Correlation inequality This question is a modified version of my previous post: Maximum correlation inequality.
Let $ X $ be a random variable and suppose $ A $ is a subset of $ \mathbb{R} $. For two functions $ f:\mathbb{R}\mapsto\mathbb{R} $ and $ g:\mathbb{R}\mapsto\mathbb{R} $, let $ \rho(f(X), g(X) | X \in A) $ denote the conditional correlation between $ f(X) $ and $ g(X) $ given $ X \in A $. I am wondering whether
$$
\sup_{f \; \text{monotone}}|\rho(f(X),g(X) | X \in A)| \geq \sup_{f \; \text{monotone}}|\rho(f(X),g(X))|.
$$
This seems intuitively true, since it becomes more difficult to find a good monotone approximation of $ g(x) $ if $ x $ belongs to a larger subset, but I'm not sure how to show it. Any thoughts?
 A: First note that if $g$ is monotone, the conjecture is trivially true by picking $f = g$ so that both sides $=1$.
If $g$ is non-monotone though, what happens outside $A$ can easily overwhelm what happens inside $A$.  Here is a non-monotone $g$ that disproves the conjecture:


*

*$X$ is continuous uniform within $S= [1, 100]$

*$A = \{1, 2, 3, \dots, 100\}$ (i.e. just the integers of $S$)

*$\forall x \in A: g(x) = (x \pmod 2)$

*$\forall x \in \mathbb{R} - A: g(x) = x$.
The RHS $=1$ by picking $f(x) = x$.  This $f$ is monotone and achieves $\rho = 1$ because $f=g$ except for set $A$ of measure $0$.
The LHS...  I'm sorry I lack the math skills to formally prove that $\rho = 1$ is unachieveable, but it seems pretty obvious.  In fact, the best "line fit" is roughly horizontal, making the best $f$ and $g$ almost independent.  Thus LHS $\rho \approx 0$.
In both this question and your previous question, the key insight is that what happens in a subset of the support can be overwhelmed by what happens in a larger subset of (or the whole) support.  Without further restrictions in the problem statement, a conditional distribution is its own entity and can be very different from the unconditioned distribution.  So if you're talking about a "global" or "aggregate" characteristic like $\rho$, there is very little you can conclude.
BTW it is easy to modify the above example to use a different continuous $g$.  The proof will be even more vague because it's no longer obvious RHS $=1$, but the intuition is just the same.
