Optimal Couplings Exist

Harley Wiltzer, May 2020

This post is a journey through the proof of a beautiful fact from the theory of optimal transportation. First let's define what a coupling is.

Given two probability spaces \((\mathcal{X},\mu), (\mathcal{Y},\nu)\), a coupling of \((\mu,\nu)\) is simply a pair (shall we say couple?) of random variables \((X, Y)\) such that \(X\sim\mu,Y\sim\nu\). On the surface, this does not seem particularly inspiring, however we will hopefully learn that couplings can dramatically simplify analysis. Constructing coupling arguments has a somewhat artistic nature, since the simplification analysis often relies on choosing a clever correlation between the couple of random variables.

Now let's define what it means for a coupling to be optimal. Suppose we have two metric spaces \(\mathcal{X},\mathcal{Y}\) and some cost function \(c\) that assigns a cost of transporting \(x\in\mathcal{X}\) to \(y\in\mathcal{Y}\). When \(\mathcal{X}=\mathcal{Y}\), an interesting and natural choice of the cost function is simply the distance function \(d\) of the metric space, perhaps raised to some power \(p\geq 1\). For probability spaces \((\mathcal{X},\mu),(\mathcal{Y},\nu)\), we say that a coupling \((X, Y)\) is optimal if it minimizes the expected transportation cost:

\( (X, Y) = \inf_{X'\sim\mu, Y'\sim\nu}\mathbf{E}c(x, y) = \inf_{X'\sim\mu, Y'\sim\nu}\int_{\mathcal{X}\times\mathcal{Y}}c(x, y)d\pi(x, y) \)

where \((X, Y)\sim\pi\). What we want to show now is that such an optimal coupling always exists under some assumptions of the probability spaces and the cost function.

Let's begin by stating out assumptions:

  1. \((\mathcal{X}, \mu), (\mathcal{Y},\nu)\) are Polish probability spaces. In other words, \(\mathcal{X},\mathcal{Y}\) are complete, separable metric spaces.
  2. There exists upper semicontinuous functions \(a: \mathcal{X}\to\mathbf{R}\cup\{-\infty\}\), \(b: \mathcal{Y}\to\mathbf{R}\cup\{-\infty\}\) where \(a\) is integrable w.r.t. \(\mu\) and \(b\) is integrable w.r.t. \(\nu\), such that the cost function \(c\) satisfies \(c(x, y)\geq a(x) + b(y)\) for each \(x, y\).
  3. The cost function \(c:\mathcal{X}\times\mathcal{Y}\to\mathbf{R}\cup\{\infty\}\) is lower semicontinuous.

To begin, we'd like to show that \(\Pi(\mu,\nu)\), the space of couplings of \((\mu,\nu)\), is compact. Prokhorov's theorem, which shows a relationship between compactness and tightness in Polish spaces, will be really handy here. The theorem states the following:

Theorem (Prokhorov): If \(\mathcal{X}\) is a Polish space, then a set \(\mathcal{P}\subset\mathscr{P}(\mathcal{X})\) is precompact (has compact closure) w.r.t. the weak topology if and only if it is tight; that is, for any \(\mu\in\mathcal{P}\) and \(\varepsilon>0\), there is a compact set \(K_\varepsilon\subset\mathcal{X}\) such that \(\mu(\mathcal{X}\setminus K_\varepsilon)\leq\varepsilon\).

Moving forward, we'd like to show that if \(\mathcal{P},\mathcal{Q}\) are subsets of \(\mathscr{P}(\mathcal{X}),\mathscr{P}(\mathcal{Y})\) respectively, then the set \(\Pi(\mathcal{P}, \mathcal{Q})\) of all couplings with marginals in \(\mathcal{P}\) and \(\mathcal{Q}\) is itself tight in \(\mathscr{P}(\mathcal{X}\times\mathcal{Y})\). Then, by Prokhorov's theorem, we will have shown that \(\Pi(\mathcal{P},\mathcal{Q})\) has a compact closure.

Fix some \(\varepsilon>0\). Since \(\mathcal{P},\mathcal{Q}\) are tight by assumption, there exist sets \(K_{\varepsilon/2},L_{\varepsilon/2}\) such that regardless of \(\mu\in\mathcal{P}\) and \(\nu\in\mathcal{Q}\), we have \(\mu(\mathcal{X}\setminus K_{\varepsilon/2})\leq\varepsilon/2\) and similarly \(\nu(\mathcal{Y}\setminus L_{\varepsilon/2})\leq\varepsilon/2\). Then, for any coupling \((X, Y)\) of \((\mu,\nu)\), we have

\( \Pr((X, Y)\not\in K_{\varepsilon/2}\times L_{\varepsilon/2})\leq\Pr(X\not\in K_{\varepsilon/2}) + \Pr(Y\not\in L_{\varepsilon/2}) = \varepsilon/2 + \varepsilon/2 = \varepsilon \)

Therefore, for any probability measure \(\pi\in\Pi(\mathcal{P}\times\mathcal{Q})\) and \(\varepsilon>0\), we have \(\pi(\mathcal{X}\times\mathcal{Y}\setminus K_{\varepsilon/2},L_{\varepsilon/2})\leq\varepsilon\), so \(\Pi(\mathcal{P}, \mathcal{Q})\) is tight and therefore precompact, by Prokhorov's theorem.

In particular, \(\{\mu\}, \{\nu\}\) are both clearly tight sets. Therefore, \(\Pi(\mu,\nu)\), the space of all couplings of \((\mu, \nu)\) is precompact. In fact, it can be shown that \(\Pi(\mu, \nu)\) is closed, so it is equal to its closure, and it is therefore compact.

Since \(\Pi(\mu,\nu)\) is compact, every sequence in \(\Pi(\mu,\nu)\) has a convergent subsequence. Let \((\pi_i)_{i\in\mathbf{N}}\) denote a convergent sequence in \(\Pi(\mu,\nu)\) such that the "transport cost" \(\int cd\pi_i\) converges to its infimum value. Then we can assume that \(\pi_i\overset{i\to\infty}{\longrightarrow}\pi\in\Pi(\mu,\nu)\). To ensure that \(\pi\) itself is the law of an optimal coupling, we need to show that \(\int cd\pi = \liminf_{k\to\infty}\int cd\pi_k\).

Let there be some upper semicontinuous function \(h\) that is integrable w.r.t. each \(\pi_i\) and \(\pi\), such that \(c\geq h\). Then \(\tilde{c} = c - h\) is a non-negative, lower semicontinuous function that is integral w.r.t. each \(\pi_i\) and \(\pi\) as well. Since \(\tilde{c}\) is a lower semicontinuous function to the nonnegtive real numbers, we construct a sequence of continuous functions \((\tilde{c}_k)_{k\in\mathbf{N}}\) that converges to \(\tilde{c}\), such that \(\tilde{c}_{k+1}\geq\tilde{c}_k\). Using the monotone convergence theorem, we have

\( \int_{\mathcal{X}\times\mathcal{Y}}\tilde{c}(x, y)d\pi = \int_{\mathcal{X}\times\mathcal{Y}}\lim_{k\to\infty}\tilde{c}_k(x, y)d\pi = \lim_{k\to\infty}\int_{\mathcal{X}\times\mathcal{Y}}\tilde{c}_k(x, y)d\pi \)

Using the monotone convergence theorem again, we have

\( \lim_{k\to\infty}\int_{\mathcal{X}\times\mathcal{Y}}\tilde{c}_k(x, y)d\pi = \lim_{k\to\infty}\int_{\mathcal{X}\times\mathcal{Y}}\tilde{c}_k(x, y)\lim_{i\to\infty}d\pi_i =\lim_{k\to\infty}\lim_{i\to\infty}\int_{\mathcal{X}\times\mathcal{Y}}\tilde{c}_k(x, y)d\pi_i \)

Ultimately, this leads us to the conclusion that

\( \int_{\mathcal{X}\times\mathcal{Y}}cd\pi = \lim_{k\to\infty}\lim_{i\to\infty}\int_{\mathcal{X}\times\mathcal{Y}}\tilde{c}_kd\pi_i \leq \liminf_{i\to\infty}\int_{\mathcal{X}\times\mathcal{Y}}\tilde{c}d\pi_i \)

So, since \(c(x, y)\geq a(x) + b(y)\), letting \(h(x, y) = a(x) + b(y)\) we see that \(\int (c - h)d\pi\leq\liminf_{i\to\infty}\int(c-h)d\pi_i\), so the coupling defined by \(\pi\) minimizes the transportation cost! Of course, this is relative to the cost function \(c-h\) rather than \(c\). According to Villani, we can usually let \(a = b = 0\) so that \(c - h = c\).

Summary (TL;DR)

Let's summarize what we just did.

  1. We first showed that \(\Pi(\mu,\nu)\), the space of couplings of \((\mu,\nu)\) is compact using Prokhorov's theorem.
  2. Since \(\Pi(\mu,\nu)\) is compact, we deduced that we can construct a convergent sequence of couplings with laws \(\pi_i\) such that the transportation cost converges to its infimum, and we say \(\pi_i\longrightarrow\pi\).
  3. Using the monotone convergence theorem, we show that \(\int cd\pi\leq\liminf_{i\to\infty}\int cd\pi_i\), which proves that the distribution \(\pi\) with marginals \(\mu,\nu\) minimizes the transportation cost.

References

I learned the ideas behind this proof from the great textbook by Villani, "Optimal Transport: Old and New" (Theorem 4.1).

Author: Harley Wiltzer

Created: 2024-07-17 Wed 21:35

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