# $2$-Wasserstein distance between mixtures

I am stuck on the following problem. I have a discrete distribution $$\mu_0$$ (it is actually an empirical distribution). I have some $$\mu_i$$ (again discrete, some empirical distribution). I have some bound on the Wasserstein distance $$W_2(\mu_0, \mu_i).$$ I now want to consider a simple mixture of $$\mu_i,$$ that is, $$\nu=\sum\limits_{i=1}^{m}\lambda_i\mu_i$$ where $$\sum\lambda_i=1, \lambda_i>0.$$

My goal is to bound $$W_2^2(\mu_0, \nu).$$ I felt that it would be easy to get a bound on $$W_2^2(\mu_0, \nu)$$ in terms of $$W_2^2(\mu_0, \mu_i),$$ but I am unable to prove anything. I want something like $$W_2^2(\mu_0, \nu)\le \sum \lambda_i^2 W_2^2(\mu_0, \mu_i).$$

This does not look terribly hard, but I am stuck. Can anyone please say if it is true or not? If anyone can give a simple demonstration of why this is true, it would be great.

$$\newcommand\Ga{\Gamma}$$ $$\newcommand\ga{\gamma}$$ $$\newcommand\la{\lambda}$$ Let $$\Ga(\mu,\rho)$$ denote the set of all measures with marginals $$\mu$$ and $$\rho$$. For each $$i$$, take any real $$c_i>W_2(\mu_0,\mu_i)^2$$, so that $$\int d(x,y)^2\ga_i(dx\times dy) for some $$\ga_i\in\Ga(\mu_0,\mu_i)$$. Let $$\ga:=\sum_i\la_i\ga_i.$$ Then $$\ga\in\Ga(\mu_0,\nu)$$ and hence $$W_2(\mu_0,\nu)^2\le\int d(x,y)^2\ga(dx\times dy) =\sum_i\la_i \int d(x,y)^2\ga_i(dx\times dy)<\sum_i\la_i c_i.$$ Letting now $$c_i\downarrow W_2(\mu_0,\mu_i)^2$$ for each $$i$$ such that $$W_2(\mu_0,\mu_i)^2<\infty$$, we get $$W_2(\mu_0,\nu)^2\le\sum_i\la_i W_2(\mu_0,\mu_i)^2.$$
The inequality you proposed, $$W_2(\mu_0,\nu)^2\le\sum_{i=1}^k\la_i^2 W_2(\mu_0,\mu_i)^2,\tag{1}$$ cannot hold in general. Indeed, suppose that for some probability measure $$\rho$$ we have $$0. Let $$\mu_i:=\rho$$ and $$\la_i:=1/k$$ for all $$i=1,\dots,k$$. Then $$\nu=\rho$$ and the left-hand side of (1) is a constant $$>0$$, whereas its right-hand side goes to $$0$$ as $$k\to\infty$$.
• Thanks! Yes, I realised that we can not hope for $\lambda_i^2$ in the right hand side. I could finally get the result with $\lambda_i$ but your argument is neater. – WhoKnowsWho Feb 12 at 2:57