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$x, y$ 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 $x$.

Please help.

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 $x$.

$x, y$ are random variables jointly distributed on $[0,1]^2$. The marginal distribution of $x$ is uniform. It is also known that $E[y]=E[x]=\frac12$, $E[x|y]=y$, so $y$ second order stochastically dominates $x$. We also know that $E[y|x]$ is non-decreasing (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 $x$.

Please help.

$x, y$ 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 $x$.

Please help.

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

I am trying to further characterize $E[y|x]$. It seems that the following might be true 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 $x$.

Please help.

$x, y$ are random variables jointly distributed on $[0,1]^2$. The marginal distribution of $x$ is uniform. It is also known that $E[y]=E[x]=\frac12$, $E[x|y]=y$, so $y$ second order stochastically dominates $x$. We also know that $E[y|x]$ is non-decreasing (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 $x$.

Please help.