4
$\begingroup$

let $(\mathcal{X},d)$ be a Polish space. For $p\geq1$ let $\mathcal{P}_p(\mathcal{X})$ be the space of all Borel probability measures $\mu$ on $\mathcal{X}$ such that

\begin{equation} \mathbb{E}_\mu\left[ d^p(X,x_0) \right]<\infty \end{equation}

On this space we can define the Wasserstien distance (turns out to be a metric) as

\begin{equation} W_{p}(\mu,\nu) = \inf_{\pi \in \Pi(\mu,\nu)}\left(\int_{\mathcal{X}\times\mathcal{X}}d^p(x,y) \pi(dx,dy) \right)^{1/p} \end{equation}

where $\Pi(\mu,\nu)$ is the set of all probability measures on the space $\mathcal{X}\times \mathcal{X}$ with marginals $\mu$ and $\nu$.

Raginsky Sason 2014 Concentration of Measure Inequalities in Information Theory, Communications and coding, page 109, states that the inf is actually attained and therefore is a minimum. Does anyone know a proof for this? or point me in the direction of one? I cant seem to find it.

$\endgroup$
1
  • 1
    $\begingroup$ Theorem 8.10.45 in Bogachev, Measure Theory, vol. 2 $\endgroup$ Feb 18, 2019 at 14:44

2 Answers 2

3
$\begingroup$

One possible reference has been given in the comments, another one is Theorem 4.1 in Villani, Optimal transport, old and new. However, the argument is simple enough to reproduce it here.

First note that the set $\Pi(\mu,\nu)$ is tight and weakly closed, hence weakly compact by Prokhorov's theorem. Moreover, since $d^p$ is the supremum of bounded continuous functions, the functional $$ I\colon \Pi(\mu,\nu)\to[0,\infty],\,I(\pi)=\int d(x,y)^p\,d\pi(x,y) $$ is weakly lower semicontinuous. Thus $I$ attains its minimum.

(More explicitly, take a minimizing sequence $(\mu_n)$. Since $\Pi(\mu,\nu)$ is weakly compact and metrizable, we can extract a weakly convergent subsequence $(\mu_{n_k})$ with limit $\mu$ in $\Pi(\mu,\nu)$. Since $I$ is weakly lower semicontinuous, we have $$ I(\mu)\leq\lim_{k\to\infty}I(\mu_{n_k}). $$ Thus $\mu$ is a minimizer.)

$\endgroup$
1
  • $\begingroup$ Good answer, thanks for the reference. Therefore I have given an up vote. I am going to also submit a slightly more detailed answer. $\endgroup$
    – Monty
    Mar 7, 2019 at 15:19
2
$\begingroup$

First we will need two lemmas from Villani Optimal Transport, old and new. Lemma 4.3, from says that the functional $F :\pi \to \int d^{p} \pi $ is lower semi continuous on the space $\mathcal{P}(\mathcal{X} \times \mathcal{X})$ with respect to the weak topology. Lemma 4.4, states that if we have two tight subsets $\mathcal{M}_1$ and $\mathcal{M}_2$ of $\mathcal{P}_p(\mathcal{X})$ then the set $\Pi(\mathcal{M}_1,\mathcal{M}_2)$ is tight in $\mathcal{P}_p(\mathcal{X} \times \mathcal{X})$.

Now, since $\{\mu \}$ is clearly convergent to $\mu$, and $\mathcal{X}$ is Polish we have that $\mu$ is tight. Similarly $\nu$ is tight. So $\Pi(\mu,\nu)$ is tight in $\mathcal{P}(\mathcal{X} \times \mathcal{X})$, and we can apply Prokhorov's Theorem $\Pi(\mu,\nu)$ is precompact (its closure is compact). Furthermore $\Pi(\mu,\nu)$ is weakly closed (which we will show at the end) hence it is in fact compact.

Let $\pi_{k}$ be a sequence of probability measures on $\Pi(\mu,\nu)$ such that $\int d^p \pi_k$ converges to \newline $ \inf_{\pi \in \Pi(\mu,\nu)} \int d^p(x,y) \pi(dx,dy) $ . Now using that $\Pi(\mu,\nu)$ is closed and compact we can say that $\pi_k \to \pi \in \Pi(\mu,\nu) $ ( extracting a sub-sequence and relabeling if necessary ). We have yet to show that $\pi$ is minimizer.

Now using that $F$ is lower semi-continuous :

\begin{align*} F(\pi) \leq & \liminf_{\pi_k \to \pi} F(\pi_k) \\ \int_{\mathcal{X} \times \mathcal{X}} d^p(x,y) \pi(dx,dy) \leq & \liminf_{\pi_k \to \pi} \int_{\mathcal{X} \times \mathcal{X}} d^p(x,y) \pi_k (dx,dy) \\ = & \inf_{\pi \in \Pi(\mu,\nu)} \int_{\mathcal{X} \times \mathcal{X}} d^p(x,y) \pi_k (dx,dy) \end{align*}

So $\pi$ is such a minimiser, and by the reasoning above one can see that $\pi$ is indeed in the set $\Pi(\mu,\nu)$. Hence the infimum exists and is in fact an minimum.

As promised, $\Pi(\mu,\nu)$ is weakly closed since : if $\pi_k ~ \underset{weakly}{\to} ~\pi$ then for all bounded continuous functions $f$

$$ \int f d\pi_k \to \int f d\pi $$ in particular

$$ \int_{\mathcal{X} \times \mathcal{X} } f(x) \pi_k(dx,dy) \to \int_{\mathcal{X} \times \mathcal{X}} f(x) \pi(dx,dy) $$ that is, since $\pi_k$ has marginal $\mu$ and we choose $f$ to only depend on one variable :

$$\int_{\mathcal{X}} f(x) \mu(x) = \int_{\mathcal{X}} f(x) \pi(dx,\mathcal{X}) $$

for all bounded continuous $f:\mathcal{X} \to \mathbb{R}$, from which it can be shown $\pi(\cdot,\mathcal{X})=\mu$. Hence $\pi\in\Pi(\mu,\nu)$. Which completes the proof.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .