# Wasserstein distance between product measures

For two probability measures $$\mu,\nu$$ on $$\mathbb{R}^n$$, let $$W_p(\mu,\nu)^p=\inf_{\pi}\int\|x-y\|^p_2\ \pi(dx\ dy)$$ denote the $$p^\text{th}$$ Wasserstein distance between $$\mu, \nu$$, where the infimum is taken with respect to all possible couplings of $$\mu,\nu$$. Consider probability measures $$\mu_1,\ldots, \mu_n$$ and $$\nu_1,\ldots, \nu_n$$ on $$\mathbb{R}$$. For $$p\ge 1$$, is it true that $$W_p(\mu_1\otimes\cdots\otimes\mu_n, \nu_1\otimes \cdots\otimes\nu_n)\le \left(\sum_{i=1}^n W^2_p(\mu_i,\nu_i)\right)^{1/2} \ \ \ \text{?}$$

• Note that section 2 of this states that this hols for $p = 2$, but it specifically calls out this requirement, e.g. it is plausible it is necessary (though I do not know).
– Mark
Sep 28 at 1:32

$$\newcommand{\de}{\delta}\renewcommand{\S}{\mathcal S}\newcommand{\T}{\mathcal T}$$The answer to your question is negative if $$p<2$$.

Indeed, let $$\nu_i=\de_0$$ for all $$i$$, where $$\de_a$$ is the Dirac measure supported on the singleton set $$\{a\}$$. Let $$X_1,\dots,X_n$$ be independent random variables (r.v.'s) with respective distributions $$\mu_1,\dots,\mu_n$$. Let $$Y_i:=X_i^2$$ for all $$i$$. Let $$\begin{equation*} \text{\mu:=\bigotimes_1^n\mu_i and \nu:=\bigotimes_1^n\nu_i.} \end{equation*}$$

Then $$W_p(\mu_i,\nu_i)=(E|X_i|^p)^{1/p}$$ and $$W_p(\mu,\nu)=(E(\sum_1^n X_i^2)^{p/2})^{1/p}$$.

So, the inequality in question becomes $$\begin{equation*} L:=\Big\|\sum_1^n Y_i\Big\|_{p/2}\overset{\text{(?)}}\le \sum_1^n \|Y_i\|_{p/2}=:R. \end{equation*}$$ Suppose now that $$n=2$$ and $$Y_1,Y_2$$ are independent r.v.'s such that $$P(Y_i=1)=t=1-P(Y_i=0)$$ for $$i=1,2$$, where $$t\downarrow0$$. Then $$L=(2t(1-t)+t^2 2^{p/2})^{2/p}\sim(2t)^{2/p}$$, whereas $$R=2t^{2/p}$$, so that the inequality $$L\le R$$ fails to hold if $$p<2$$ and $$t$$ is small enough. $$\quad\Box$$

I have just gotten up during the night and it occurred to me how to show, very simply, that the inequality in question holds for $$p\ge2$$, and then I saw the answer by the OP :-), which is correct, except it had to be mentioned there that the pairs $$(X_i,Y_i)$$ may be assumed to be independent.

Actually, we have the following somewhat more general fact. Let $$p\in[2,\infty)$$. For each $$i\in[n]:=\{1,\dots,n\}$$, let $$(S_i,\S_i,\mu_i)$$ and $$(T_i,\T_i,\nu_i)$$ be probability spaces. Let $$(S,\S,\mu):=\bigotimes_{i\in[n]}(S_i,\S_i,\mu_i)$$ and $$(T,\T,\nu):=\bigotimes_{i\in[n]}(T_i,\T_i,\nu_i)$$. For each $$i\in[n]$$, let $$\Pi(\mu_i,\nu_i)$$ denote the set of all probability measures on the measurable space $$(S_i\times T_i,\S_i\otimes\T_i)$$ with marginals $$\mu_i$$ and $$\nu_i$$. Let $$\Pi(\mu,\nu)$$ denote the set of all probability measures on the measurable space $$(S\times T,\S\otimes\T)$$ with marginals $$\mu$$ and $$\nu$$. For each $$i\in[n]$$, let $$f_i\colon S_i\times T_i\to[0,\infty)$$ be a measurable function. Let $$f(s,t):=\sqrt{\sum_{i\in[n]} f_i(s_i,t_i)^2}$$ for all $$s=(s_1,\dots,s_n)\in S= S_1\times\cdots\times S_n$$ and $$t=(t_1,\dots,t_n)\in T=T_1\times\cdots\times T_n$$. For each $$i\in[n]$$, let $$\begin{equation*} W_p(\mu_i,\nu_i):=\inf_{\pi_i\in\Pi(\mu_i,\nu_i)} \Big(\int_{S_i\times T_i}f_i(s_i,t_i)^p\,\pi_i(ds_i,dt_i)\Big)^{1/p}. \end{equation*}$$ Let $$\begin{equation*} W_p(\mu,\nu):=\inf_{\pi\in\Pi(\mu,\nu)} \Big(\int_{S\times T}f(s,t)^p\,\pi(ds,dt)\Big)^{1/p}. \end{equation*}$$ Then $$\begin{equation*} W_p(\mu,\nu)\overset{\text{(?)}}\le \Big(\sum_{i\in[n]} W_p(\mu_i,\nu_i)^2\Big)^{1/2}. \tag{1}\label{1} \end{equation*}$$

Indeed, take any real $$w_1,\dots,w_n$$ such that $$W_p(\mu_i,\nu_i) (if such $$w_1,\dots,w_n$$ exist; otherwise, inequality \eqref{1} is trivial). Then for each $$i\in[n]$$ there exists some $$\pi_i\in\Pi(\mu_i,\nu_i)$$ such that $$\begin{equation*} \Big(\int_{S_i\times T_i}f_i(s_i,t_i)^p\,\pi_i(ds_i,dt_i)\Big)^{1/p} Let $$\pi:=\bigotimes_{i\in[n]}\pi_i.$$ Then $$\pi\in\Pi(\mu,\nu)$$. Moreover, for each $$i\in[n]$$ $$\begin{equation*} \Big(\int_{S_i\times T_i}f_i(s_i,t_i)^p\,\pi_i(ds_i,dt_i)\Big)^{2/p} =\|g_i\|_{L^{p/2}(\pi)}, \end{equation*}$$ where $$g_i(s,t):=f_i(s_i,t_i)^2$$ for for all $$s=(s_1,\dots,s_n)\in S= S_1\times\cdots\times S_n$$ and $$t=(t_1,\dots,t_n)\in T=T_1\times\cdots\times T_n$$, and $$\begin{equation*} \Big(\int_{S\times T}f(s,t)^p\,\pi(ds,dt)\Big)^{2/p} =\|g\|_{L^{p/2}(\pi)}, \end{equation*}$$ where $$g:=f^2=g_1+\cdots+g_n$$.

So, by the condition $$p\in[2,\infty)$$, Minkowski's inequality, and \eqref{2}, $$\begin{multline*} W_p(\mu,\nu)^2\le \Big(\int_{S\times T}f(s,t)^p\,\pi(ds,dt)\Big)^{2/p} =\|g\|_{L^{p/2}(\pi)} \\ \le\sum_{i\in[n]} \|g_i\|_{L^{p/2}(\pi)} =\sum_{i\in[n]} \Big(\int_{S_i\times T_i}f_i(s_i,t_i)^p\,\pi_i(ds_i,dt_i)\Big)^{2/p} <\sum_{i\in[n]}w_i^2. \end{multline*}$$ Letting now $$w_i\downarrow W_p(\mu_i,\nu_i)$$ for each $$i\in[n]$$, we get \eqref{1}. $$\quad\Box$$

• Also, I think in your example, the inequality should read as $\|\sum_{1}^n X^2_i\|_{p/2}\le (\sum_{1}^n \|X_i\|^2_p)^{1/2}$ Sep 28 at 3:30
• @Ribhu : The inequality in your comment is not homogeneous in the $X_i$'s. On the other hand, the inequality $L\le R$ in my answer seems to be the right one. Sep 28 at 3:36

From the previous answer, it follows that the inequality is true when $$p\ge 2$$. Let $$X_i, Y_i$$ be the optimal choice of random variables for which $$\|X_i-Y_i\|_p=W_p(\mu_i,\nu_i) \ \forall i=1,\ldots, n.$$ Then, for $$p\ge 2$$, using the triangle inequality we have $$\left\|\sum_{i=1}^n (X_i-Y_i)^2\right\|_{p/2}\le \sum_{i=1}^n \|(X_i-Y_i)^2\|_{p/2}=\sum_{i=1}^n W^2_p(\mu_i,\nu_i).$$ Minimizing the left hand side with respect to all possible couplings $$(X_i,Y_i)_{i=1}^n$$ yields the result.