I have a question about the convexity of an Wasserstein ambiguity set.


Let $W\_1(\mu, \nu)$ be the Wasserstein distance of order 1 between $\mu$ and $\nu$, defined as 
$$W\_1(\mu, \nu) := \min\limits\_{\gamma \in \Gamma(\mu, \nu)} \bigg \{ \int\_{\Xi \times \Xi} d(\xi, \zeta) \gamma(d\xi, d\zeta) \bigg \}, $$ where $\Gamma(\mu, \nu)$ denotes the set of all probability measures on $\Xi \times \Xi$ with marginals $\mu$ and $\nu$.


Let $\nu$ be the empirical distribution. The Wasserstein ambiguity set $\mathcal{M}$ is defined by $$\mathcal{M} := \{ \mu \in \mathcal{P}(\Xi) : W\_1(\mu, \nu) \leq \theta \},$$ where $\theta$ is a given radius.


I am curious about whether the set $\mathcal{M}$ is convex. I notice that Wasserstein distance satisfies the triangle inequality, but I'm not sure that the set $\mathcal{M}$ is convex.


Is the Wasserstein ambiguity set of order 1 always convex?



Is an ambiguity set with Wasserstein distance of order 1 is convex?
**Edit:** (This question may be better suited for math.stackexchange.)


This is true. It suffices to show the map $\mu\rightarrow W\_{1}(\mu,\nu)$ is convex. Let $\mu\_{i}\in\mathcal{P}(\Xi)$ and $\psi^{\*}\_i\in\Gamma(\mu\_{i},\nu)$ be optimal transport plans between $\mu\_{i}$ and $\nu$ for each $i=1,2$. Then $\lambda\psi\_{1}^{\*}+(1-\lambda)\psi\_{2}^{\*}\in\Gamma(\lambda\mu\_{1}+(1-\lambda)\mu\_{2},\nu)$ and so


$W\_{1}(\lambda\mu\_{1}+(1-\lambda)\mu\_{2},\nu)=\inf\limits\_{\gamma\in\Gamma(\lambda\mu\_{1}+(1-\lambda)\mu\_{2},\nu)} \int d(x,y) d\gamma(x,y)\leq \\ \int d(x,y) d(\lambda\psi\_{1}^{\*}+(1-\lambda)\psi\_{2}^{\*})(x,y) = \lambda W\_{1}(\mu\_{1},\nu)+(1-\lambda)W\_{1}(\mu\_{2},\nu)$


which gives convexity. Thus, $$W\_{1}(\lambda\mu\_{1}+(1-\lambda)\mu\_{2},\nu)\leq \lambda W\_{1}(\mu\_{1},\nu)+(1-\lambda)W\_{1}(\mu\_{2},\nu)\leq \theta.$$


When restrictions are made such as replacing $\mathcal{P}(\Xi)$ with some other set of probability measures, as is done in the paper [here](https://papers.nips.cc/paper/8067-wasserstein-distributionally-robust-kalman-filtering.pdf), the answer can be no:



> 
> [For $\mathcal{P}(\Xi)$ replaced with $\mathcal{N}(\Xi)$, Gaussian distributions]...the Wasserstein ambiguity set $P$ is nonconvex (as mixtures of normal distributions are
> generically not normal).
> 
> 
> 


