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Given $X,Y \in L^2(\Omega,\mathscr{F},\Bbb{P})$ such that

$\mathbb{E}[X|Y] = Y$ a.s.

$\mathbb{E}[Y|X] = X$ a.s.

show that $\Bbb{P}(X = Y ) = 1.$

$Attempt: $

I can see that $\mathbb{E}[X|Y] = Y$ means

$$ \int_{\{Y\in{A}\}}X \, d\Bbb{P} = \int_{\{Y\in{A}\}}Y \, d\Bbb{P} $$ for any $A \subset{\Bbb{R}} $ Borel, so

$$ \int_{\{Y\in{A}\}}X - Y \, d\Bbb{P} = 0 $$ over any such sets. Similarly for sets of the form $\{X \in A \}$.

Now if I could show that $ \int_U X - Y \, d\Bbb{P} = 0$ for any set $U \subset \Omega$ of the form $\{X \lt a, Y \lt b\}$ for $a,b \in \Bbb{R}$ then I'd be done, because sets of that form are a $\pi$-system that generates $\sigma(X,Y)$, so I'd have (by a lemma) that the integral of $X-Y$ vanishes on all $\sigma(X,Y)$ sets - so of course, it would be zero. But I can't work out how to do that.

I'm given the hint that $$ \int_{ \{ X \gt c, Y \le c \} } X - Y \, d\Bbb{P} + \int_{ \{ X \le c, Y \le c \} } X - Y \, d\Bbb{P} = 0 \; \text{for all} \; c $$

because that's the integral over $\{Y\le c\}$, a set of the above form. With the condition that $X$ and $Y$ are integrable, this is exercise 9.2 in Williams' "Probability with Martingales". Driving me up the wall to get so stuck on what seems such a simple exercise!

Did
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Latimer Leviosa
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  • Williams asks for the case when X and Y are integrable, not necessarily square integrable. The method to solve this case is quite different from the one used in the accepted answer and, indeed, fully uses the hint provided in the book and recalled in your question. – Did Apr 15 '15 at 19:38
  • Damn you're right. I don't know why I read $L^2$ , I guess I've been doing too much Hilbert spaces. – Latimer Leviosa Apr 15 '15 at 20:43
  • Related: https://math.stackexchange.com/questions/34101/conditional-expectation-ea-mid-b-b-and-eb-mid-a-a-implies-a-b – StubbornAtom Jun 23 '19 at 20:12
  • https://math.stackexchange.com/q/666843/321264 – StubbornAtom Mar 09 '20 at 10:45

1 Answers1

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Since

$\mathbb E[X^2] = \mathbb E[X\mathbb E[Y|X]] = \mathbb E[\mathbb E[XY|X]] = \mathbb E[XY]= \mathbb E[\mathbb E[XY|Y]] = \mathbb E[Y\mathbb E[X|Y]] = \mathbb E[Y^2], $

observe that $$ \mathbb E[(X-Y)^2]=\mathbb E[X^2+Y^2-2XY]=0, $$

which implies that $X$ and $Y$ differ on a set of measure zero.

For the weaker condition where $X$ and $Y$ are merely integrable we proceed as follows. Choose an arbitrary rational number $c\in\mathbb Q$. Note that

$$ \mathbb E[(X-Y){\bf 1}_{Y\leqslant c}]=\mathbb E[(X-Y){\bf 1}_{Y\leqslant c}{\bf 1}_{X\leqslant c}]+\mathbb E[(X-Y){\bf 1}_{Y\leqslant c}{\bf 1}_{X> c}]\tag{1} $$ Reversing the roles of $X$ and $Y$ gives $$ \mathbb E[(X-Y){\bf 1}_{X\leqslant c}]=\mathbb E[(X-Y){\bf 1}_{X\leqslant c}{\bf 1}_{Y\leqslant c}]+\mathbb E[(X-Y){\bf 1}_{X\leqslant c}{\bf 1}_{Y> c}]\tag{2} $$ The lhs of each of (1) and (2) is zero, since $\mathbb E[(X-Y){\bf 1}_{X\leqslant c}]=\mathbb E[(X-\mathbb E[X|Y]){\bf 1}_{X\leqslant c}]=$$\mathbb E[X{\bf 1}_{X\leqslant c}]-\mathbb E[X{\bf 1}_{X\leqslant c}]=0$, for instance. This, in turn, implies from (1) and (2) that $$ \mathbb E[(X-Y){\bf 1}_{X\leqslant c}{\bf 1}_{Y> c}]=\mathbb E[(X-Y){\bf 1}_{Y\leqslant c}{\bf 1}_{X> c}],\tag{3} $$

a proposition that equates a surely non-positive number to a surely non-negative number. Thus, the lhs and rhs of (3) are surely zero. The events $[X\leqslant c]\cap[Y> c]$ and $[Y\leqslant c]\cap[X> c]$, therefore, must each be null events. But this holds for all $c$, so that (by countable unions) the events

$$[X<Y]=\bigcup_{c\in\mathbb Q}[X\leqslant c]\cap[Y> c]\ \ \text{and}\ \ [Y<X]=\bigcup_{c\in\mathbb Q}[Y\leqslant c]\cap[X> c]$$

are also null events. That is, $[X\ne Y]$ is a null event. This completes the proof.

ki3i
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  • Absolutely wonderful, thank you. Didn't need to faff around with integrals over this set or that one. – Latimer Leviosa Apr 15 '15 at 19:16
  • @LatimerLeviosa, You are welcome. – ki3i Apr 15 '15 at 19:22
  • @Did, I haven't looked at Williams; I simply used square-integrability as that was what the OP indicated. Yes, for $\mathbb L^1(P)$ the proof is indeed different. – ki3i Apr 15 '15 at 19:42
  • I just realised that I did want $L^1$ after all, feeling foolish. I'm gonna change the question and I guess I have to dis-accept your answer, but I've still given a +1 so hope that's okay! If you or anyone else could give the $L^1$ case that'd be great... – Latimer Leviosa Apr 15 '15 at 20:50
  • @LatimerLeviosa Not the thing to do, leave the $L^2$ condition and, if you want to see Williams' exercise solved, ask another question. – Did Apr 15 '15 at 20:51
  • What should I do then? Ask the question again with L1 instead? – Latimer Leviosa Apr 15 '15 at 20:53
  • @Did please see edits – ki3i Apr 15 '15 at 20:54
  • Thanks so much for editing the answer! Will be more careful in future. – Latimer Leviosa Apr 15 '15 at 20:59
  • @ki3i Indeed this is the idea of the $L^1$ proof. Note however that $\mathbb E[(X-Y){\bf 1}{Y\leqslant c}{\bf 1}{X> c}]>0$ is not guaranteed, only $\mathbb E[(X-Y){\bf 1}{Y\leqslant c}{\bf 1}{X> c}]\geqslant0$. And later on, the step [$\mathbb E[(X-Y){\bf 1}{Y\leqslant c}{\bf 1}{X\leqslant c}]<0$ for every $c$ implies $X<Y$ almost surely] is unclear. – Did Apr 15 '15 at 21:00
  • @Did, I will improve the weaker argument, thanks – ki3i Apr 15 '15 at 21:44
  • @Did, improvements made. Please see edits. Thank you. – ki3i Apr 15 '15 at 22:35
  • This is the correct proof, +1. – Did Apr 16 '15 at 07:17
  • One quick edit. To see that $\displaystyle \int_{X\geq c, Y<c}(X-Y);d\mathbb{P}=0\rightarrow \mathbb{P}(X\geq c>Y) =0$, we need the following line. Notice $0 \leq \mathcal{X}{X\geq c,Y\leq c-\frac{1}{n}}\nearrow \mathcal{X}{X\geq c,Y<c}$. Hence, by Monotone convergence Theorem, $0\geq \int_{X\geq c,Y\leq c-\frac{1}{n}}(X-Y);d\mathbb{P}\geq\frac{1}{n}\mathbb{P}(X\geq c\geq Y+\frac{1}{n})$, hence the last set(s) indexed with $n$ must all have measure $0$. – TBTD Jul 08 '17 at 16:26
  • @Aaron Not necessary at all, simply note that $$(X-Y)\mathbf 1_{X\ge c>Y}\geqslant0$$ almost surely, and that if $P(Z\geqslant0)=1$ and $E(Z)=0$ then $P(Z=0)=1$ almost surely. – Did Jul 10 '17 at 00:18