Timeline for Is there a good explanation for this fact on pairwise independent variables?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jul 12 at 19:44 | answer | added | jlewk | timeline score: 0 | |
| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jun 3, 2011 at 19:05 | answer | added | Jason Swanson | timeline score: 1 | |
| Jun 3, 2011 at 5:26 | vote | accept | Ewan Delanoy | ||
| Jun 2, 2011 at 21:22 | answer | added | Ori Gurel-Gurevich | timeline score: 4 | |
| Jun 2, 2011 at 20:29 | answer | added | Did | timeline score: 3 | |
| Jun 2, 2011 at 20:21 | history | edited | Ewan Delanoy | CC BY-SA 3.0 |
added update
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| Jun 2, 2011 at 18:18 | comment | added | Ewan Delanoy | @Michael : Sorry for the unclear English in the OP. By "one" I mean "at least one", not "exactly one". | |
| Jun 2, 2011 at 18:15 | comment | added | Nate Eldredge | @Ewan: Algebra failure. | |
| Jun 2, 2011 at 18:15 | comment | added | Nate Eldredge | @Michael Hardy: I think the desired conclusion could be written as $P(\vec{X} \in \{(0,0,0,0), (0,0,0,1), (0,0,1,0) \} ) > 0$. | |
| Jun 2, 2011 at 18:15 | comment | added | Ewan Delanoy | @Nate : why was it wrong? | |
| Jun 2, 2011 at 18:14 | history | edited | Ewan Delanoy | CC BY-SA 3.0 |
corrected "at least one" instead of "one"
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| Jun 2, 2011 at 18:12 | comment | added | Nate Eldredge | I deleted my answer for now since it was wrong, and I didn't see how to salvage it. | |
| Jun 2, 2011 at 16:08 | comment | added | Ewan Delanoy | @Stanley Yao Xiao : The third value is (0,0,1,0), do you have trouble reading it on your screen? | |
| Jun 2, 2011 at 14:46 | comment | added | Stanley Yao Xiao | What is the third value? | |
| Jun 2, 2011 at 14:26 | history | asked | Ewan Delanoy | CC BY-SA 3.0 |