Timeline for The largest Wasserstein distance to uniform distribution among all probability distributions with uniform marginals
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 13, 2017 at 3:13 | vote | accept | O. Richard | ||
| Mar 13, 2017 at 0:10 | comment | added | O. Richard | Yes, you're right. In the meantime, I'm not sure if this can be extended to $K>2$, since the set of extreme points are undecided. | |
| Mar 12, 2017 at 23:59 | comment | added | Steve | Well, this was just the intuition for the proof. It is not needed. But what I meant is that if $\hat{\pi}$ is any optimal coupling for the monotonic distribution, then we should have $\hat{\pi}_{(i,i),(i',j')} = 0$ if $||(i,i)-(i',j')||_1 > ||(i',i')-(i',j')||_1$. I.e. if we cannot go from (i,i) to (i',j') by always going away from the diagonal, then $\hat{\pi}_{(i,i),(i',j')}$ should be zero. | |
| Mar 12, 2017 at 23:51 | comment | added | O. Richard | How do you describe "strictly away" rigorously? | |
| Mar 12, 2017 at 23:42 | comment | added | Steve | The coupling $\pi$ is only optimal for the monotonic/comonotonic case. As you mentioned, this only works with the $l_1$ norm. For the $l_1$ norm, we can visualize going from the diagonal to the other lattice points in $\{1,...,N\}^2$ in horizontal/vertical steps, and the $l_1$ norm is just the number of steps. The idea was, if we are going over from the monotonic distribution to the uniform distribution, any coupling that moves mass strictly away from the diagonal is going to be optimal, and the simplest of those is just $\pi$. | |
| Mar 12, 2017 at 23:11 | comment | added | O. Richard | I think if we can prove that the optimal coupling between uniform and comonotonic distribution is indeed given by $\pi$, then combining with your answer we can obtain a proof. I know one optimal coupling between uniform and comonotonic distribution is given by the monotone coupling which is different from $\pi$, but maybe due to the specialty of $\ell_1$-norm, $\pi$ is also an optimal coupling. I have tested on $N=4$ and they give the same value. | |
| Mar 12, 2017 at 22:54 | comment | added | O. Richard | Thanks for your answer. It seems that in your optimal coupling $\pi$ for computing $W_1$, the probability mass on $(i,\sigma(i))$ is only allowed to split in the same row. Why? This is not true for $\ell_2$-norm, but may be correct for $\ell_1$. | |
| Mar 12, 2017 at 22:13 | review | First posts | |||
| Mar 12, 2017 at 22:15 | |||||
| Mar 12, 2017 at 22:12 | history | answered | Steve | CC BY-SA 3.0 |