}_1]=E[E[XY|\mathcal{F_2}]|\mathcal{F}_1]=E[XE[Y|\mathcal{F_2}]|\mathcal{F}_1]. $$