Timeline for Expected length of longest stick in a stick snapping process
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 20, 2022 at 17:01 | comment | added | Max Alekseyev | @JacobManaker: Wouldn't $F_n(t)=\sqrt t\,\mathbb{P}\left[{X_n\leq\frac{1}{\sqrt t}}\right]$ do the job by any chance? | |
| Sep 19, 2022 at 4:42 | history | edited | Jacob Manaker | CC BY-SA 4.0 |
Removed useless computations; highlighted flaw
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| Sep 19, 2022 at 4:19 | comment | added | Jacob Manaker | @MaxAlekseyev: Ah, crap. That is the error, and I don't see a way to fix it. Thanks for wading through my equations. | |
| Sep 15, 2022 at 3:05 | comment | added | Jacob Manaker | @DavidESpeyer: Yes; I should think the error is in my calculation, not yours. (I've screwed up problems like this before.) But I know of analogous problems for which the structure of the argument is sound; there should just be some fortuitous cancellations that end up with an integral that is $\approx n^{-1/4}$. | |
| Sep 15, 2022 at 3:02 | comment | added | David E Speyer | Even if the $n^{-1/4}$ bound is wrong, we certainly have a lower bound of $n^{-1}$, since that is the average length of a piece! | |
| Sep 15, 2022 at 0:57 | history | answered | Jacob Manaker | CC BY-SA 4.0 |