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Total variation, Wasserstein, and Prokhorov generate the same topology on the space of probability measures on a finite and discrete space.

I'm curious about a countably infinite space. When do they generate the same topology in this case? It would seem that a sufficient condition is the metric taking only the values 0 and 1, i.e., the space is discrete. Is there a nice necessary and sufficient condition?

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  • $\begingroup$ They don't even all live on the same space, do they? The total variant and Prokhorov metrics make sense on the space of all measures on your state space $X$, but the Wasserstein metric is only defined for measures of finite $p$th moment, for some fixed $p$. As soon as $X$ is unbounded, there are measures with infinite $p$th moment, and so these spaces of measures are no longer the same. $\endgroup$
    – Nate Eldredge
    Commented Apr 18, 2016 at 21:02
  • $\begingroup$ By the way, it's easy to see that if the total variation and weak (Prokhorov) topologies agree, then the topology on the state space $X$ must be discrete. Prove the contrapositive: suppose $X$ has a limit point $x_0$. Then there is a sequence $x_n \to x_0$ with all the $x_n$ different from $x_0$. Consider the sequence of Dirac measures $\delta_{x_n}$. These converge to $\delta_{x_0}$ weakly but not in total variation, so the two topologies differ. $\endgroup$
    – Nate Eldredge
    Commented Apr 18, 2016 at 21:28
  • $\begingroup$ OK, great. Guess that answers my question (provided I toss out Wasserstein). I suppose in thinking about Wasserstein, for measures with no pth moment, they would live at infinity and be distance infinity from every other measure. So they would be included, but topologically isolated. $\endgroup$
    – D.R.
    Commented Apr 19, 2016 at 1:18
  • $\begingroup$ Well, that will certainly give you a different topology than the other two, provided there exist any measures with infinite $p$th moments, which happens iff your space $X$ is unbounded. So I guess all three topologies will agree iff $X$ is bounded and has the discrete topology. $\endgroup$
    – Nate Eldredge
    Commented Apr 19, 2016 at 3:39

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