Timeline for For $r>s\geq1$, convergence in $s^{\text{th}}$ mean does not imply convergence in $r^{\text{th}}$ mean
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 10, 2022 at 15:12 | vote | accept | CrunchySia24 | ||
| Feb 10, 2022 at 15:11 | comment | added | CrunchySia24 | Ok thank you so much | |
| Feb 10, 2022 at 15:09 | comment | added | Christoph Hanck | Yes, for $r\geq a$ actually, as $n^0=1\neq0$. | |
| Feb 10, 2022 at 15:00 | comment | added | CrunchySia24 | Ok so basically, $E|X_n-X|^r=n^{(r-a)}$ and for $r>a$, it doesn't converge to 0. Is that how it works @Cristoph ? | |
| Feb 10, 2022 at 14:48 | history | answered | Christoph Hanck | CC BY-SA 4.0 |