Covariance inequality for left skewed distributions

Question

Consider a left skewed random variable $X$ with mean $1$, median $>1$ and support on $[0,2)$. Suppose we have a class of functions $\mathbf{G}$ and each of it's members satisfy $G(x): [0,\infty) \to (-\infty,0]$ which is increasing on $[0,1)$, decreasing on $(1,\infty)$ and such that $G(1) = 0$. Finally, we know that $|G'(1-x)| > |G'(1+x)|$ for $x \in (0,1)$; an illustration of a particular $G$ is shown in the figure.

I'm interested in proving or providing conditions such that $\mathrm{cov}[1(X \le 1), G(X)] \le 0$. For multiple reasons, this inequality seems plausible, but I am not able to prove it for general left-skewed distributions $X$ with mean = 1 < median.

To see why the claim is plausible, consider $\Gamma(K) = \mathrm{cov}[1(X \le K), G(X)]$, then $\Gamma'(K) = f(K)[G(K) - \mathbb{E}(G(X))]$, where $f$ is the PDF of $X$ and it's easy to prove $\Gamma'(K) < 0$ for $K$ small enough, but unfortunately $\Gamma'(1) = f(1)[G(1) - \mathbb{E}(G(X))] = -f(1)\mathbb{E}(G(X)) > 0$.

Second, since $X$ is negatively skewed, it has large outliers close to zero, where $G$ is supposed to be very negative, so it seems intuitively plausible that the sufficient condition $\mathbb{E}[1(X \le 1) G(X)] \le \mathbb{E}[1(X > 1) G(X)]$ should hold, but again I'm not able to prove this for general $G \in \mathbf{G}$.

Are there any results/ sufficient conditions I can impose on the distribution of $X$ such that my claim holds? Notice that simple tricks like Chebyshev sum inequality will not help since $G$ is not monotone.

Answer 1

$\newcommand\G{\mathbf G}$Your inequality is almost never true for all $G\in\G$.

Indeed, let $X$ be any random variable with values in $[0,2)$, mean $1$, c.d.f. $F$, p.d.f. $f$ continuous at $1$ with $f(1)>0$, and with all medians of $X$ being $>1$. So, $X$ will satisfy all your conditions.

Let us now show that then the inequality $$Cov(1(X\le1),G(X))\le0 \tag{1}\label{1}$$ cannot hold for all $G\in\G$. To do this, suppose the contrary. Then, by approximation, \eqref{1} will hold for for all functions $G=-1_{(1+v,\infty)}-1_{(0,1-v)}$, with any $v\in(0,1)$. That is, with $\bar F:=1-F$, for all $v\in(0,1)$ we will have \begin{align}0\le d(v)&:=Cov(1(X\le1),1(X>1+v)+1(X<1-v)) \tag{2}\label{2}\\ &=E1(X\le1)(1(X>1+v)+1(X<1-v)) \\ &\ \ \ -E1(X\le1)\,E(1(X>1+v)+1(X<1-v)) \\ &=F(1-v)-F(1)(\bar F(1+v)+F(1-v)) \\ &=F(1-v)\bar F(1)-F(1)\bar F(1+v). \end{align} The condition that all medians of $X$ are $>1$ implies that $F(1)<1/2<\bar F(1)$. So, $d'(0)=-f(1)\bar F(1)+F(1)f(1)=f(1)(F(1)-\bar F(1))<0$. Also, $d(0)=0$. So, $d(v)<0$ for all $v$ in a right neighborhood of $0$, which contradicts inequality \eqref{2}. $\quad\Box$