# Is the minimum of the Doksum ratio always unbounded?

Consider $\mathcal{F}$ and $\mathcal{G}$ are two continuous distributions and that $\mathcal{G}$ is more dispersed than $\mathcal{F}$ in the sense of satisfying:

$$(1)\quad F^{-1}(\beta)-F^{-1}(\alpha)\leq G^{-1}(\beta)-G^{-1}(\alpha),\quad \forall 0< \alpha < \beta <1.$$

Then, we know (Kochar (2012), eq. 2.16) that (1) is equivalent to

$$(2)\quad\frac{F^\prime(x)}{G^\prime G^{-1}F(x)}\geq 1.$$ (I call the left hand side of (2) the Doksum ratio after (Doksum 1969))

Now consider the ratio:

$$(3)\quad\gamma(\mathcal{F},\mathcal{G})=\underset{x:F^\prime(x)>0}{\min}\;\frac{F^\prime(x)}{G^\prime G^{-1}F(x)}$$

For many distributions I find that, for fixed $\mathcal{F}$, $\gamma(\mathcal{F},\mathcal{G})$ can be made arbitrary large by shifting/rescaling $\mathcal{G}$.

For example, for the normal, student t, Weibull, gamma I find that $\gamma(\mathcal{F},\mathcal{G})$ is an increasing function of $\text{Var}(\mathcal{G})$.

My question is this: Is it always so that for fixed $\mathcal{F}$, $\gamma(\mathcal{F},\mathcal{G})$ can be made arbitrarily large by re-scaling/shifting $\mathcal{G}$?

It is geometrically obvious that shifting $G$ changes nothing and that rescaling will rescale the denominator by the same factor--it's merely a matter of keeping track of units of measurement. The following is a rigorous algebraic demonstration. Your conclusion follows immediately.

Let $a\gt 0$ and $b$ be real numbers and define $G_{(a,b)}$ to be the $a$-scaled, $b$-shifted version of $G$:

$$G_{(a,b)}(x) = G(ax + b).$$

Let $0 \le \alpha \le 1$. Compute that

$$(G_{(a,b)})^{-1}(\alpha) = \frac{G^{-1}(\alpha) - b}{a}$$

and (via the Chain Rule)

$$\frac{d}{dx}G_{(a,b)}(x) = a \left(\frac{d}{dx} G\right)(ax + b).$$

Thus

$$\left(\frac{d}{dx}G_{(a,b)}\right)\left(\left(G_{(a,b)}\right)^{-1}(\alpha)\right) = a \left(\frac{d}{dx}G\right)\left(G^{-1}(\alpha)\right).$$

This shows that shifting does not change the denominator of the Doksum ratio and scaling multiplies the denominator by $a$. As $a\to 0^{+}$, the ratio will grow without bound provided its numerator $F^\prime(x)$ exists and is positive.