Showing $\int_{\mathbb R} \mid F(x)-G(x)\mid dx = \int_0^1 \mid F^{-1}(u)-G^{-1}(u)\mid du$ with $F$, $G$ CDF functions

Question

Let's $X$ and $Y$ have CDF functions admitting moment of order $1$. Let's be $F$ cdf of $X$ and $G$ cdf of $Y$.

I want to show that $$\int_{\mathbb R} \mid F(x)-G(x)\mid dx = \int_{0}^{1} \mid F^{-1}(u)-G^{-1}(u)\mid du\,.$$

Answer 1

If $\mu$ is 2D Lebesgue measure, then interpreting the integral as the unsigned area$^*$ between $F$ and $G$, $$\int |F(x) - G(x)| dx = \mu\big[(x,y)\in\mathbb R\times [0,1]:G(x)\le y<F(x)\big] + \mu\big[(x,y)\in\mathbb R\times [0,1]:F(x)\le y<G(x)\big ] $$

Then note that

$$G(x) \le y < F(x) \iff x \le G^{-1}(y) , F^{-1}(y)<x \iff F^{-1}(y)<x \le G^{-1}(y)$$ and similarly $ F(x) \le y < G(x) \iff G^{-1}(y) < x \le F^{-1}(y)$. thus

$$\int |F(x) - G(x)| dx = \mu\big[(x,y)\in\mathbb R\times [0,1]:F^{-1}(y)<x \le G^{-1}(y)\big] + \mu\big[(x,y)\in\mathbb R\times [0,1]:G^{-1}(y) < x \le F^{-1}(y)\big ] $$

returning to the 1D integral notation, this is saying that $$ \int_{\mathbb R} |F(x) - G(x)| dx = \int_0^1 |F^{-1}(y) - G^{-1}(y) | dy $$

Finally a graph - this indicates that the result should be true even for some functions without an inverse. ( desmos link )

$^*$ For a positive function $f$, $\int_A f(x) dx = \int_A \int_0^{f(x)} dydx = \mu( (x,y) \in A\times \operatorname{im}f : 0\le y\le f(x)).$ Appropriate case analysis leads to the above expression.

Answer 2

Since you use the notation $F^{-1}$, resp. $G^{-1}$, for the quantile function you tacitly assume that $F$ and $G$ are continuous and strictly increasing on some interval $J\subset{\mathbb R}$. I suggest you draw a figure showing two reasonable such functions. The left hand side of the claimed formula then represents the unsigned area enclosed between the graphs of $F$ and $G$. Turning the figure $90^\circ$ you then can verify that the right hand side of the claimed formula is the same area.

This means that one has to prove that $$A:=\bigl\{(x,u)\in J\times[0,1]\bigm|\min\{F(x),G(x)\}\leq u\leq\max\{F(x),G(x)\}\bigr\}$$ and $$A':=\bigl\{(x,u)\in J\times[0,1]\bigm|\min\{F^{-1}(u),G^{-1}(u)\}\leq x\leq\max\{F^{-1}(u),G^{-1}(u)\}\bigr\}$$ are in fact the same sets. This is "pure logic"; one just has to go through the motions.

Answer 3

This answer is inspired by Calvin Khor's solution. Here, we do not assume that $F$ and $G$ possess inverse functions. In this answer, we define $T^{-1}:(0,1)\to \mathbb{R}$ as $$T^{-1}(u):=\sup\big\{v\in\mathbb{R}\,|\,T(v)\leq u\big\}$$ for any cumulative distribution function $T:\mathbb{R}\to[0,1]$. Since $T^{-1}$ is nondecreasing, it is a measurable function. We first note that, if $T$ admits the first moment, then $$\int_{-\infty}^0\,T(x)\,\text{d}x+\int_0^{+\infty}\,\big(1-T(x)\big)\,\text{d}x=\int_\mathbb{R}\,|x|\,\text{d}T(x)<\infty\,,$$ so we have $$\int_{-\infty}^0\,T(x)\,\text{d}x<\infty\text{ and }\int_0^{+\infty}\,\big(1-T(x)\big)\,\text{d}x<\infty\,.\tag{*}$$

Now, because $F$ and $G$ admit the first moments, the integral $$I:=\int_\mathbb{R}\,\big|F(x)-G(x)\big|\,\text{d}x$$ is finite due to (*). From Calvin Khor's answer, we have $$I=\mu(E^+)+\mu(E^-)\,,$$ where $\mu$ is the Lebesgue measure on $\mathbb{R}^2$, $$E^+:=\big\{(x,y)\in\mathbb{R}\times (0,1)\,|\,G(x)\leq y<F(x)\big\}\,,$$ and $$E^-:=\big\{(x,y)\in\mathbb{R}\times (0,1)\,|\,F(x)\leq y<G(x)\big\}\,.$$ Observe that $$E^+\subseteq S^+:=\big\{(x,y)\in\mathbb{R}\times (0,1)\,|\,F^{-1}(y)\leq x \leq G^{-1}(y)\big\}$$ and $$E^-\subseteq S^-:=\big\{(x,y)\in\mathbb{R}\times (0,1)\,|\,G^{-1}(y)\leq x \leq F^{-1}(y)\big\}\,.$$ Note that $$S^+\setminus E^+\subseteq \big\{(x,y)\in\mathbb{R}\times (0,1)\,|\,x\text{ is the unique solution to }F(x)=y\big\}$$ and $$S^-\setminus E^-\subseteq \big\{(x,y)\in\mathbb{R}\times (0,1)\,|\,x\text{ is the unique solution to }G(x)=y\big\}$$ are of Lebesgue measure $0$. Therefore, $$I=\mu(S^+)+\mu(S^-)=\int_0^1\,\big|F^{-1}(u)-G^{-1}(u)\big|\,\text{d}u\,.$$