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Suppose $x, y$ are random variables jointly distributed on $[0,1]^2$. The marginal distribution of $x$ is uniform. It is also known that $E[y]=E[x]=\frac12$ and $E[x|y]=y$, so $y$ second-order stochastically dominates $x$. We also know that $E[y|x]$ is non-decreasing in $x$ (not sure whether this is helpful.)

I am trying to further characterize $E[y|x]$. It seems that the following might be true for any $c\in[0,1]$ but I have no idea how to prove it:

$\int_0^c E[y|x]dx\ge \int_0^c xdx$.

Imagine if $x$ and $y$ are independent, then $LHS=\frac12c\ge\frac12 c^2=RHS$. Imagine if $y$ reveals $x$ completely, then $E[y|x]=E[E[x|y]|x]=x$. Moreover, the inequality seems to be true if $y$ is a function of $x$.

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  • $\begingroup$ @Minkov, this must be one of the most useless edits on MO, and removing the thanks at the end of the post is even rude. I bet those who approved it didn't think too much about it. $\endgroup$ – Alex M. May 11 at 16:46
  • $\begingroup$ @AlexM. I am sorry that you felt this way. I have retracted my revision. As a moderator, you may find this useful: mathoverflow.net/help/someone-answers ("Please do not add a comment on your question or on an answer to say "Thank you"".) $\endgroup$ – Minkov May 13 at 6:35
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Your conjecture is true. Indeed, take any $c\in[0,1]$. Then $$\int_0^c E(y|x)\,dx=Ey1_{x\le c}\quad\text{and}\quad Ex1_{x\le c}=\int_0^c x\,dx=c^2/2, $$ because $x$ is uniform on $[0,1]$. Next, $$E(x-y)1_{y\le c}=0, $$ because $E(x|y)=y$. Also (which is the key point), $$(y-x)(1_{x\le c}-1_{y\le c})\ge0. $$ So, \begin{align} &\int_0^c E(y|x)\,dx-c^2/2+0 \\ &=Ey1_{x\le c}-Ex1_{x\le c}+E(y-x)1_{y\le c} \\ &=E[(y-x)(1_{x\le c}-1_{y\le c})]\ge0, \end{align} whence $$\int_0^c E(y|x)\,dx\ge c^2/2, $$ as claimed.

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  • $\begingroup$ Thank you! Just one typo: in the third-to-last equation, it should be "...+E(x-y)1..." $\endgroup$ – Lemma1 May 14 at 7:38
  • $\begingroup$ Well, I don't see a typo. The last term in the three-line display should be (and is) written as $+E(y-x)1_{y\le c}$, for the last equality in that display to be obvious. However, this term indeed equals its opposite, $E(x-y)1_{y\le c}$, because the latter is $0$, as shown in the second display in the answer. Do you have any other comments or concerns about the answer? $\endgroup$ – Iosif Pinelis May 14 at 17:26

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