X is stochastically increasing in Y $\implies$ $E\left[Y| X\right]$ increasing in X I have two random variables, $X$ and $Y$.  I know:
$\text{pr}\left(X \le u | Y\right)$ is a decreasing function of $Y$ for all $u$.
Does this imply that: $\mathbb{E}\left[Y | X\right]$ is increasing in $X$?
Intuitively this makes sense.  If large values of $Y$ imply we're more likely to see large values of $X$, seeing large values of $X$ should make us expect larger values of $Y$.
Things I have been able to figure out:
I know the covariance of $X$ and $Y$ is positive (so the statement is true in the case of linear expectations, $\mathbb{E}\left[Y|X\right] = \alpha + \beta x$).
Of course, $\mathbb{E}\left(X|Y\right)$ is increasing in $Y$.
A specific application of the linear result: If $X$ is binary and $\text{pr}\left(X = 1 | Y\right) = Y$, then $\mathbb{E}\left[Y|X = 1\right] \ge \mathbb{E}\left[Y | X=0\right]$.
 A: No: $\mathbb{E}(Y|X)$  does not have to be an increasing function.
Suppose $X$ and $Y$ can take on the values $0,1,2$, with the following probabilities for $(X,Y)$:
$$\Pr(0,0)=1/3;\ \Pr(0,1)=\Pr(2,1)=\Pr(2,2)=1/6;\ \Pr(0,2)=\Pr(1,2)=1/12.$$
From this we readily compute the conditional distributions of $X$ and $Y$, verifying they meet the conditions.  The only things to check are:


*

*$\Pr(X|0)$ rises from $0$ to $1$ at $X=0$.

*$\Pr(X|1)$ rises from $0$ to $1/2$ at $X=0$ and then jumps to $1$ when $X=2$.

*$\Pr(X|2)$ rises from $0$ to $1/4$ at $X=0$, jumps to $1/2$ when $X=1$, then leaps to $1$ when $X=2$.
At each stage, the values are never greater than at the previous stage.
However, the conditional expectations of $Y$ are 
$$\mathbb{E}(Y|X=0) = 1;\ \mathbb{E}(Y|X=1)=2;\ \mathbb{E}(Y|X=2)=3/2.$$
It's not even a monotonic progression.
The idea underlying this counterexample is that some probability can be pushed up to focus on the highest values of $Y$ in the middle of the range of $X$ (in this case, when $X=1$) without destroying the condition on the conditional distributions of $X$.  You only need to move a very tiny bit of probability around to completely distort the conditional distribution of $Y$ at these values of $X$.
