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Consider two random variables $X,Y$. Given some metric $d(X,Y)$, the Wasserstein distance, with respect to $d$, is $$d_W(X,Y)=\inf_{\text{couplings}}\mathbb{E}(d(X,Y))$$

where the infimum is over all couplings of $X,Y$.

How does the expected value vary with respect to different couplings? That is, if $X',Y'$ are any random variables such that $X,Y$ are the marginals of the random vector $(X'Y')$ (i.e., $(X',Y')$ is a coupling of $X,Y$) isn't $\mathbb{E}(d(X',Y')=\mathbb{E}(d(X,Y))$ since $X',Y'$ have the same distributions as $X,Y$, respectively?

Iced Palmer
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1 Answers1

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The distribution of the distance $d(X,Y)$ depends on the joint distribution of $X$ and $Y.$ Not just on the marginals. This is the reason for which you take the infimum over all possible joint distributions. For example, take a random variable distributed \begin{align*} X &= 1 \text{ with probability } 1/2 \\ X &= {-}1 \text{ with probability } 1/2 \end{align*} Then of course $d_W(X,X) = 0,$ since you can choose the couple $(X,X)$ which has the correct marginals and $$\mathbb{E}[d(X,X)] = \mathbb{E}[0] = 0$$

On the other side you can see that $-X$ has the same distribution as $X$ and $$d(X, -X) = d(1,-1) = 2$$

So that for this particular coupling $$\mathbb{E}[d(X,{-}X)] = \mathbb{E}[2] = 2.$$

Kore-N
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