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Suppose that $X$ and $Y$ are random variables and suppose that for 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?

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?

Suppose that $X$ and $Y$ are random variables and suppose that for 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 $$ |\operatorname{Cov}(cX,cY)|\le c^2\big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}. $$

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

Suppose that $X$ and $Y$ are random variables and suppose that for 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?

Suppose that $X$ and $Y$ are random variables and suppose that for 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 $$ |\operatorname{Cov}(cX,cY)|\le c^2\big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}. $$ On the other hand, applying \eqref{1} to $Z=cX$ and $W=cY$ (with $f(x)=g(x)=x$), one gets $$ |\operatorname{Cov}(Z,W)|\le \big(1+\|Z\|_p\|W\|_p\big)\alpha^{1-\frac{2}{p}}, $$ which implies that $$ |\operatorname{Cov}(X,Y)|\le \big(c^{-2}+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}, $$ and we're back to \eqref{3}.

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

Suppose that $X$ and $Y$ are random variables and suppose that for 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 $$ |\operatorname{Cov}(cX,cY)|\le c^2\big(1+\|X\|_p\|Y\|_p\big)\alpha^{1-\frac{2}{p}}. $$

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