Two distributions with inverse distance closer than distance

Question

Let $F$ and $G$ are two distributions with a support, say $[0,1]$. Assume they are continuous and increasing. When we use the Kolmogorov distance $$\operatorname{dist}(F,G)=\sup_{x\in[0,1]}|F(x)-G(x)|,$$ What kind of relationship can we find between the distance and inverse distance? $$\operatorname{dist}(F,G) \text{ and } \operatorname{dist}(F^{-1},G^{-1})$$ Can there be an example that shows inverses are a lot closer than the just the distance of $F$ and $G$?

Answer 1

As you assume $F$ and $G$ are both continuous functions from $\left[0,1\right]$ to $\left[0,1\right]$, there is no guaranteed order relation between $\text{dist}(F,G)$ and $\text{dist}(F^{-1},G^{-1})$.

Consider a simply example, where $$ F(x)=\left\{ \begin{array}{ll} kx,&x\in\left[0,\alpha\right]\\ \frac{1-k\alpha}{1-\alpha}\left(x-\alpha\right),&x\in\left(\alpha,1\right] \end{array} \right.\quad\text{and}\quad G(x)=\left\{ \begin{array}{ll} kx,&x\in\left[0,\beta\right]\\ \frac{1-k\beta}{1-\beta}\left(x-\beta\right),&x\in\left(\beta,1\right] \end{array} \right.. $$ Here the parameters $k>0$, $0<\alpha,\beta<1$ are chosen such that both $F$ and $G$ are increasing on $\left[0,1\right]$.

With this example, take $\alpha$ and $\beta$ close to $1$, and $k$ close to $0$. The image of $F$ and $G$ would then be similar to the following figure, where $\text{dist}(F,G)$ is obviously very large but $\text{dist}(F^{-1},G^{-1})$ is kind of small.

likewise, take $\alpha$ and $\beta$ close to $0$, and $k$ much larger than $1$. In this case, the image of $F$ and $G$ would be close to the following figure, where $\text{dist}(F,G)$ is kind of small but $\text{dist}(F^{-1},G^{-1})$ is obviously very large.