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Oct 14, 2022 at 2:32 history edited YCor
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Jan 29, 2013 at 18:21 answer added Günter Rote timeline score: 5
May 8, 2012 at 7:55 vote accept Fedor Petrov
May 7, 2012 at 22:50 comment added Deane Yang Yes, I was using the wrong topology. Sorry about that.
May 7, 2012 at 20:25 comment added alvarezpaiva @Fedor: sorry, I was thinking of something else.
May 7, 2012 at 20:18 history edited Fedor Petrov CC BY-SA 3.0
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May 7, 2012 at 20:10 answer added Anton Petrunin timeline score: 17
May 7, 2012 at 19:50 comment added Fedor Petrov @Deane: you mean something like "if $K_1$ and $K_2$ are on Hausdorff distance $c$, then the probability [for a random point in $K_1$ have first coordinate less then $x$] does not exceed the probability [for a random point in $K_2$ have first coordinate less then $x+c$]"? I probably should be able to prove such things, but alas, just now I feel myself disabled.
May 7, 2012 at 19:47 comment added Fedor Petrov @alvarezpaiva: sorry, what volumes are you saying about? Bodies may even have different dimensions.
May 7, 2012 at 19:40 comment added alvarezpaiva @Deane: You're right, but I think you're implicitly using that the Hausdorff distance and the distance defined as the volume of the symmetric difference of two convex bodies define the same topology on the space of convex compact sets. It may be worthwhile to point this out.
May 7, 2012 at 19:25 comment added Deane Yang Yes, the barycenter depends continuously on the Hausdorff metric. You should be able to show this directly from the definition of the barycenter as the expected value of a point in $R^n$ with respect to the uniform probability density on $K$.
May 7, 2012 at 19:07 history asked Fedor Petrov CC BY-SA 3.0