Wasserstein$-k$ distance between two probability measures $\mu,\nu$ on $\mathbb{R}^d$ is defined as:

$$W_k(\mu,\nu)=\left(\inf_{(X,Y)\in\mathcal{C}(\mu,\nu)}\mathbb{E}\left[\|X-Y \|^k \right]\right)^{1/k}$$

where $\mathcal{C}(\mu,\nu)$ is the set of all couplings of $\mu,\nu$.

Total variation distance between the same probability measures is defined as:

$$\|\mu-\nu\|_{\rm{TV}}=\max_{A\subset \mathbb{R}^d}|\mu(A)-\nu(A)|=\inf \{\mathbb{P}(X\neq Y): (X,Y)\,\text{is a coupling of $\mu$ and $\nu$.}\}$$

where the $\inf$ is again over all couplings of $\mu,\nu$.

I am trying to check if the total variation distance is smaller than Wasserstein-$1$ distance for any two probability measures.

$$\|\mu-\nu\|_{\rm{TV}}\leq \mathbb{P}(X\neq Y)$$

where $(X,Y)$ is any coupling and then I was trying to apply Markov's inequality but did not succeed. Any ideas?

  • 1
    $\begingroup$ A sketch of a counterexample: Consider $S_n=\sum_{i=1}^n X_i$ for $X_i$, $i=1,...,n$ i.i.d. with $\mathbb{P}(X_i=1)=\mathbb{P}(X_i=-1)=1/2$. Let $Z \sim \mathcal{N}(0,1)$. Then $\|\frac{S_n}{\sqrt{n}}-Z\|_{TV}=1$ for all $n$ (because $S_n$ is a discrete random variable), but $\mathcal{W}(S_n/\sqrt{n},Z)\rightarrow 0$ for $n \rightarrow \infty$. If it is too sketchy, I will write it down properly as an answer. $\endgroup$
    – lukanz
    Commented Mar 9, 2022 at 15:02

1 Answer 1


As @lukanz mentioned, what you are trying to prove is not true. Here is an even simpler counterexample: consider the sequence $\delta_{\frac{1}{n}}$ (Dirac measures centred at $1/n$). It is easy to check that \begin{align} \lVert\delta_0 -\delta_{1/n} \rVert_{\mathrm{TV}}=1\, , \quad \forall \, n \geq 1 \, . \end{align} On the other hand, choosing the coupling $(X,X+1/n)$ where $X\sim \delta_0$, one has that \begin{align} W_1\left(\delta_0,\delta_{\frac1n}\right) \leq \frac1n\, . \end{align} So your inequality cannot hold true. The inequality the other way does hold true but with a constant and using a weighted version of the total variation distance. You can find the statement and proof in Chapter 6 of Villani's book Optimal Transport: Old and New.

  • $\begingroup$ Just to confirm, you are considering $d=1$ in $\mathbb{R}^d$, right? $\endgroup$
    – user621937
    Commented Mar 10, 2022 at 3:37
  • $\begingroup$ That is correct. $\endgroup$
    – user621937
    Commented Mar 10, 2022 at 3:58
  • $\begingroup$ Edit (latex typo): Yes. But you can tweak the example to work in any dimension by centering the measures at $(1/n,0,\dots,0)$ in higher dimensions. $\endgroup$ Commented Mar 10, 2022 at 5:29

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