Timeline for Linking Wasserstein and total variation distances
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2016 at 13:10 | answer | added | Arnak | timeline score: 1 | |
| Feb 3, 2016 at 12:43 | history | edited | Guillaume Dehaene | CC BY-SA 3.0 |
equation was wrong
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| Jan 31, 2016 at 22:35 | comment | added | Christian Clason | @GuillaumeDehaene That's why I only wrote a comment, not an answer. I'd expect that co-measurability alone would not be sufficient, and you'd need to use more specific properties of the two marginals you wish to compare (i.e., prove the bound for the concrete case you're interested in, rather than apply a general theorem). | |
| Jan 31, 2016 at 18:35 | comment | added | Guillaume Dehaene | You misunderstood me slightly (because I explained poorly \^\^). I mean to bind the total-variation distance between the marginals at time t, and not the distribution of the paths. @ChristianClason : wouldn't the claim in theorem 4 of Gibbs et al not apply only to trivial examples like: $p_1$ is a binomial (only rational values) while $p_2$ is a gaussian. For such an example, the wasserstein can tend to 0, while TV convergence can't happen because these aren't co-measurable. Couldn't a co-measurability condition of some kind save me from such boring counter-examples ? | |
| Jan 31, 2016 at 18:26 | history | edited | Guillaume Dehaene | CC BY-SA 3.0 |
Clarified the question
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| Jan 31, 2016 at 14:59 | comment | added | Dan | If $X_1$ and $X_2$ both solve the same SDE you've written but are started from distinct initial positions, then the laws (on the path space) of these processes are singular. In this case, there's no (nontrivial) bound on the total variation. If the initial positions are random, and if the law $\mu_1$ of $X_1(0)$ is absolutely continuous to the law $\mu_2$ of $X_2(0)$, then the relative entropy on the path space $H(Law(X_1)|Law(X_2))$ is equal to the relative entropy of the first marginals $H(\mu_1|\mu_2)$. Then use Pinsker's inequality to get a good bound on the total variation. | |
| Jan 31, 2016 at 13:44 | comment | added | Christian Clason | On an infinite set, no such bound is possible; see Theorem 4 in Gibbs, A. L. and Su, F. E. (2002), On Choosing and Bounding Probability Metrics. International Statistical Review, 70: 419–435 . | |
| Jan 31, 2016 at 13:33 | history | asked | Guillaume Dehaene | CC BY-SA 3.0 |