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Dec 28, 2020 at 16:56 comment added Gabe K In many situations the lack of invariance for the Wasserstein metric is a feature, not a bug. If you are comparing categorical datasets (e.g. draws from a multinomial distribution), then it makes sense to use the Fisher metric or something "entropic" which is invariant under sufficient statistics. However, in many applications the sample space has an underlying metric to take into account and if you choose a divergence with strong invariance properties, you will ignore the underlying geometry.
Dec 28, 2020 at 15:42 comment added Catologist_who_flies_on_Monday Yes true, but the Wasserstein distance has a nice dual representation which has "robust/adversarial" interpretations...this does not seem to be the case for the Fisher metric.
Dec 28, 2020 at 15:19 history answered Frederic Barbaresco CC BY-SA 4.0