Timeline for How to check for bivariate Gaussianity without the use of regression?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2012 at 1:03 | vote | accept | CommunityBot | ||
| Apr 1, 2011 at 2:25 | comment | added | cardinal | @H_S, please don't simultaneously cross-post. It's impolite. | |
| Apr 1, 2011 at 0:00 | history | tweeted | twitter.com/#!/StackStats/status/53607592743804929 | ||
| Mar 31, 2011 at 23:54 | comment | added | whuber♦ | No, that's the whole point: univariate checks are manifestly not regression methods. Neither is taking a linear combination of variables. | |
| Mar 31, 2011 at 23:53 | comment | added | user1102 | @whuber: I guess what you are suggesting is regression, and I have stated in my question that without the use of regression | |
| Mar 31, 2011 at 23:50 | comment | added | whuber♦ | Good point. But it seems like those replies generalize in some ways. For instance, they suggest that one way to detect bivariate normality is to check various linear combinations of the two variables separately for normality. This reduces it to a set of univariate tests. | |
| Mar 31, 2011 at 22:53 | comment | added | user1102 | @whuber: Thanks for the link. However, that question talks about univariate gaussianity, and nothing about bivariate gaussianity. | |
| Mar 31, 2011 at 22:49 | comment | added | whuber♦ | It is difficult to see how the variogram applies. It summarizes a single multivariate observation and requires strong additional assumptions in order to estimate variances and covariances. | |
| Mar 31, 2011 at 22:05 | answer | added | schenectady | timeline score: 4 | |
| Mar 31, 2011 at 21:58 | history | edited | user1102 | CC BY-SA 2.5 |
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| Mar 31, 2011 at 21:48 | history | asked | user1102 | CC BY-SA 2.5 |