Wasserstein distance between a distribution of a random variable and the distribution of its projection onto a subset of its sample space

Question

Consider a random variable $x$ with a distribution $p_x$ supported on whole of $\mathbb{R}^n$ ($n$ being a natural number). Let $S \subset \mathbb{R}^n$. Let $y = {\rm proj}_S(x)$ denote the orthogonal projection of $x$ on $S$ (in terms of the euclidean norm). Let $p_y$ denote the distribution of $y$. Is it true that the square of the Wasserstein 2-distance between $p_x$ and $p_y$ is $$\mathbb{E}_{x\sim p_x}\|x - {\rm proj}_S(x) \|_2^2?$$ Intuitively, it does seem true, because you would want to pile on all the joint probability on points where the L2 norm is the minimum.

Answer 1

For the orthogonal projection to be well defined I would expect additional assumption on $S$, for instance nonempty closed and convex.

Then it is obviously true. because for any random variable $X$ and $Y$ such that $X$, resp. $Y$ is distributed according to $p_x$, resp, $p_y$, we have that $$\Vert X-\textrm{proj}_S(X)\Vert^2_2\le\Vert X-Y\Vert^2_2,$$ hence $$\mathbb E[\Vert X-\textrm{proj}_S(X)\Vert^2_2]\le\mathbb E[\Vert X-Y\Vert^2_2].$$

We can take the infimum over $Y\sim p_y$ in the right-hand side, and then take the infimum over $X\sim p_X$ in both sides to deduce that $$ \mathbb E[\Vert X-\textrm{proj}_S(X)\Vert^2_2]\le W_2^2(p_x,p_y). $$

Of course the other inequality holds by definition of the Wasserstein distance.