# Relation between Wasserstein distance and distribution convergence

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 ?

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:

Theorem:

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

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