Let's have a succession $X_n$ of real value random variable and another real value random variable X, then $$ X_n \overset{d}{\xrightarrow{}} X \iff \lim_{n \to \infty}{}d_K(X_n,X) = 0 $$ where $d_K(X,Y)$ is the Kolmogorov-Smirnov distance between X and Y.

Another result tells us that: $$ d_K(X,Y) \leq C\cdot\sqrt{d_{W_1}(X,Y)} $$ where C is a constant and $d_{W_1}(X,Y)$ is the 1-Wasserstein distance between X and Y.

From the two statements above is trivial to prove that $$d_{W_1}(X_n,X) \xrightarrow{} 0 \implies X_n \overset{d}{\xrightarrow{}} X$$

But what can we say about the other implication? Can we say that if $X_n$ converges in distribution to $X$ then the 1-Wasserstein distance converges to $0$ or we can't ?


1 Answer 1


For the reverse direction, we need a slightly stronger assumption. To see this, first note that weak convergence $X_n\stackrel{d}{\Rightarrow} X$ is defined for general real-valued random variables, regardless of moment conditions. However, the same thing is not true for convergence w.r.t. the Wasserstein distance. Note that $W_1(X_n,X)$ only makes sense, i.e. is well-defined and finite, if and only if $X_n$ and $X$ are integrable, i.e. $X_n,X\in L_1$. Hence, the least we will have to further assume is that $X_n,X\in L_1$.

This is not quite enough, but if we further assume that the moments converge, i.e. $$\mathbb E|X_n|\rightarrow\mathbb E|X|$$ together with $X_n\stackrel{d}{\Rightarrow}X$, then we have convergence of with respect to the Wasserstein distance, $W_1(X_n,X)\to 0$.

To see this, note that we can without loss of generality assume that $X_n\to X$ almost-surely (by Skorokhod's Theorem ) and if $\mathbb E|X_n|\to\mathbb E|X|$ then $X_n$ also converges to $X$ in $L_1$, by Vitali's Theorem. Hence we have $$W_1(X_n,X)=\inf_{\pi}\mathbb E_{(X_n,X)\sim\pi}\big|X_n-X\big|\to 0 \hspace{1cm}n\to\infty.$$

Altogether we get the following result:


Assume $X_n,X\in L_1$ are integrable random variables. Then $$W_1(X_n,X)\Leftrightarrow X_n\stackrel{d}{\Rightarrow} X \text{ and } \mathbb E|X_n|\to \mathbb E|X|$$

  • $\begingroup$ And, just to mention the obvious, everything is easier when $X_n, X$ share compact support. $\endgroup$ Jun 2, 2023 at 16:45

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