Let $\mathcal{F}_1\subseteq\mathcal{F}_2\subseteq\mathcal{F}_3$ be a filtration, and X, Y two random variables, where X and Y are both $\mathcal{F}_3$-measurable.
Can the expression:
$$\mathbb{E}[(X\mathbb{E}[Y|\mathcal{F}_2])^+|\mathcal{F_1}]$$ be rewritten as
$$\mathbb{E}[(\mathbb{E}[XY|\mathcal{F}_2])^+|\mathcal{F_1}]$$ knowing that we condition to $\mathcal{F}_1$? Please motivate your answer.
No.
Take for example $\mathcal F_1=\mathcal F_2=\{\emptyset,\Omega\}$, $\mathcal F_3=\mathcal F$ (the $\sigma$-algebra which endows the universe $\Omega$) and $X=Y\sim\mathcal N(0,1)$.
On the one hand, $$ \mathbb{E}[(X\mathbb{E}[Y|\mathcal{F}_2])^+|\mathcal{F_1}]=\mathbb{E}[(X\mathbb{E}[Y])^+]=0, $$ but on the other hand, $$ \mathbb{E}[(\mathbb{E}[XY|\mathcal{F}_2])^+|\mathcal{F_1}]=\mathbb{E}[(\mathbb{E}[XY])^+]=\mathbb E[1]=1. $$