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It is well-known fact that if we have two DFs F and G with finite second moments, then one can calculate the Wasserstein distance between them using this formula: $$ W_2^2(F,G) = \inf E(ξ-η)^2 = \int_0^1|F^{-1}(t)-G^{-1}(t)|^2 \, dt , $$ where infimum is taken over all pairs $(ξ,η)$ of random variables with distributions $F$ and $G$.

How to prove the last equality?

Michael Hardy
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Ilya68
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  • If $F$ and $G$ are continuous, then the infimum should be realized by $\xi=F^{-1}(T)$ and $\eta=G^{-1}(T)$, where $T\sim\mathrm{Uniform}(0,1)$. If they're not continuous I wouldn't be surprised if that or something close enough to it is still true, but I'd have to think about it. More later, maybe. ${}\qquad{}$ – Michael Hardy Jun 09 '15 at 20:39
  • @Michael Yes, You are right. The pair $(F^{-1}(T),G^{-1}(T))$ will have common distribution $\min(F(x), G(y))$. But why the integral $\iint(x-y)^2dH(x,y) $ attains infimum with $H(x,y) = \min(F(x), G(y))$ – Ilya68 Jun 09 '15 at 21:08

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Note that you have to show that $\mathbb{E}(XY)$ is maximized if $\mathbb{P}(X\leq x,Y\leq y)=\min\{F(x),G(y)\}$. For bounded nonnegative $X\leq a$,$Y\leq b$ this follows from integrating the equality $$\mathbb{P}(X>x,Y>y)=\mathbb{P}(X>x)+\mathbb{P}(Y>y)-1+\mathbb{P}(X\leq x,Y\leq y)$$ over $[0,a]\times [0,b]$, and the obvious inequality $\mathbb{P}(X\leq x, Y\leq y)\leq \min\{F(x),G(y)\}$. From here it extends to bounded rvs and finally arbitrary rvs. See C.L. Mallows, Ann. Math. Statist., 43 (1972), 508-515

esg
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