Timeline for Expected length of longest stick in a stick snapping process
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2022 at 0:55 | comment | added | David E Speyer | Thanks! At this point, I think I have found a way to fix the argument that I am satisfied with, so don't feel a need to rush on my account. Maybe I'll add it as a comment if I get some time. | |
| Sep 15, 2022 at 19:25 | comment | added | Timothy Budd | Yes, I have been skipping corners especially on the lower bound. When I find some time I will add some details. Essentially I am just translating the proof of Corollary 1.4 in the book to the discrete setting, so in the meantime if you wish (and can access the book) you could have a look there for the proper arguments. | |
| Sep 15, 2022 at 16:33 | comment | added | David E Speyer | Some of this is probably my discomfort with probabilistic language. When you say $\lim \text{sup} \frac{\log X_n}{\log n}$ is less than something, or $\lim \text{inf} \frac{\log X_n}{\log n}$ is more than something, you mean with probability $1$, right? As compared to $X_{n,0}^{\epsilon} \geq \frac{\sum_k X_{n,k}^p}{\sum_k X_{n,k}^{p-\epsilon}}$, which is literally true everywhere in the probability space? | |
| Sep 15, 2022 at 16:31 | comment | added | David E Speyer | I am confused about the lower bound. You have $X_{n,0}^{\epsilon} \geq \frac{\sum_k X_{n,k}^p}{\sum_k X_{n,k}^{p-\epsilon}}$. How do you go from there to the lower bound on $\frac{\log X_{n,0}}{\log n}$. At first I thought you were saying that $\mathbb{B}(X_{n,0}^{\epsilon}) \geq \frac{\mathbb{E}(\sum_k X_{n,k}^p)}{\mathbb{E}(\sum_k X_{n,k}^{p-\epsilon})}$, but I don't see how to take the division outside the expectation. Also, you say the next line is only good for $0 < p < \overline{p}$, which suggests you are doing something else; could you please spell out what it is? | |
| Sep 15, 2022 at 15:48 | vote | accept | Nate River | ||
| Sep 15, 2022 at 15:48 | comment | added | Nate River | I think this is as close to a complete answer as one can hope for - asymptotics in $n$. So I have accepted the answer, though others are still very welcome to contribute. Thank you for the solution! | |
| Sep 15, 2022 at 11:56 | history | edited | Timothy Budd | CC BY-SA 4.0 |
minus sign fix
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| Sep 15, 2022 at 11:29 | history | edited | Timothy Budd | CC BY-SA 4.0 |
Added a detail to the lower bound
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| Sep 15, 2022 at 10:11 | history | answered | Timothy Budd | CC BY-SA 4.0 |