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Let $X \in L^1(\Omega, F, P)$ r.r.v., $Y \in L^1(\Omega, F, P)$ r.r.v. and $G \subset F$ be a sub-algebra. (r.r.v. Real Random Variable)

$X = E[X|G] $ in law $\Rightarrow$ $X = E[X|G]$ a.s.

$Y = E[X|Y] $ and $X = E[Y|X]$ $\Rightarrow$ $Y=X$ a.s.

For the first property, I think I should prove that $X$ is $G$-measurable but I could not find a way to use the given information of distribution equality.

For the second property, I have an idea but I would like to be sure of it. We can see that for any set $A$ $X$-measurable and $Y$-measurable, $E[X1_A ] = E[Y1_A]$. Suppose that $A = \{Y>X\}$, we get $E[(Y-X)1_A]=0$ as $A$ is $X$-measurable and $Y$-measurable. As $Y>X$ over $A$ we necessarily have $1_A = 0$. Thus, $Y\leq X$ a.s.. In the same way, $X\leq Y$ a.s..

Thank you in advance

jean
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  • The second one has been discussed before: https://math.stackexchange.com/q/666843/321264. – StubbornAtom Feb 10 '22 at 15:40
  • @StubbornAtom ok I will check that, Thanks. What do you think of my answer ? – jean Feb 10 '22 at 15:44
  • What does r.r.v. mean? – William M. Feb 10 '22 at 16:50
  • @WilliamM. Real Random Variable – jean Feb 10 '22 at 17:16
  • How do you show ${Y > X}$ is $X$-measurable? – Brian Moehring Feb 10 '22 at 17:59
  • To elaborate, you seem to be mixing up being $(X,Y)$-measurable with being both $X$-measurable and $Y$-measurable. It is obvious that ${Y>X}$ is $(X,Y)$-measurable, but it feels like begging the question to claim it's either $X$-measurable or $Y$-measurable. – Brian Moehring Feb 10 '22 at 18:29
  • @BrianMoehring I totally agree with you. What if I say : We can see that for any set $A$ $X$-measurable and $Y$-measurable, $E[X1_A ] = E[Y1_A]$. ( Thus this is true for $A$ (X,Y)-measurable ? ) Suppose that $A = {Y>X}$, we get $E[(Y-X)1_A]=0$ as $A$ is (X,Y)-measurable. As $Y>X$ over $A$ we necessarily have $1_A = 0$. Thus, $Y\leq X$ a.s.. In the same way, $X\leq Y$ a.s.. – jean Feb 13 '22 at 10:28
  • You would need to prove that it's true for all $(X,Y)$-measurable $A$. Just saying it's so is effectively begging the question. In other words, you have a condition you know to be true for all $A \in\mathcal{F}X\cap \mathcal{F}_Y$ and you want to show it's true for all $A\in\mathcal{F}{(X,Y)}=\mathcal{F}_X\vee\mathcal{F}_Y.$ Without an argument for why this is true (e.g. $\mathcal{F}_X=\mathcal{F}_Y$) there's no reason for this to be valid. – Brian Moehring Feb 13 '22 at 10:47
  • What does it even mean that $X = E(X \mid \mathscr{G})$ in law? You start with two random variables $X, Y$ on the same probability space $(\Omega, \mathscr{F}, \mathbf{P})$ and then you consider the image measures (a.k.a. distributions) of $\mathbf{P}$ by $X$ and $Y,$ namely $F_X = X(\mathbf{P}):A \mapsto P(X \in A)$ and similarly, $F_Y.$ To say that "$X = Y$ in law" means that $F_X = F_Y$ on the Borel sets. However, $X$ and $\mathbf{E}(X \mid \mathscr{G})$ live on different spaces. Can you please clarify? It may very well be that the question is nonesensical. – William M. Feb 25 '22 at 20:03

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