Covariance inequality with Lipschitz functions

Question

Suppose that $X$ and $Y$ are random variables and suppose that for all Lipschitz functions $f$ and $g$ s.t. $f(X),g(Y)\in L^p$, $p>2$, $$ |\operatorname{Cov}(f(X),g(Y))|\le \big(\operatorname{Lip}(f)\operatorname{Lip}(g)+\|f(X)\|_p\|g(Y)\|_p\big)\alpha^{1-\frac{2}{p}}.\tag{1}\label{1} $$ When $f$ and $g$ are the identity functions, the inequality becomes $$ |\operatorname{Cov}(X,Y)|\le \big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}.\tag{2}\label{2} $$ Now, multiplying $X$ and $Y$ by a constant $c>0$, we get $$ |\operatorname{Cov}(X,Y)|\le \inf_{c>0}\big(c^{-2}+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}=\|X\|_p\|Y\|_p\alpha^{1-\frac{2}{p}}.\tag{3}\label{3} $$ This result seems weird because the inequality in \eqref{1} is clearly "scale invariant", that is, applying \eqref{1} directly (with $f(x)=g(x)=cx$) yields $$ |\operatorname{Cov}(cX,cY)|\le c^2\big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}.\tag{4}\label{4} $$

Is the inequality in Equation \eqref{3} actually correct?

Answer 1

This just looks to me like an example where fixing something prematurely (or at all) gives you a sub-optimal estimate. Specifically, fixing $f=g=\mathrm{id}$ immediately leaves you unable to do much about the first term on the right hand side of (1), so it's not that surprising that doing so gives you a worse form of your estimate than if you don't. Unless I'm missing something, it looks like you've deduced that your hypotheses imply that

$$ |\mathrm{Cov}(X,Y)|\leq (c^{-2} + \Vert X \Vert_p\Vert Y \Vert_p)\alpha^{1-2/p} $$

for any $c>0$ (i.e. (3)), discovering along the way that fixing $f=g=\mathrm{id}$ only gives you the $c=1$ case.