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Given the wasserstein distance of order two, $$d(\mu_{1},\mu_{2})^{2}=inf_{p\in P(\mu_{1},\mu_{2})}\int_{R^{n}\times R^{n}}|x-y|^{2}p(dxdy)$$ Is there $d(\mu_{1},\mu_{2})=d(\mu_{2},\mu_{1})$ or $d(\mu_{1},\mu_{2})=-d(\mu_{2},\mu_{1})$?

Sha
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1 Answers1

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One can show that the Wasserstein is a distance/metric on the space of probability measures, therefore we have symmetry, i.e. $d(\mu_1, \mu_2) = d(\mu_2, \mu_1)$. See for instance this reference.

Sharon Puthuparambil
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