Timeline for minimum of two probability densities
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 20, 2012 at 8:49 | vote | accept | Alekk | ||
| Sep 20, 2012 at 1:40 | answer | added | George Lowther | timeline score: 5 | |
| Sep 18, 2012 at 19:51 | comment | added | Alekk | @George: great! I can't believe that I missed that. I did notice that the whole thing was invariant by rearrangement but did not see that $E \|X\|^d$ was decreasing. Many thanks! | |
| Sep 18, 2012 at 18:56 | comment | added | Mark Meckes | @George: very nice! I'm kicking myself for not thinking of it. @Alekk: the buzzwords for what George is describing are "symmetric decreasing rearrangement". See Chapter 3 of Analysis by Lieb and Loss for more information. | |
| Sep 18, 2012 at 18:36 | comment | added | George Lowther | The answer must be no, and in fact you must have the integral bounded by the volume of a ball of radius $\lVert X\lVert_d$. This is because you can swap regions of equal volume in $R^d$ about to move the large probability regions closer to the origin, which doesn't change the integral but can only decrease $\lVert X\lVert_d$. So, it reduces to the radially decreasing density case. | |
| Sep 18, 2012 at 18:18 | answer | added | Arthur B | timeline score: 0 | |
| Sep 18, 2012 at 17:54 | answer | added | Mark Meckes | timeline score: 2 | |
| Sep 18, 2012 at 13:52 | history | asked | Alekk | CC BY-SA 3.0 |