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Let $X\in L^2(\Omega,\mathcal{F},P)$ and $\mathcal{G}\subset\mathcal{F}$ be sub-$\sigma$ fields and define the random variable $Y=E(X\mid\mathcal{G})$, which is to say $Y$ is a $\mathcal{G}$-measurable random variable $\int_{A}YdP\mid_{\mathcal{G}}=\int_{A}XdP\mid_{\mathcal{G}},\forall A\in \mathcal{G}$.

If $X\overset{d}{=}Y$, is it true that $X\overset{a.s.P}{=}Y$?

What if $X\in L^1(\Omega,\mathcal{F},P)$?

What if $X\in L^q(\Omega,\mathcal{F},P)$ for general $q$?

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    $\begingroup$ It is true for $X \in L^1$, hence for all $q$, see math.stackexchange.com/questions/1374777/… $\endgroup$ – clark Jun 23 '17 at 0:07
  • $\begingroup$ @clark: That reference only shows that $X$ and $Y$ have the same sign, a.s. $\endgroup$ – John Dawkins Jun 23 '17 at 17:15
  • $\begingroup$ @JohnDawkins Yes the next step is to notice that $X-q$ and $Y-q$have the same sign a.s. for every $q\in \mathbb{Q}$ $\endgroup$ – clark Jun 23 '17 at 20:38
  • $\begingroup$ Excellent! $\phantom{bbbbb}$ $\endgroup$ – John Dawkins Jun 23 '17 at 20:42
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Assuming only that $X\in L^1(\mathcal F)$, there is an even, strictly convex, function $\Phi:\Bbb R\to(0,\infty)$ such that $\Bbb E[\Phi(X)]<\infty$. [This is a special case of a uniform integrability criterion due to Ch. de la Vallée-Poussin.] By the conditional form of Jensen's inequality, $$ \Bbb E[\Phi(X)|\mathcal G]\ge \Phi\left(\Bbb E[X|\mathcal G]\right)=\Phi(Y), $$ almost surely. Taking expectations and using the fact that $Y$ has the same distribution as $X$: $$ \Bbb E[\Phi(X)]=\Bbb E[\Bbb E[\Phi(X)|\mathcal G]]\ge\Bbb E[\Phi(Y)]=\Bbb E[\Phi(X)], $$ forcing $\Bbb E[\Phi(X)|\mathcal G]=\Phi(Y)$ a.s. Because $\Phi$ is strictly convex, we must have $\Bbb P[X=Y|\mathcal G]=1$ a.s., hence $\Bbb P[X=Y]=1$.

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  • $\begingroup$ Why is even $\Phi$ actually needed? For strictly convex $\Phi$ the equality may fail so I think it shall be convex... $\endgroup$ – Henry.L Jun 23 '17 at 13:47
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Hint for the $L^2$ case: Because $Y$ is orthogonal to $X-Y$, you have $E[X^2] = E[Y^2]+E[(X-Y)^2]$.

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  • $\begingroup$ yep the real interesting case is $L^q$, probably can be easily derived, but I am wondering if the statement itself will hold. $\endgroup$ – Henry.L Jun 22 '17 at 21:09

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