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 $\forall p\in[0,0.5]$:


  • [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.
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    $\begingroup$ I suppose there is some error in the last inequality - if it holds $\forall p \in [0,1]$, symmetry would imply equality $\frac{F^{-1}_X(p)}{F^{-1}_Y(p)}=\frac{F^{-1}_X(1-p)}{F^{-1}_Y(1-p)}$, which in turn would be symmetric w.r.t. $X$ vs. $Y$. $\endgroup$ – Juho Kokkala Jun 8 '15 at 11:22
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    $\begingroup$ Note that there is $\alpha \in (0,\frac{1}{2})$ after equation (6) of [2]. $\endgroup$ – Juho Kokkala Jun 8 '15 at 12:08
  • $\begingroup$ you are correct. My bad. I fix this now. $\endgroup$ – user603 Jun 8 '15 at 12:08

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 $q>\bar q$.

  • $\begingroup$ Could you please add some explanations to this answer? It is a bit short for our standards! $\endgroup$ – kjetil b halvorsen Dec 16 '16 at 15:20

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:


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 $F_X(x)$)


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