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Let $p \geq1$ and consider space $\mathcal{P}(\mathbb{R})$ of Borel probability measures on $\mathbb{R}$. The $p$-order Wasserstein distance is defined by $$ W_p(\mu, \nu) = \left( \inf_{\pi \in \Pi(\mu,\nu)} \int_{\mathbb{R}^2} |x-y|^p d\pi(x,y) \right)^{1/p}, $$ where $\Pi(\mu,\nu)$ denotes the set of all Borel probability measures on $\mathbb{R}^2$ with marginals $\mu$ and $\nu$ in $\mathcal{P}(\mathbb{R})$.

Let's define the expectation distance between $\mu$ and $\nu$ by

$$ d(\mu,\nu) = \bigg|\int_{\mathbb{R}} x d\mu(x) - \int_{\mathbb{R}} x d\nu(x)\bigg|. $$

While it is easy to prove that $d(\mu, \nu) \leq W_p(\mu, \nu), \forall \mu,\nu \in \mathcal{P}(\mathbb{R})$ (e.g., see https://math.stackexchange.com/a/2269491/253451), is there any existing result about upper bounding $W_p$ by $d$? E.g., Does there exist $C > 1$ such that $W_p(\mu, \nu) \leq C d(\mu,\nu), \forall \mu,\nu \in \mathcal{P}(\mathbb{R})$? Any counterexample?

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Short answer: No.

Simple counterexample:

$$ W_2\left(\mathcal{N}(m_1, \sigma_1), \mathcal{N}(m_2, \sigma_2) \right) = \sqrt{(m_1 - m_2)^2 + (\sigma_1 - \sigma_2)^2}. \\ d\left(\mathcal{N}(m_1, \sigma_1), \mathcal{N}(m_2, \sigma_2) \right) = |m_1 - m_2|. $$

Thus, $W_2\left(\mathcal{N}(m_1, \sigma_1), \mathcal{N}(m_2, \sigma_2) \right)$ is arbitrarily larger than $d\left(\mathcal{N}(m_1, \sigma_1), \mathcal{N}(m_2, \sigma_2) \right)$; thus such $C$ does not exist.

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