# Can we always rewrite a right skewed distribution in terms of composition of an arbitrary and a symmetric distribution?

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  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).

•  van Zwet, W.R. (1979).
Mean, median, mode II (1979).
Statistica Neerlandica. Volume 33, Issue 1, pages 1–5.

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 , 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 $$g_Y\approx g_Z/2$$).

•  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 .

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
$$w_Y=\sqrt{w_Z/w_X}$$.

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