Let $(X,d)$ be a metric space, $x\_1,\ldots,x\_N\in X$ and $x\_1',\ldots,x\_N'\in X$ be atoms, and $G=\sum\_{i=1}^Np\_i\delta\_{x\_i}$, $G'=\sum\_{i=1}^Np\_i'\delta\_{x\_i}$, and $G''=\sum\_{i=1}^Np\_i'\delta\_{x\_i'}$ be mixing measures. In words: $G$ and $G'$ have the same atoms, but different weights. $G'$ and $G''$ have different atoms, but the same weights.


Assuming $G'\ne G''$ (i.e. there is at least one distinct atom), is it true that $W\_p(G,G')\le W\_p(G,G'')$? Here, $W\_p$ is the usual $p$th Wasserstein distance between the measures $G$ and $G'$.


In other words, if two discrete measures have the same support, does "perturbing" the atoms in one of the measures always increase the Wasserstein distance? Or is it possible to move the atoms in one measure in such a way to decrease the Wasserstein distance?



Effect of perturbing the atoms of a measure on the Wasserstein distance
There is a nontrivial counterexample for $N=2$, $p=1$, and $X=\mathbb{R}$. Pick $x\_1=-2$, $x\_2=2$, $x'\_1=-1$, $x'\_2=1$ and $p\_1=4/5$ and $p\_1'=1/5$. Then $2.2=W\_1(G,G'')<W\_1(G,G')=2.4$. (I hope I did not mess up the calculation).


The intuition seems clear:


In the counterexample, you have to move $1/5$ of the total mass from $-2$ to $-1$ and $3/5$ of the total mass from $-2$ to $1$, while moving another $1/5$ of the total mass from $2$ to $1$. This is cheaper than moving $3/5$ of the mass all the way from $-2$ to $2$.


