A Pair Maximising a Quadratically Penalised Difference of Law-Invariant Functions Within Its Laws Realises an Optimal Coupling
lemmaAnalysisProbabilitylem:penalised-maximiser-optimal-coupling-2026aOn a rich probability space, if a pair of square-integrable random vectors maximises U(X) - V(Y) - alpha times the squared mean-square distance among pairs with the same laws, U and V law-invariant, then its mean-square distance equals the Wasserstein distance of the laws and the law of the pair is an optimal coupling.
In the setting of The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation, assume that is rich. Let be the space of classes of square-integrable random vectors, and the Wasserstein space, so that for every by The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation §law-map. Let be law-invariant, let be positive, and let satisfy
for all with and . Then the following hold.
1. (The penalty is the Wasserstein distance)¶
2. (The pair realises an optimal coupling)¶ The law of the pairing of any representatives of and , a random vector in by Basic Properties of Random Vectors: Coordinates, Borel Images and Arithmetic, Change of Variables, Almost Sure Equality and Pairs §pair, is an optimal coupling of and .
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