Conditional expectation with mean-preserving spread I have a question which I believe relates to stochastic dominance. I have a random variables $Z \in \lbrace 0, 1 \rbrace$ and $X \in \mathbb{R}$, where $X \sim F_X$ is continuous. I use these to construct a random variable $A = f(Z,X)$, where $f$ is some continuous function such that $f(1,x) \geq f(0,x)$ for all $x$. 
I have another random variable $B = A + E$, where $E \sim F_E$ is some mean zero noise. I want to know if the following is true in general and, if not, under what conditions it is true.
$E[F_A(A) \mid Z = 1] - E[F_A(A) \mid Z = 0] \geq E[F_B(B) \mid Z = 1] - E[F_B(B) \mid Z = 0]$
So far I've observed that $B$ is a mean-preserving spread of $A$ and so $A$ should second-order stochastically dominate $B$. Can I use stochastic dominance to tackle this problem? Thanks very much for your thoughts. 
 A: Not a (full) answer / too long for a comment.
If $Z, X$ can be dependent, then the conjecture can be false, because between $X$ and $f$ it is easy to make $Z=1 \implies$ a lower $A$.  Here is a specific counter-example:


*

*$Z \in \{0, 1\}$ with equal prob $\frac12$

*$X = -2Z$

*$A = f(Z, X) = Z + X = -Z \in \{-1, 0\}$ with equal prob $\frac12$

*$F_A(a) = P(A \le a) = 0, \frac12, 1$ in the regions $(-\infty, -1), [-1, 0), [0, +\infty)$ respectively.
$$LHS = F_A(-1) - F_A(0) = \frac12 - 1 = -\frac12$$


*

*$E \in \{-10, +10\}$ with equal prob $\frac12$

*$B = A + E \in \{-11, -10, 9, 10\}$ with equal prob $\frac14$, and $F_B$ has $4$ equal sized steps at those points.
$$RHS = {F_B(-11) + F_B(9) \over 2} - {F_B(-10) + F_B(10) \over 2} = \frac12 (\frac14 + \frac34 - \frac12 - 1) = -\frac14$$
$$Gap = RHS - LHS = \frac12 > 0$$
Note that for simplicity I used discrete variables above, but you can easily add a tiny bit of "fuzz" around $X$ and/or $E$ e.g. $X = -2Z + Uniform(0,0.0001)$ and it would perturb the numbers a bit, but the gap should only be slightly perturbed from $\frac12$ and still be $> 0$.

Further thoughts: My gut feel is the conjecture is true if $Z,X$ are independent, but I haven't been able to prove that.
