Conditional independence of random variables and $\mathbf{E}[XY]=\mathbf{E}[X]\cdot\mathbf{E}[Y]$

Suppose $$X,Y$$ are integrable random variables on some probability space $$\newcommand{\calF}{\mathscr{F}}\newcommand{\pp}{\mathbb{P}}(\Omega,\calF,\pp)$$ and suppose $$\newcommand{\calG}{\mathscr{G}}\calG\subset\calF$$ is a $$\sigma$$-subalgebra. Neither of the two following conditions implies the other:

1. $$X$$ and $$Y$$ are conditionally independent w.r.t. $$\calG$$, i.e. $$X\;\underset{\calG}{\perp \!\!\!\!\perp}\;Y$$,
2. $$\newcommand{\E}{\mathbf{E}}\E[X\mid \calG]\perp \!\!\!\!\perp\E[Y\mid\calG]$$.

(recall that conditional independence of random variables means that for all $$A\in\sigma(X)$$ and all $$B\in\sigma(Y)$$, $$\E[1_A1_B\mid\calG]=\E[1_A\mid\calG]\cdot\E[1_B\mid\calG]$$.) Indeed this question shows that (2) does not imply (1), and this question shows that (1) does not imply (2).

My question concerns the analogue of one of a useful properties of independence: whenever $$X,Y$$ are integrable and independent, their product is automatically integrable and $$\E[XY]=\E[X]\cdot\E[Y].$$ Let's consider $$X$$ and $$Y$$ nonnegative (for simplicity) integrable real random variables conditionally independent w.r.t. $$\calG$$. It's easy to see that $$\E[XY\mid\calG]=\E[X\mid\calG]\cdot\E[Y\mid\calG]$$.

Question. does it hold that $$XY$$ integrable and $$\E[XY]=\E[X]\cdot\E[Y]$$?

EDIT. I now realize that my question makes no sense: if one takes $$\calG=\calF$$ then $$\E[X\mid\calG]=X$$ $$\pp$$-almost surely (and similarly for $$Y$$) and there is no reason at all why $$XY$$ should be integrable. I guess the question then becomes: what makes $$\calG_0=\{\emptyset,\Omega\}$$ special among $$\sigma$$-subalgebras of $$\calF$$? Is it simply the fact that conditional expectation w.r.t. $$\calG_0$$ produces (almost surely) constant functions?

Proof of $$\,\E[XY\mid\calG]=\E[X\mid\calG]\cdot\E[Y\mid\calG]$$. Let us set $$\left\{\begin{array}{rcl} \displaystyle X_n & = & \sum_{0\leq k< n2^n}\frac{k}{2^n}1_{A_{k,n}},\quad A_{k,n}=\Big[k2^{-n}\leq X< (k+1)2^{-n}\Big]\\ \displaystyle Y_n & = & \sum_{0\leq k< n2^n}\frac{k}{2^n}1_{B_{k,n}},\quad B_{k,n}=\Big[k2^{-n}\leq Y< (k+1)2^{-n}\Big] \end{array}\right.$$ Since $$X,Y$$ are assumed integrable they are almost surely finite and so $$0\leq X_n\nearrow X$$, $$0\leq Y_n\nearrow Y$$ as well as $$0\leq X_nY_n\nearrow XY$$ almost surely. By the conditional monotone convergence theorem we get almost sure convergences $$\left\{\begin{array}{ccccc} \displaystyle 0 & \leq & \E[X_n\mid\calG] & \nearrow & \E[X\mid\calG]\\ \displaystyle 0 & \leq & \E[Y_n\mid\calG] & \nearrow & \E[Y\mid\calG]\\ \displaystyle 0 & \leq & \E[X_nY_n\mid\calG] & \nearrow & \E[XY\mid\calG] \end{array}\right.$$ Now for all $$n$$, $$X_nY_n=\sum_{0\leq k < n2^n}\sum_{0\leq l < n2^n}kl 1_{A_{k,n}}1_{B_{l,n}}$$ and so, by conditional independence w.r.t. $$\calG$$, $$\begin{array}{rcl} \E[X_nY_n\mid\calG] & = & \displaystyle\sum_{0\leq k < n2^n}\sum_{0\leq l < n2^n}kl\cdot\E[1_{A_{k,n}}1_{B_{l,n}}\mid\calG]\\ & \overset{\pp\text{-a.s.}}= & \displaystyle\sum_{0\leq k < n2^n}\sum_{0\leq l < n2^n}kl\cdot\E[1_{A_{k,n}}\mid\calG]\cdot\E[1_{B_{l,n}}\mid\calG]\\ & = & \displaystyle \E[X_n\mid\calG]\cdot\E[Y_n\mid\calG] \end{array}$$ Letting $$n\to+\infty$$ we get $$\E[XY\mid\calG]=\E[X\mid\calG]\cdot\E[Y\mid\calG]~$$ $$\pp$$-almost surely.

• "their product is automatically integrable"... Why is that?
– user140541
Aug 29, 2021 at 16:07
• @d.k.o. Independence is equivalent to the pushforward measure $(X,Y)_*\mathbb{P}$ splitting as $X_*\Bbb{P}\otimes Y_*\Bbb{P}$. You can then apply Fubini's theorem to get the integrability of the product $XY$ and the fact that its expectation is the product of the expectations of $X$ and $Y$. Aug 29, 2021 at 16:15
• Alternatively I prove it in the proof following my question - consider the case where $\mathscr{G}$ is the trivial $\sigma$-algebra $\{\emptyset,\Omega\}$. Aug 29, 2021 at 16:18
• OK. $X$ and $Y$ are independent...
– user140541
Aug 29, 2021 at 16:18

If $$X,Y\in\mathrm L^1(\Omega,\mathscr F,\mathbf P)$$ are independent w.r.t. $$\mathscr G$$, then it is true that $$XY\in\mathrm L^1(\Omega,\mathscr F,\mathbf P(\cdot\mid\mathscr G))$$ with $$\mathbf E[XY\mid\mathscr G]=\mathbf E[X\mid\mathscr G]\,\mathbf E[Y\mid\mathscr G]$$.
That this implies $$\mathbf E[XY]=\mathbf E[X]\,\mathbf E[Y]$$ is not true:
Suppose $$X,U$$ are independent uniform r.v., and let $$Y:=XU$$. Given $$\mathscr G:=\sigma(X)$$, $$X$$ and $$Y$$ are independent (because $$X$$ is “a constant”; we have $$\mathbf E[f(X)g(Y)\mid\mathscr G]=f(X)\int_0^1g(Xu)\,\mathrm du=\mathbf E[f(X)\mid\mathscr G]\,\mathbf E[g(Y)\mid\mathscr G]$$ for any measurable functions $$f,g\ge0$$.) Nonetheless, $$\mathbf E[XY]=\mathbf E[X^2]\,\mathbf E[U]=\frac16,$$ while $$\mathbf E[X]\,\mathbf E[Y]=\mathbf E[X]^2\,\mathbf E[U]=\frac18.$$