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

Question

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.)?

Answer 1

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.