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.

  • $\begingroup$ "their product is automatically integrable"... Why is that? $\endgroup$
    – user140541
    Aug 29, 2021 at 16:07
  • $\begingroup$ @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 $Y$. $\endgroup$ Aug 29, 2021 at 16:15
  • $\begingroup$ Alternatively I prove it in the proof following my question - consider the case where $\mathscr{G}$ is the trivial $\sigma$-algebra $\{\emptyset,\Omega\}$. $\endgroup$ Aug 29, 2021 at 16:18
  • $\begingroup$ OK. $X$ and $Y$ are independent... $\endgroup$
    – user140541
    Aug 29, 2021 at 16:18

1 Answer 1


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 $$\mathbf E[X]\,\mathbf E[Y]=\mathbf E[X]^2\,\mathbf E[U]=\frac18.$$

  • $\begingroup$ Thanks, I like your counter example. $\endgroup$ Aug 29, 2021 at 18:21

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