# Is the Wasserstein distance to the empirical measure minimized by the underlying distribution?

Let $$S$$ be a metric space and denote the set of probability measures on $$S$$ by $$\mathcal{P}(S)$$. Fix $$\mu\in \mathcal{P}(S)$$ and denote the law of $$N\geq 1$$ i.i.d samples $$X=(X_1,\ldots,X_N)$$ from $$\mu$$ as $$\mu_N$$. Do we have the following for $$p\geq 1$$: \begin{align} \inf_{\nu \in \mathcal{P}(S)} \mathbb{E}_{X\sim \mu_N}[\mathcal{W}_p(\frac{1}{N}\sum_{i=1}^N \delta_{X_i}, \nu)] = \mathbb{E}_{X\sim \mu_N}[\mathcal{W}_p(\frac{1}{N}\sum_{i=1}^N \delta_{X_i}, \mu)] \,? \end{align} In plain words: Is the empirical distribution, on average, closest to its underlying distribution?

So far, I did neither succeed in proving this nor in finding some reference, even though there are many articles studying the asymptotics of the r.h.s.. Does anyone know whether this is true, at least in some simplified scenario ($$S=[0,1], p=1$$ e.g.)?

No. E.g., let $$N=1$$ and suppose that $$X:=X_1$$ has a nondegenerate zero-mean distribution $$\mu$$ such that $$E|X|^p<\infty$$. Let $$Y$$ be an independent copy of $$X$$.
Then the expected $$\mathcal W_p$$-distance from the empirical distribution to $$\mu$$ is
$$E\mathcal W_p(\delta_X,\mu)^p=E|X-Y|^p>E|X|^p=E\mathcal W_p(\delta_X,\delta_0)^p;$$ the inequality here is an instance of a strict version of Jensen's inequality, which holds because the distribution $$\mu$$ is nondegenerate.
• @joemrt : Condition on $X$ and then apply Jensen's inequality to the zero-mean random variable $Y$. At least for $p>1$, the inequality will be strict, because then the function $|\cdot|^p$ is strictly convex. I am sure the inequality will be strict even for $p=1$ (and nondegenerate $\mu$), but cannot prove this at the moment. Commented Jan 30, 2023 at 16:31
• @joemrt : As is now shown, the only exception to the strict inequality is when the support of $\mu$ consists of at most two points: mathoverflow.net/a/439727/36721 Commented Jan 30, 2023 at 20:00