Timeline for An inequality for two independent identically distributed random vectors in a normed space
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 3, 2015 at 11:17 | comment | added | Bill Johnson | I added the missing minus sign. | |
| Jun 3, 2015 at 11:16 | history | edited | Bill Johnson | CC BY-SA 3.0 |
added 1 character in body
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| Jun 3, 2015 at 3:12 | vote | accept | Iosif Pinelis | ||
| Jun 3, 2015 at 3:11 | vote | accept | Iosif Pinelis | ||
| Jun 3, 2015 at 3:11 | |||||
| Jun 3, 2015 at 3:10 | comment | added | Iosif Pinelis | This is great! As far as I am concerned, this completes the answer. Thank you! | |
| Jun 3, 2015 at 1:01 | comment | added | Bill Johnson | Oh, there should be a minus sign in front of the fourth sample. Sorry for the typo. | |
| Jun 3, 2015 at 0:21 | comment | added | Iosif Pinelis | Alas, for dimension $3$ my calculations (with Mathematica) suggest that $E\|X-Y\|=\frac{18}{16}$ and $E\|X+Y\|=\frac{22}{16}>E\|X-Y\|$. The Mathematica notebook and its pdf copy can be found at dropbox.com/s/z69emx0royqdjjt/2iidVectors.nb?dl=0 and dropbox.com/s/6fcl07z3evdpjw6/2iidVectors.pdf?dl=0 , respectively. | |
| Jun 2, 2015 at 23:15 | history | edited | Bill Johnson | CC BY-SA 3.0 |
added 613 characters in body
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| Jun 2, 2015 at 20:14 | comment | added | Iosif Pinelis | This is very nice, thank you Bill! Now only dimension 3 remains. In fact, your construction works for any dimension $\ge4$, and it seems to be the simplest construction so far! | |
| Jun 2, 2015 at 18:52 | history | answered | Bill Johnson | CC BY-SA 3.0 |