Timeline for What is the exact definition of a sharp transition?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2022 at 19:24 | vote | accept | apg | ||
| S Aug 25, 2022 at 19:01 | history | suggested | apg | CC BY-SA 4.0 |
Added clarity over what "requries" means here.
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| Aug 25, 2022 at 13:00 | review | Suggested edits | |||
| S Aug 25, 2022 at 19:01 | |||||
| Jul 23, 2022 at 1:11 | comment | added | Yuval Peres | The theorem "every monotone graph property has a sharp threshold" is describing a weaker type of sharp threshold, where one just requires $\delta_n \to 0$. This is appropriate when $p_n$ are bounded away from 0 and 1. | |
| Jul 23, 2022 at 0:30 | comment | added | apg | Also, just to clarify something, since every monotone graph property has a sharp threshold, does that mean $A_n$ is not a monotone graph property (since there is no sharp threshold)? It would seem $A_n$ was monotone, as adding edges can only preserve the non-isolated nature of a vertex. Perhaps I am missing something about the definition of monotone property. | |
| Jul 22, 2022 at 23:49 | comment | added | Yuval Peres | Exactly, this $A_n$ is an example of a local event that does not have a sharp threshold. | |
| Jul 22, 2022 at 21:48 | comment | added | apg | i.e. c=1 is the only case where $\delta_n / p_n$ would need to go to zero with $n$, so the condition is indeed satisfied. Overall, I need to evaluate the width of the critical window to prove we have a sharp transition, not just that you can go arbitrarily close to the transition point from either side, and keep the zero/one nature of the order parameter. I think that's what is meant by the condition $\delta_n / p_n \to 0$ which you clarify. | |
| Jul 22, 2022 at 21:19 | comment | added | apg | What explicitly are $p_n$ and $\delta_n$ in this case of $A_n$ (i.e. that a vertex is not isolated)? With $f_{n}(c/n)$, can I instead write the equivalent $f_{n}(1/n + (c-1)/n)$, so that $p_n = 1/n$ and $\delta_n = (c-1)/n$? But, then we have $\delta_n / p_n = c-1$, which is not going to zero with $n$. Unless c=1. | |
| Jul 22, 2022 at 17:24 | history | answered | Yuval Peres | CC BY-SA 4.0 |