Timeline for Covariance and independence?
Current License: CC BY-SA 4.0
9 events
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| Mar 23, 2023 at 2:21 | comment | added | whuber♦ | @Dilip That's why I specified that the absolute third moment must be finite. | |
| Mar 22, 2023 at 19:34 | comment | added | whuber♦ | @Dilip Note, though, that the zero third moment is an immediate consequence of symmetry and a zero mean, making this example more general than it might appear. A distribution with these properties can be generated from any starting distribution with finite absolute third moment simply by symmetrizing it, giving arbitrarily rich examples. | |
| S Feb 12, 2022 at 14:55 | history | suggested | Lycanthropeus | CC BY-SA 4.0 |
Improved formatting by adding square brackets.
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| Feb 12, 2022 at 3:57 | review | Suggested edits | |||
| S Feb 12, 2022 at 14:55 | |||||
| Feb 16, 2012 at 8:25 | comment | added | mpiktas | @DilipSarwate, thanks, I've edited my answer accordingly. When I wrote it I though about normal variables, for them zero third moment follows from zero mean. | |
| Feb 16, 2012 at 8:23 | history | edited | mpiktas | CC BY-SA 3.0 |
added 1 characters in body
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| Feb 16, 2012 at 3:09 | comment | added | Dilip Sarwate | +1 but as a minor nitpick, you do need to assume that $E[X^3] = 0$ separately (it does not follow from the assumption of symmetry of the distribution or from $E[X] = 0$), so that we don't have issues such as $E[X^3]$ working out to be of the form $\infty - \infty$. And I am queasy about @ocram's assertion that "a N(0,1) rv and a chi2(1) rv are uncorrelated." (emphasis added) Yes, $X \sim N(0,1)$ and $X^2 \sim \chi^2(1)$ are uncorrelated, but not any $N(0,1)$ and $\chi^2(1)$ random variables. | |
| Jul 15, 2011 at 8:21 | comment | added | ocram | I like that example too. As a particular case, a N(0,1) rv and a chi2(1) rv are uncorrelated. | |
| Jul 15, 2011 at 8:18 | history | answered | mpiktas | CC BY-SA 3.0 |