Timeline for Are most random variables trivially sub-gaussian? [closed]
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
|
|
| Jul 30, 2015 at 18:44 | history | closed |
Brendan McKay Douglas Zare Nate Eldredge Joonas Ilmavirta Yoav Kallus |
Not suitable for this site | |
| Jul 30, 2015 at 18:19 | answer | added | Igor Rivin | timeline score: 2 | |
| Jul 30, 2015 at 14:24 | comment | added | Guillaume Dehaene | Thank you very much for the insight usul. I now realize what I got wrong. | |
| Jul 30, 2015 at 13:59 | review | Close votes | |||
| Jul 30, 2015 at 18:44 | |||||
| Jul 30, 2015 at 13:57 | comment | added | usul | As Brendan points out, but to put a more fine point on it, the condition that $\mathbb{E} e^{a X^2}$ exists for some $a > 0$ is a quite strong condition that the tails of $X$ be "light". To put it another way, let $Y = \exp(a X^2)$; the condition is that the expectation of $Y$ exists. So conversely, if we have a $Y$ whose expectation exists and we want to guarantee to get a subgaussian $X$, we need to first take its logarithm and then its square root, which really "shrinks" the tails. | |
| Jul 30, 2015 at 13:43 | comment | added | Brendan McKay | I don't know what "most" random variables are, but you can take anything whose tail is merely exponential, like the exponential distribution, and $E(\exp(aX^2)$ won't even exist for $a>0$. Ans this is not a research-level question. | |
| Jul 30, 2015 at 13:34 | comment | added | Budenn | A random variable with Cauchy distribution is not sub-gaussian, although this might fit the 'absurdly miss-behaved' part. Same for slash distribution. | |
| Jul 30, 2015 at 12:57 | history | asked | Guillaume Dehaene | CC BY-SA 3.0 |