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Consider a twice differentiable and symmetric distribution $\mathcal{F}_X$. Now consider a second twice differentiable distribution $\mathcal{F}_Z$ rigth skewed in the sense that:

$$(1)\quad\mathcal{F}_X\preceq_c\mathcal{F}_Z.$$

where $\preceq_c$ is the convex ordering of van Zwet [0] so that $(1)$ is equivalent to:

$$(2)\quad F^{-1}_ZF_X(x)\text{ is convex $\forall x\in\mathbb{R}.$}$$

Consider now a third twice differentiable distribution $\mathcal{F}_Y$ satisfying:

$$(3)\quad\mathcal{F}_Y\preceq_c\mathcal{F}_Z.$$

My question is: can we always find a distribution $\mathcal{F}_Y$ and a symmetric distribution $\mathcal{F}_X$ to rewrite any $\mathcal{F}_Z$ (all three defined as above) in terms of a composition of $\mathcal{F}_X$ and $\mathcal{F}_Y$ as:

$$F_Z(z)=F_YF_X^{-1}F_Y(z)$$

or not?

Edit:

For example, if $\mathcal{F}_X$ is the Weibull with shape parameter 3.602349 (so that it is symmetric) and $\mathcal{F}_Z$ is the Weibull distribution with shape parameter 3/2 (so that it is right skewed), I get

$$\max_z|F_Z(z)-F_YF_X^{-1}F_Y(z)|\approx 0$$

by setting $\mathcal{F}_Y$ as the Weibull distribution with shape parameter 2.324553. Note that all three distributions satisfy:

$$\mathcal{F}_{-X}=\mathcal{F}_X\preceq_c\mathcal{F}_Y\preceq_c\mathcal{F}_Z,$$ As required. I wonder if this is true in general (under the stated conditions).

  • [0] van Zwet, W.R. (1979). Mean, median, mode II (1979). Statistica Neerlandica. Volume 33, Issue 1, pages 1--5.
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No!

A simple counter-example is provided by the Tukey $g$ distribution (the special case for $h=0$ of the Tukey $g$ and $h$ distribution).

For example, let $\mathcal{F}_X$ be the Tukey $g$ with parameter $g_X=0$ and $\mathcal{F}_Z$ be the Tukey $g$ with parameter $g_Z>0$ and $\mathcal{F}_Y$ a Tukey $g$ distribution for which $g_Y\leq g_Z$. Since $h=0$, theses three distributions satisfy:

$$\mathcal{F}_{-X}=\mathcal{F}_X\preceq_c\mathcal{F}_Y\preceq_c\mathcal{F}_Z.$$

(the first one comes from the definition of the Tukey $g$ which is symmetric if $g=0$, the next ones from [0], Theorem 2.1(i)).

For example, for $g_Z=0.5$, we have that:

$$\min_{g_Y\leq g_Z}\max_z|F_Z(z)-F_YF^{-1}_XF_Y(z)|\approx0.005>0$$

(for some reason, the minimum seems to always be near $g_Y\approx g_Z/2$).

  • [0] H.L. MacGillivray Shape properties of the g-and-h and Johnson families. Comm. Statist.—Theory Methods, 21 (5) (1992), pp. 1233–1250

Edit:

In the case of the Weibull, the claim is true:

Let $\mathcal{F}_Z$ be the Weibull distribution with shape parameter $w_Z$ (the scale parameter doesn't affect convex ordering so we can set it to 1 without loss of generality). Likewise $\mathcal{F}_Y$, $\mathcal{F}_X$ and $w_Y$ and $w_X$.

First note that any three Weibull distributions can always be ordered in the sense of [0].

Next, note that: $$\mathcal{F}_X=\mathcal{F}_{-X}\implies w_X=3.602349.$$

Now, for the Weibull:

$$F_Y(y)=1-\exp((-y)^{w_Y}),\;F_Y^{-1}(q)=(-\ln(1-q))^{1/w_Y},$$

so that

$$F_YF_X^{-1}F_Y(z)=1-\exp(-z^{w_Y^2/w_X}),$$

since

$$F_Z(z)=1-\exp(-z^{w_Z}).$$

Therefore, the claim can always be satisfied by setting $w_Y=\sqrt{w_Z/w_X}$.

  • [0] van Zwet, W.R. (1979). Mean, median, mode II (1979). Statistica Neerlandica. Volume 33, Issue 1, pages 1--5.
  • [1] Groeneveld, R.A. (1985). Skewness for the weibull family. Statistica Neerlandica. Volume 40, Issue 3, pages 135–140.
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