Suppose $(\Omega, \mathcal F, \mathbb P)$ is a probability space. Suppose $X, X', Y, Y'$ are random variables.
Denote $W_1$ the Wasserstein-2 distance between $\mathbb P_X$ and $\mathbb P_{X'}$
and $W_2$ the Wasserstein-2 distance between $\mathbb P_{X,Y}$ and $\mathbb P_{X',Y'}$
$$W_1:= \inf_{\pi\in \Gamma_1}(\int |x-x'|^2d\pi(x,x'))^{1/2}$$
$$W_2:= \inf_{\gamma\in \Gamma_2}(\int |x-x'|^2+|y-y'|^2d\gamma((x,y),(x',y')))^{1/2}$$
where $\Gamma_1$ are all the couplings with marginals $\mathbb P_X$ and $\mathbb P_{X'}$ and $\Gamma_2$ are all the couplings with marginals $\mathbb P_{X,Y}$ and $\mathbb P_{X',Y'}$
Is it true that $W_1\leq W_2$?
My guess is yes. I'm trying to show that for each ${\gamma\in \Gamma_2}$ there exists a $\pi\in \Gamma_1$ such that $$\int |x-x'|^2d\pi(x,x')= \int |x-x'|^2d\gamma((x,y),(x',y'))$$
Or is there any example such that $W_1> W_2$?
Best Answer
Yes.
Let $\gamma$ be a coupling between $\mathbb P_{X,Y}$ and $\mathbb P_{X',Y'}$. Let $\pi$ denote its marginal distribution on the $(x,x')$ coordinates, so that $\pi$ is a coupling between $\mathbb P_X$ and $\mathbb P_{X'}$. Then $$ \begin{align} \int\left(\vert x-x'\vert^2+\vert y-y'\vert^2\right)\,d\gamma((x,y),(x',y')) &\ge\int\vert x-x'\vert^2\,d\gamma((x,y),(x',y'))\\ &=\int\vert x-x'\vert^2\,d\pi(x,x')\\ &\ge W_1^2. \end{align} $$
By taking the infimum over coupling $\gamma\in\Gamma_2$ you get that $W_2^2\ge W_1^2$.