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.


1 Answer 1


$\newcommand\Ga{\Gamma}$ $\newcommand\ga{\gamma}$ $\newcommand\la{\lambda}$ 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 $$\int d(x,y)^2\ga_i(dx\times dy)<c_i$$ 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 $0<W_2(\mu_0,\rho)<\infty$. 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 $k\to\infty$.

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    $\begingroup$ 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. $\endgroup$
    – Raghav
    Feb 12, 2020 at 2:57
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    $\begingroup$ Also note that this argument works for any cost function! $\endgroup$
    – Dirk
    Feb 12, 2020 at 8:21
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    $\begingroup$ @Dirk : That's right. $\endgroup$ Jun 18, 2020 at 0:14
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    $\begingroup$ @Zorn'sLama : That's right. $\endgroup$ Jun 18, 2020 at 0:14
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    $\begingroup$ @Elbebe : No, I don't know any reference to this. You can cite MathOverflow questions and answers by click on the "cite" buttons below them. $\endgroup$ Oct 23, 2020 at 18:38

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