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

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)

• Yes, it is true and it can be proved by approximating $X$ and $Y$ by bounded random variables. $X_n=X1_{|X|<n}$ and $Y_n=Y1_{|Y|<n}$ will work. – Georgii Riabov May 26 '18 at 18:31

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.