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The pull out property of conditional expectations ($E(XY| \xi) = X E(Y|\xi)$-a.s if $X$ is $\xi$-measurable) is formulated (as least in my book by Bauer) either for $X \ge 0, Y \ge 0$ or for $X \in \mathcal L^p(P), Y \in \mathcal L^q(P)$ where $p \in [1, + \infty]$ and $\frac{1}{p} + \frac{1}{q} = 1$.

The case $X \ge 0, Y \ge 0$ can be easily proved using the property:

$\int ZX d P = \int Z E(X| \xi) dP, \forall Z \ge 0, \xi$-measurable, $X \ge 0$.

But I do not see how to prove the second case (Bauer says "analogously").

Edit

I will give it a shot as P.Jo suggested. If $X \in \mathcal L^p(P)$ and $Y \in \mathcal L^q(P)$, then so $Y^+:=\max(Y,0),Y^-:=-\min(Y,0),X^+:=\max(X,0),X^--\min(X,0)$ because $|X| \ge X^+$ and $|X| \ge X^-$ (the same for Y). Thus $X^+Y^-,X^+Y^+,X^-Y^-,X^-Y^+$ are integrable.

Now using the fact that the pull out property works for non-negative random variables we can show that

$(1)$ $E(X^-Y^+|\xi)=X^-E(Y^+|\xi)$-a.s

$(2)$ $E(X^-Y^-|\xi)=X^-E(Y^-|\xi)$-a.s

$(3)$ $E(X^+Y^+|\xi)=X^+E(Y^+|\xi)$-a.s

$(4)$ $E(X^+Y^-|\xi)=X^+E(Y^-|\xi)$-a.s

If we now calculate $[(1)-(3)]-[(4)-(2)]$ using the fact that $X=X^+-X^-,Y=Y^+-Y^-$, then we get the pull-out property.

The same way the pull-out property can be proved for $X,Y \in \mathcal L^1(P)$ if $X*Y$ is integrable.

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I guess you could decompose X as well as Y into their positive and negative parts - then use the fact that each product term is in $L^1$ (according to Hölder-inequality) and the additivity of the conditional expectation.

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  • $\begingroup$ I gave it a try above. $\endgroup$
    – zesy
    Sep 28, 2015 at 19:38

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