$X,Y$ are two independent random variables, with $Y$ distributed uniformly on $[0,1]$ and $X$ distributed uniformly on $[0,\bar{x}]$ with $\bar{x} \leq 1$. Let $Z \equiv (X,Y)$. I have to find the conditional expectation $E(Z|(Y-EY)\geq h(X-EX))$, as a function of $h$ and $\bar{x}$, for $h \in (0,\infty)$.
My approach: First find conditional expectation of $Y$ given the condition and $X$, and then, using the law of iterated expectations, take unconditional expectation w.r.t. $X$. So, $E(Y|(Y-EY)\geq h(X-EX),X)=\frac{1+EY+h(X-EX)}{2}$. Now take unconditional expectation w.r.t. $X$, so the second term vanishes, and we get, $E(Y|(Y-EY)\geq h(X-EX),X)=\frac{1+EY}{2}=\frac{3}{4}$, which is independent of $h$! Does this sound right to you? I'm not sure because the expectation depends neither on $h$ nor on $\bar{x}$.
Edit: OK I think I have discovered my mistake. Let's take the example of $\bar{x}=1,h=1$. The part I missed is, conditional on $Y \geq X$ (the condition in this case), $X$ is no longer distributed uniformly on $[0,1]$! It is, in fact, distributed according to the density $2(1-x), x\in[0,1]$. And this gives us the conditional expectation of $Y = \frac{2}{3}$.
Analogously, we can calculate the density of $X$ conditional on the condition as $\frac{1-2h(x-\frac{\bar{x}}{2})}{\bar{x}}$ and calculate the rest of the conditional expectations.