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The wikipedia link for Wasserstein metric is defined for $p\in[1,\infty)$. https://en.wikipedia.org/wiki/Wasserstein_metric Given some data the distance can be calculated using an optimization routine. In the routine one can choose any value for $0<p<1$ and obtain a solution. The question is can we interpet the solution as a valid distance, with proper meaning?

Dan Asimov
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Creator
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1 Answers1

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Note that for $0<p<1$, $d_p(x,y)\equiv (d(x,y))^{p}$ is a metric on your original space. Thus, you can define a Wasserstein $p$ metric for $(d(x,y))^{p}$ as the Wasserstein $1$ metric for $d_p(x,y)$ in the usual setting.

Haoqing Yu
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  • Do you mean Wasserstein metric 1 is valid for p<1? Further even for p<1 it is called as Wasserstein metric 1? – Creator Jan 02 '24 at 23:15
  • I would say it is the most natural way to define the Wasserstein $p$ metric for $p<1$. I have not seen anywhere where Wasserstein $p$ metric is explicitly defined for $p<1$ (In the computational optimal transport by Peyré and Cuturi, it is just mentioned in remark 2.15). In a lot of cases, the convexity of the cost function is needed to produce many results in optimal transport but $|x-y|^{p}$ is non-convex for $p<1$. I don't know anything other than that. – Haoqing Yu Jan 03 '24 at 02:08
  • Thank you so much for your reference. – Creator Jan 04 '24 at 21:37
  • Would you mind to explain why $(d(x,y))^p)$ is a metric in the original space only when $p<1$. I am not sure and still thinking of it. – Creator Jan 06 '24 at 19:03
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    It basically follows from the fact that $d(x,y)$ is a metric and $x \mapsto x^p$ is a strictly increasing function for nonnegative $x$. For triangular inequality, note the following result – Haoqing Yu Jan 06 '24 at 20:11
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    However, the inequality result doesn't hold when $p>1$ so it is a metric ONLY when $p\leqslant1$ – Haoqing Yu Jan 06 '24 at 20:19