E[X|Y] and E[Y|X]

Question

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 $x$.

Answer 1

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.