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Sep 12, 2021 at 14:36 vote accept GJC20
Sep 12, 2021 at 14:27 history edited YCor
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Sep 12, 2021 at 11:52 answer added user95282 timeline score: 3
Sep 11, 2021 at 17:51 comment added Dirk Yeah, Wasserstein-1 would be without boundedness of f. This variant is also know as Kantorovich-Rubinstein metric.
Sep 11, 2021 at 10:30 comment added GJC20 Yes. While $\rho$ is not the Wasserstein metric (unless $\mu,\nu$ have bounded supports) but is the metrization of the weak convergence (which can be found in Billingsley's book)
Sep 11, 2021 at 9:10 comment added Dirk I don't know the answer of the top of my head, but isn't the $d$ the Prokhorov metric and $\rho$ the 1-Wasserstein metric (also known as bounded Lipschitz metric and under other names).
Sep 11, 2021 at 8:27 history asked GJC20 CC BY-SA 4.0