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Timeline for Brenier's theorem

Current License: CC BY-SA 2.5

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Nov 28, 2010 at 4:04 comment added Deane Yang Any statement of uniqueness is always "except for sets of measure zero". You can always transport arbitrary sets of measure zero to anywhere you want without changing the cost.
Nov 28, 2010 at 1:46 comment added WhitAngl mmmm.. finally I still have the problem with the gaussians : I agree the gaussians can be split and are not fully transported. But if we just look at the center of each gaussian, they can move in a non unique way similarly to the dirac example, isn't it ? The theorem shouldn't be stated as : "there is uniqueness, except for a set of point of zero measure" ?
Nov 27, 2010 at 3:26 comment added Deane Yang First, what I wrote above is correct only if the target distribution are point masses at (0,1) and (1,0). Otherwise the details are more complicated. I am not an expert on this but it seems to me that nonuniqueness arises when there are symmetric relationships between the source and target distributions. And this can definitely arise with discrete distributions. But you can always try to break the symmetries by moving the point masses a little or better using a continuous source distribution.
Nov 27, 2010 at 2:39 comment added WhitAngl I mean : for uniqueness issues, if the histograms that I am transporting are 2 gaussians in 2D but discretized in bins, is it similar to the 0-measure case (each bin is considered as a dirac) or to the continuous case (after all they both represent gaussians) ?
Nov 27, 2010 at 2:07 comment added WhitAngl ok, great, it makes sense. Thanks! Since I'm using a discrete formulation (a transportation simplex) to solve the mass transport, I'm wondering whether such non uniqueness issues can come up in the discrete setting (each "bin" of my histograms could be considered as dirac mass, isn't it?).
Nov 27, 2010 at 2:05 vote accept WhitAngl
Nov 27, 2010 at 1:30 history answered Deane Yang CC BY-SA 2.5