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In a book I found the following exercise:

Let $F,G$ be two cumulative distribution functions. Then

$$\int_0^1 \vert F^{-1} (t) - G^{-1} (t)\vert \text d t = \int_\Bbb R \vert F(x) - G(x) \vert \text d x$$ where $H^{-1} (t) := \inf \{ x \in \Bbb R : H(x) > t\}$, for a CDF $H$.

I tried to write $\vert F(x) - G(x) \vert = \int_0^1 1_{(-\infty , \vert F(x) - G(x) \vert]} (t) \text d t$ and use Fubini afterwards, in order to rewrite the set $\{x : t\leq \vert F(x) - G(x) \vert\}$ for fixed $t$ and obtaining a set with mass $\vert F^{-1}(t) - G^{-1}(t) \vert$.

But I fail with the last part. Maybe this is even the wrong approach.

Falrach
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    This appears to be a statement about two congruent domains in ${\Bbb R}^2$ - one is transformed into the other by a symmetry wrt to the diagonal $y=x$ - having the same areas (Lebesgue measures). – sds Oct 11 '19 at 14:19
  • Note that the domain of $|F-G|$ is $\mathbb R$ while its codomain is $[0,1]$, while these statements are reversed for $|F^{-1}-G^{-1}|$. – Math1000 Oct 11 '19 at 14:36
  • Ah, yes the picture in mind is clear to me. Due to the symmetry the area between $F,G$ and between $F^{-1} , G^{-1}$ is the same. I will make this rigorous. – Falrach Oct 11 '19 at 15:06

1 Answers1

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Clearly, $$\int_\Bbb R \vert F(x) - G(x) \vert \text d x = \int_\Bbb R \int_{\min \{F(x),G(x)\}}^{\max\{F(x),G(x)\}} \text d t \text d x \\ = \text{Lebesgue} (\{(x,t)\in \Bbb R\times [0,1] : \min \{F(x),G(x)\} < t <\max\{F(x),G(x)\} \})$$ But if w.l.o.g. $F(x) < t < G(x)$, then by definition holds: $$x\geq G^{-1}(t) \quad \text{and since $F^{-1}$ is non-decreasing:} \quad F^{-1}(t) \geq F^{-1}(F(x)) \geq x$$ Conversly, if $F^{-1}(t) > x > G^{-1}(t)$, then by definition $$F(x) \leq t \quad \text{and since $G^{-1}$ is non-decreasing:} \quad G(x) \geq G(G^{-1}(t)) \geq t$$ Thus, the quantity above equals by Fubini $$\text{Lebesgue}(\{(x,t)\in \Bbb R\times [0,1] : \min \{F^{-1}(t),G^{-1}(t)\} < x <\max\{F^{-1}(t),G^{-1}(t)\} \}) \\ = \int_0^1 \int_{\min \{F^{-1}(t),G^{-1}(t)\}}^{\max\{F^{-1}(t),G^{-1}(t)\}} \text d x \text d t = \int_0^1 \vert F^{-1} (t) - G^{-1} (t) \vert \text dt$$

Falrach
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