I have two variables $X_1$ and $X_2$ $\in [0,1]$, where $X_2$ (strictly) first order stochastically dominate (FOSD) $X_1$, i.e., $F_{X_2}(x) \leq F_{X_1}(x)$ for all $x$.
Then I have, $Y$ and $Z$, two other variables which are strictly increasing functions of $X_1$ and $X_2$ with: $Y_1 = \phi(X_1)$, $Y_2 = \phi(X_2)$, $Z_1 = \gamma(X_1)$, $Z_2 = \gamma(X_2)$, where $\phi(\cdot)$ and $\gamma(\cdot)$ are strictly increasing for all $x$. Because of the strict FOSD, we have that $\mathbb{E}(X_1) < \mathbb{E}(X_2)$, and thus, with the strictly increasing functions, $\mathbb{E}(Y_1) < \mathbb{E}(Y_2)$, and $\mathbb{E}(Z_1) < \mathbb{E}(Z_2)$.
Finally, and that's where the difficulty comes in, I also have a binary variable, $D$, with the specificity that $\textrm{Pr}(D=1 | X_1 = x) = \textrm{Pr}(D=1 | X_2 = x)$ for all $x$. Define, $D(X_1)$ (and similarly $D(X_2)$) such that:
\begin{align*}
\mathbb{E}(D(X_1)) = \int_0^1 \textrm{Pr}(D=1 | X_1 = x) f_{X_1}(x) dx.
\end{align*}
An alternative definition would be that we have an underlying variable $A=1$ or $=2$. $X_1 = X | A=1$ and $X_2 = X | A=2$, and then, $D(X_1)$ is simply $D|A=1$ and $D(X_2)$ is $D|A=2$. The equality of the probabilities mean that $D \perp A | X$.
Remark that, in general, the equality of the probabilities with respect to $x$, does not necessarily imply that $\mathbb{E}(D(X_1)) = \mathbb{E}(D(X_2))$ since both variables don't have the same density. Also note that $\textrm{Pr}(D=1 | X_1 = x)$ is not a strictly increasing function of $X_1$, so I don't have the usual FOSD property anymore.
Objective: I would like to show (I'm not sure it's true) that it is not possible to have, simultaneously: \begin{align} \mathbb{E}(D(X_1)) &= \mathbb{E}(D(X_2)), \\ \mathbb{E}(Y_1 D) &= \mathbb{E}(Y_2 D), \\ \mathbb{E}(Z_1 (1-D)) &= \mathbb{E}(Z_2 (1-D)). \end{align}
So far, I have been unable to prove it because, with a general $D$, I don't have a strictly increasing function anymore and I lose the properties of the FOSD in a sense.
However, simply satisfying jointly: $\mathbb{E}(D(X_1)) = \mathbb{E}(D(X_2))$ and $\mathbb{E}(Y_1 D) = \mathbb{E}(Y_2 D)$ is quite difficult. The only example I found satisfying this requires $\mathbf{f_{X_1|D=1}(x) = f_{X_2|D=1}(x)}$ for all $x$.
If this is truly the only possibility, my proof becomes trivial. Indeed, in this case,
$\mathbb{E}(Z_1 D) = \mathbb{E}(Z_2 D)$ as well. Moreover, $\mathbb{E}(Z_j) = \mathbb{E}(Z_j D) + \mathbb{E}(Z_j (1-D))$, so, it is not possible that $\mathbb{E}(Z_1 D) = \mathbb{E}(Z_2 D)$ and $\mathbb{E}(Z_1(1-D)) = \mathbb{E}(Z_2(1-D))$, because, by FOSD, we need that $\mathbb{E}(Z_1) < \mathbb{E}(Z_2)$.
However, I have been unable to prove that simply satisfying $\mathbb{E}(D(X_1)) = \mathbb{E}(D(X_2))$ and $\mathbb{E}(Y_1 D) = \mathbb{E}(Y_2 D)$ requires $\mathbf{f_{X_1|D=1}(x) = f_{X_2|D=1}(x)}$ for all $x$. Maybe it is not true. But then, do you have a counterexample where it holds without the conditional densities equality at every $x$? Or a general example for which the 3 conditions can hold together? I am stuck: unable to prove it's not true but unable to build any counterexample yet... My intuition was that, if the weights ($Pr(D=1)$) are going to give a zero, then the other variable weighted by $1-$these weights cannot be simultaneously zero, because BOTH transformations are strictly increasing. But it's much more complicated than I thought.
EDIT: I guess a subquestion would be to know, in the problem above, is it possible that:
\begin{align}
\mathbb{E}(D(X_2)) - \mathbb{E}(D(X_1)) &= \int^1_0 \textrm{Pr}(D=1 | X=x) ( f_{X_2}(x) - f_{X_1}(x) ) dx \\
\mathbb{E}(Y_2 D) - \mathbb{E}(Y_1 D) &= \int_{Y} y \ \textrm{Pr}(D=1 | Y=y) ( f_{Y_2}(y) - f_{Y_1}(y) ) dy \\
&= \int^1_0 \phi(x) \frac{d\phi^{-1}(y(x))}{dx} \textrm{Pr}(D=1 | X=x) ( f_{X_2}(x) - f_{X_1}(x) ) dx,
\end{align}
without imposing $\textrm{Pr}(D=1 | X=x) ( f_{X_2}(x) - f_{X_1}(x)) = 0$ for all $x$?
Answer: this is indeed possible.
However, in the example I have, then it becomes impossible that $\mathbb{E}(Z_2 (1-D)) - \mathbb{E}(Z_1 (1-D)) = 0$ too.