Timeline for Why are isotropic random vectors called isotropic if they aren't? [closed]
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 2, 2020 at 16:08 | vote | accept | lamlame | ||
| Nov 30, 2020 at 20:18 | history | closed |
Joonas Ilmavirta Desiderius Severus Eric Peterson Mark Wildon Alexandre Eremenko |
Needs details or clarity | |
| Nov 22, 2020 at 23:43 | answer | added | R W | timeline score: 2 | |
| Nov 22, 2020 at 22:49 | comment | added | Michael Engelhardt | I still don't get why you assert that "isotropic random vectors don't have the property of isotropy". The contradiction doesn't seem to be between these notions of isotropy, but between the given definition of an isotropic random vector and a mistaken interpretation of that definition in a post on math.stackexchange.com. | |
| Nov 22, 2020 at 21:11 | answer | added | Carlo Beenakker | timeline score: 1 | |
| Nov 22, 2020 at 21:11 | comment | added | lamlame | Okay thank you. In the other post the OP asked if the specific translation X-EX is isotropic. This would be true if EX = 0, but since the answer was this translation did not preserve isotropy presumably EX=/=0. So there isn't "sameness in all directions", even with respect to the origin, unless their answer was wrong. | |
| Nov 22, 2020 at 21:06 | review | Close votes | |||
| Nov 30, 2020 at 20:18 | |||||
| S Nov 22, 2020 at 20:59 | history | suggested | RobPratt |
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| Nov 22, 2020 at 20:30 | comment | added | Malkoun | Well, I think they mean "isotropic" as "the same in all directions", roughly speaking, but with respect to the origin only. In the other post, the OP asked whether translation preserves isotropy and it turns out the answer is no. But $\mathbb{R}^n$ is not just an affine space, it also has a "special point", namely the origin. Your question was about terminology, so I hope my somewhat vague comment helps. | |
| Nov 22, 2020 at 20:27 | review | Suggested edits | |||
| S Nov 22, 2020 at 20:59 | |||||
| Nov 22, 2020 at 20:23 | comment | added | lamlame | I added a link. If a random vector was to some amount uniform in different directions, it should at the least have expectation $0$. (If it wasn't rotate the vector to get a contradiction). | |
| Nov 22, 2020 at 20:19 | history | edited | lamlame | CC BY-SA 4.0 |
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| Nov 22, 2020 at 20:08 | comment | added | Michael Engelhardt | It's hard to make sense of your question. You assert that an isotropic object doesn't have the property of isotropy, without explaining what you mean by "property of isotropy". | |
| Nov 22, 2020 at 19:58 | review | First posts | |||
| Nov 22, 2020 at 20:08 | |||||
| Nov 22, 2020 at 19:53 | history | edited | lamlame | CC BY-SA 4.0 |
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| Nov 22, 2020 at 19:48 | history | asked | lamlame | CC BY-SA 4.0 |