Timeline for How to sample pairwise independent gaussians
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2010 at 16:34 | vote | accept | Anindya De | ||
| Nov 16, 2010 at 18:05 | comment | added | András Salamon | @Kevin: $m = \omega(k)$ in this context means that $m$ must grow faster than $k$, so $m = \sqrt{k}$ would not be fine. $m = 0.0001 k \log k$ would be fine. | |
| Nov 16, 2010 at 17:17 | answer | added | Shai Covo | timeline score: 9 | |
| Nov 16, 2010 at 16:19 | comment | added | Kevin O'Bryant | @Kaveh, I think you have the question right, except that it isn't necessarily more than $k$. Perhaps he/she would be satisfied with $m=\sqrt{k}$. Perhaps the question is really ``how large can we take $m$''? In any case, all of these variations seem to be answered by Nate below. | |
| Nov 16, 2010 at 6:49 | comment | added | Kaveh | I think the question is: we have $k$ i.i.d. standard Gaussian random variables ($X_i$) and using them we want to generate more than $k$ pairwise independent standard Gaussian random variables ($Y_i$) (and we want to do it an efficient way). Cross posted on cstheory.SE: cstheory.stackexchange.com/questions/3034/… | |
| Nov 16, 2010 at 5:53 | comment | added | zhoraster | See Devroye - Non-Uniform Random Variate Generation. | |
| Nov 16, 2010 at 4:51 | answer | added | Nate Eldredge | timeline score: 10 | |
| Nov 16, 2010 at 4:09 | comment | added | Mike Spivey | @Sleepless: i.i.d. means "independent and identically distributed." | |
| Nov 16, 2010 at 4:00 | comment | added | Gjergji Zaimi | I changed the tag since independence-results refers to a different concept of independence. | |
| Nov 16, 2010 at 3:59 | history | edited | Gjergji Zaimi |
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| Nov 16, 2010 at 3:56 | comment | added | sleepless in beantown | What does i.i.d stand for? independently something-something? | |
| Nov 16, 2010 at 3:45 | history | asked | Anindya De | CC BY-SA 2.5 |