Timeline for Distribution of infinity-norm over the unit sphere
Current License: CC BY-SA 4.0
50 events
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| Dec 11, 2022 at 22:08 | review | Low quality posts | |||
| Dec 12, 2022 at 0:27 | |||||
| Dec 11, 2022 at 8:01 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 11, 2022 at 7:54 | comment | added | Amit Portnoy | @YemonChoi . I understood the comment and found it to be extremely rude. Instead of asking users to change their writing and editing capabilities (just be better at noticing their small typos etc.,), which is quite arrogant IMHO, a more community-inclusive approach would be to inquire with stack overflow meta team on how to throttle notifications on small edits. | |
| Dec 11, 2022 at 5:35 | comment | added | Yemon Choi | Responding to @StefanKohl's advice with "go and set your alert notifications issues and preferences differently" seems to be missing the point of the modeator's comment. | |
| Dec 10, 2022 at 15:59 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 15:57 | comment | added | Amit Portnoy | @მამუკაჯიბლაძე Marsaglia (1972), en.m.wikipedia.org/wiki/… | |
| Dec 10, 2022 at 15:46 | comment | added | მამუკა ჯიბლაძე | Why is normalization of an i.i.d. vector of standard normals uniformly distributed on the unit sphere? | |
| Dec 10, 2022 at 15:40 | review | Low quality posts | |||
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| Dec 10, 2022 at 15:22 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 15:18 | comment | added | Amit Portnoy | Please be aware that there is meta.stackoverflow.com where you can inquire into your alert notifications issues and preferences. | |
| Dec 10, 2022 at 15:09 | comment | added | Stefan Kohl♦ | Please be aware that every edit of a question or of one of its answers bumps the thread to the front page. This has happened for this thread already more than 20 times, and this is a nuisance for other users. Please refrain from unnecessary edits to your posts. -- Usually, the vast majority of minor edits can be avoided by writing and proofreading a question or an answer (or any update to such) carefully before posting it. | |
| Dec 10, 2022 at 11:18 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 10:33 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 10:20 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 9:54 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 9:29 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 3:14 | comment | added | Amit Portnoy | It is possible to run simulations to verify this. I'm not even sure what the value is, but the logic overall seems OK. Are you sure it less than 1 for $r \ge 1$? | |
| Dec 10, 2022 at 3:05 | history | rollback | Amit Portnoy |
Rollback to Revision 19
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| Dec 10, 2022 at 3:02 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 2:42 | comment | added | fedja | Yes, the entire formula: the last one in the chain. If $r\ge 1$, the whole sphere is there, so the output should be $1$ regardless of $r$. | |
| Dec 10, 2022 at 2:41 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 2:40 | comment | added | Amit Portnoy | I'm not sure which part becomes 1? The entire formula? In any case, the integral is the law of total probability over the scale that transforms it into the sphere. $F_{||Z||_\infty}$ is defined over the Normal vector not the n-1-sphere, | |
| Dec 10, 2022 at 2:32 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 2:32 | comment | added | fedja | The integral is defined over the normalization factor to get to the n-sphere so it is fine I don't understand the meaning of this phrase. Neither do I understand how it becomes constant $1$ for large $r$ (I mean the integral with $erf$, of course) | |
| Dec 10, 2022 at 2:29 | comment | added | Amit Portnoy | Added a simplifying update above. No need to calculate the integral directly. I think. | |
| Dec 10, 2022 at 2:27 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 10, 2022 at 1:55 | comment | added | Amit Portnoy | The integral is defined over the normalization factor to get to the n-sphere so it is fine. | |
| Dec 9, 2022 at 23:54 | comment | added | fedja | It cannot be true for all values of $r$ because the last integral is certainly analytic when $r>0$, while the answer definitely isn't. | |
| Dec 9, 2022 at 16:19 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| S Dec 9, 2022 at 16:09 | review | First answers | |||
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| S Dec 9, 2022 at 16:09 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 15:55 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| S Dec 9, 2022 at 14:01 | review | First answers | |||
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| S Dec 9, 2022 at 14:01 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 13:39 | review | First answers | |||
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| Dec 9, 2022 at 13:38 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| S Dec 9, 2022 at 13:30 | review | First answers | |||
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| S Dec 9, 2022 at 13:30 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 11:27 | review | First answers | |||
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| Dec 9, 2022 at 11:25 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 10:55 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 10:25 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 10:20 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 10:09 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 9:26 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 9:20 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 9:14 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| Dec 9, 2022 at 9:07 | history | edited | Amit Portnoy | CC BY-SA 4.0 |
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| S Dec 9, 2022 at 8:57 | review | First answers | |||
| Dec 9, 2022 at 11:25 | |||||
| S Dec 9, 2022 at 8:57 | history | answered | Amit Portnoy | CC BY-SA 4.0 |