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Transport on a finite chain

Example

Let $n \geq 1$ and let $X = {0, 1, \dots, n}$ with metric $d (i, j) = ∣ i - j ∣$ . Let $p = (p_{i})_{i = 0}^{n}$ and $q = (q_{i})_{i = 0}^{n}$ be probability vectors. Put $R_{k} = \sum_{i = 0}^{k} (p_{i} - q_{i})$ for $0 \leq k < n$ . Define $sgn (t) = 1$ for $t > 0$ , $sgn (t) = - 1$ for $t < 0$ , and $sgn (0) = 0$ . Then

W_{1} (p, q) = \sum_{k = 0}^{n - 1} ∣ R_{k} ∣ .

An attaining normalized potential is given recursively by $f_{0} = 0$ and $f_{k + 1} = f_{k} - sgn (R_{k})$ for $0 \leq k < n$ .

Example references

Peyré and Cuturi, Computational Optimal Transport, §6.1

Santambrogio, Introduction to Optimal Transport Theory, §1.3

Facts & Assumptions

Given: The chain, probability vectors, and cumulative differences in the Example.

[F1]

The usual metric on $R$ is $d_{R} (s, t) = ∣ s - t ∣$ , and its restriction to ${0, \dots, n}$ is a metric (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded).

[F2]

For the ordered points $x_{0}, \dots, x_{n}$ , finite-metric transport equals the maximum of $\sum_{i = 0}^{n} (p_{i} - q_{i}) f_{i}$ over 1-Lipschitz vectors normalized by $f_{0} = 0$ ; this is the duality theorem's first-coordinate normalization after relabeling (Finite-metric transport duality and complementary slackness).

[F3]

A probability vector has nonnegative entries summing to one (Finite probability vectors).

Verification

1.1givenF1F3algebra

Write $r_{i} = p_{i} - q_{i}$ for $0 \leq i \leq n$ , set $R_{- 1} = 0$ and $R_{n} = 0$ , and use $\sum_{i} r_{i} = 0$ from [F3]. Then $r_{i} = R_{i} - R_{i - 1}$ for every $i$ , and expanding the finite sums gives $\sum_{i = 0}^{n} r_{i} f_{i} = \sum_{k = 0}^{n - 1} R_{k} (f_{k} - f_{k + 1})$ . For any feasible 1-Lipschitz vector, $∣ f_{k} - f_{k + 1} ∣ \leq 1$ , so each term is at most $∣ R_{k} ∣$ and the dual objective is at most $\sum_{k = 0}^{n - 1} ∣ R_{k} ∣$ .

2.1givenF1step 1.1algebra

The recursively defined potential has $∣ f_{k + 1} - f_{k} ∣ = ∣ sgn (R_{k}) ∣ \leq 1$ on each adjacent pair. For $i < j$ , the triangle inequality on real absolute values gives $∣ f_{i} - f_{j} ∣ \leq \sum_{k = i}^{j - 1} ∣ f_{k + 1} - f_{k} ∣ \leq j - i = d (x_{i}, x_{j})$ ; also $f_{0} = 0$ . Thus it is feasible. Its objective, by step 1.1, is $\sum_{k = 0}^{n - 1} R_{k} sgn (R_{k}) = \sum_{k = 0}^{n - 1} ∣ R_{k} ∣$ , so it attains the dual maximum.

3.1step 1.1step 2.1F2∎

The dual maximum is therefore $\sum_{k = 0}^{n - 1} ∣ R_{k} ∣$ by steps 1.1–2.1. Applying [F2] identifies this value with $W_{1} (p, q)$ and proves the formula and the asserted attaining potential.

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Sources

Gabriel Peyré and Marco Cuturi, Computational Optimal Transport, §6.1 (standard reference, not scraped)
Filippo Santambrogio, Introduction to Optimal Transport Theory, §1.3, PDF pp. 5–6 (standard reference, not scraped)