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Question

Assume that for integrable random variables $X$ and $Y$ we have $E(X|Y)=Y$ and $E(Y|X)=X$.

Show that for $x\in \mathbb{R}$ we have

$0\leq E[(X-Y)1(Y\leq x< X)]=E[(Y-X)1(X>x,Y>x)]$ $0\geq E[(X-Y)1(X\leq x< Y)]=E[(Y-X)1(X>x,Y>x)]$

Conclude that $X=Y$ a.s.

PS: 1 stands for indicator variable

Attempt Pick $x\in \mathbb{R}$

\begin{align} E[(X-Y)1(Y\leq x\leq X)]&=E[X1(Y\leq x<X)]-E[Y1(Y\leq x<X)]\\ &=E[X1(Y\leq x<X)]-E[Y1(Y\leq x<X)]\\ &=E[X1(X>x)]-E[Y1(Y\leq x)]\\ &=E[X1(X>x)]+E[Y1(Y>x)]\\ &=E[X1(X>x)]+E[Y1(Y>x)]\\ &=E[(X+Y)1(X>x,Y>y)] \end{align}

Clearly i did something wrong? Any thoughts/hints? I don't even understand where i could use the assumptions..

Edit

In the post Show that $X = Y$ almost surely he show how to prove $X=Y$ a.s. from our assumptions.

José Duarte de Azevedo e Cunha
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1 Answers1

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Actually I think $X=Y$ a.s. can be proved simpler by properties of conditional expectation(CE). You see CE is basically $L^2$-projections, so the condition roughly says $X$ after two projections remains the same. Since projection is shrinking in $L^2$ sense, this implies the following:

\begin{align*} EX^2 &= E\left(E\left[E[X\mid Y]\mid X\right]\right)^2\\ &\leq E[E[(E[X\mid Y])^2\mid X]] \\ &= E[(E[X\mid Y])^2]\\ &\leq EX^2, \end{align*}

so the equality by Jensen implies $Y=E[X\mid Y]=X$ a.s. on $\sigma(Y)$. Similarly we can conclude $X=Y$ a.s. on $\sigma(X)$, and we are done.

As for the approach suggested in context, note that $$1(X>x\geq Y)=1(X>x)-1(X>x,Y>x),$$

so basically for the first equality you only need to show $$E[X1(X>x)]=E[Y1(X>x)].$$

This is easily done by the condition given: $$E[Y1(X>x)]=E[E[Y \mid X]1(X>x)]=E[X1(X>x)].$$

MikeG
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  • That $X = Y$ when they both belong to $\mathscr{L}^2$ is relatively easy as one only needs to expand out $E((X-Y)^2)$ by conditioning on $X$ and on $Y.$ However, when $X$ or $Y$ do not belong to $\mathscr{L}^2,$ but only $\mathscr{L}^1,$ then you need to do other tricks. – William M. Apr 12 '22 at 18:44
  • I see. That makes sense. – MikeG Apr 12 '22 at 18:46
  • In fact, out of my memory, the proof of the $\mathscr{L}^1$ case, I don't know. I've seen it here in MSE but I don't recall right now. – William M. Apr 12 '22 at 18:48
  • I think I just wrote it in OP's manner which works in $L^1$ case. I would admire if you could check whether it is correct. – MikeG Apr 12 '22 at 18:52
  • That is correct, ${X > x}$ is $X$-measurable, so $E(X \mathbf{1}{{X > x}}) = E(Y \mathbf{1}{{Y > x}})$ by definition of CE. – William M. Apr 12 '22 at 19:00