When $ \mathbb{E}(XY|\mathcal{A}) = \mathbb{E}(X|\mathcal{A})Y? $

Question

Let $\mathcal{A}\subseteq\mathcal{B}$ be a sub $\sigma$-algebra on the probability space $(\Omega,\mathcal{B},\mu)$, $X,Y$ real random variables being $X$ $\mathcal{B}$-measurable and $Y$ $\mathcal{A}$-measurable. It is true that $$ \mathbb{E}(XY|\mathcal{A}) = \mathbb{E}(X|\mathcal{A})Y? $$ I know it is true for bounded $Y$ and integrable $X$, but I think it also is true whenever $X$ and $XY$ are integrable (we only need $\mathbb{E}(X),\mathbb{E}(XY)$ to be defined)

Answer 1

The mentioned statement is true. After having decomposed $X$ and $Y$ into positive and negative part, it suffices to treat the case where $X$ and $Y$ are both non-negative, by linearity of conditional expectation. Define $Y_n$ by giving the value $Y$ if $Y\leqslant n$ and zero otherwise. Then $Y_n$ is a bounded $\mathcal A$-measurable random variable hence $$ \tag{*}\mathbb{E}(XY_n|\mathcal{A}) = \mathbb{E}(X|\mathcal{A})Y_n $$ Since $Y_n\leqslant Y_{n+1}$ almost surely, $Y_n\to Y$ a.s. and $X$ is non-negative, the monotone convergence theorem for conditional expectation gives that the left hand side of (*) converges to $\mathbb{E}(XY|\mathcal{A})$ almost surely.

Answer 2

Yes you are right. It is true whenever X and XY are integrable, i.e., their expectation is well-defined.