Does convex ordering imply right tail dominance? Given two continuous distributions $\mathcal{F}_X$ and $\mathcal{F}_Y$, It is not clear to me whether the relation of convex dominance among them:
$$(0)\quad \mathcal{F}_X <_c \mathcal{F}_Y$$
implies that 
$$(1)\quad F_Y^{-1}(q) \leq F_X^{-1}(q),\quad \forall q\in[0.5,1]$$
holds or if some further 
 hypothesis are needed if $(1)$ is to hold?

Definition of Convex dominance.
If two continuous distributions $\mathcal{F}_X$ and $\mathcal{F}_Y$ satisfy:
$$(2)\quad F_Y^{-1}F_X(x)\text{ is convex in } x$$ 
[0] then we write:
$$F_X <_c F_Y$$
and say that $\mathcal{F}_Y$ is more right skewed than $\mathcal{F}_X$. Because $F_X$ and $F_Y$ are probability distributions, $(2)$ also implies that the derivative of $F_Y^{-1}F_X(x)$ is monotonically non decreasing and non-negative [1], that  $F_Y^{-1}F_X(x)-x$ is convex [2], that $F_X$ and $F_{aY+b}$ cross each other at most twice $\forall a>0,b\in\mathbb{R}$ [2] and that [2], for $\forall p\in[0,0.5]$:
$$\frac{F^{-1}_X(p)}{F^{-1}_Y(p)}\geq\frac{F^{-1}_X(1-p)}{F^{-1}_Y(1-p)}.$$


*

*[0] Zwet, W.R. van (1964).
Convex Transformations of Random Variable. (1964). 
Amterdam: Mathematish Centrum.

*[1] Oja, H. (1981).
On Location, Scale, Skewness and Kurtosis of Univariate Distributions.
Scandinavian Journal of Statistics. Vol. 8, pp. 154--168 

*[2] R.A. Groeneveld and G. Meeden. (1984). Measuring skewness and kurtosis. The Statistician. 33:391-399.

 A: Ok, I think this can be solved like so (comments welcome):
Denoting $\mathcal{F}_X$ and $\mathcal{F}_Y$ the distributions of $X$ and $Y$ and recalling that
$$\mathcal{F}_X <_c \mathcal{F}_Y$$
implies (Oja, 1981) that $\exists z^*\in\mathbb{R}$ such that:
$$F_Y(z)<F_X(z),\forall z>z^*.$$
Since shifting does not affect convex ordering, we can assume 
 without loss of generality that $X$ has been shifted 
 so that:
$$z^*\leqslant\min(F^{-1}_X(0.5),F^{-1}_Y(0.5))$$
so that 
$$F_Y^{-1}(q) \leqslant F_X^{-1}(q),\quad \forall q\in[0.5,1].$$
So, it seems that yes, convex ordering of $\mathcal{F}_X<_c \mathcal{F}_Y$ implies right tail dominance of $F_Y(y)$ over $F_X(x)$ (or to be precise some version $F_{X+b}(x),\;b\in\mathbb{R}$ of $F_X(x)$)
A: In general it is not true. 
Consider for example the $\mu=\frac{3}{8} \delta_{-1}(x)+\frac{1}{4}\delta_0(x)+\frac38 \delta_1(x)$ and $\nu=\frac12\delta_{-\frac12}(x)+\frac12\delta_{\frac12}(x)$.
You can immediately see that $\nu\leq_{cx}\mu$. However $F_\mu^{-1}(0.6)=0<\frac12 =F_\nu^{-1}(0.6) $. It is however true that from a certain $\bar{q}$ on, $F_\mu^{-1}(q)<F_\nu^{-1}(q)$ for all $q>\bar q$.
