# Wasserstein distance inequality

Suppose $$(\Omega, \mathcal F, \mathbb P)$$ is a probability space. Suppose $$X, X', Y, Y'$$ are random variables. Denote $$W_1$$ the Wasserstein-1 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 $$W_1> W_2$$?

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 $$W_2^2\ge W_1^2$$.