Timeline for Return probabilities for random walks on infinite Schreier graphs
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2012 at 7:06 | comment | added | Andreas Thom | Sure, more details would be great! | |
| Mar 29, 2012 at 0:31 | comment | added | fedja | Actually, you can get $n^{-1/2}$ with the technique I described. It requires just one more trick. If you are interested, I'll post the details :). | |
| Mar 28, 2012 at 23:01 | comment | added | YCor | sorry, for $\mathbf{Z}$ the probability of return is $n^{-1/2}$ (exercise using Stirling's formula). So the behavior of $\delta_n$ should be between $n^{-1/2}$ and $n^{-1/3}$. | |
| Mar 28, 2012 at 5:44 | comment | added | Andreas Thom | It seems to me that the random walk on $\mathbb Z$ is the one where it is most difficult to get lost; $\delta_n$ should be maximal. However, I do not know if this intuition is misleading. | |
| Mar 28, 2012 at 5:42 | history | edited | Andreas Thom | CC BY-SA 3.0 |
added 3 characters in body; edited body
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| Mar 27, 2012 at 20:48 | comment | added | YCor | Nice! I tried this Chabauty argument but I missed the point that $\delta_{2n}$ is decreasing and couldn't conclude. Anyway, this does not give any explicit bound. Do you expect $\delta_n$ to be in $1/n$? Fedja's argument (that I didn't check) shows it's at most $1/n^{1/3}$ and the case of $\mathbf{Z}$ shows it's at least $1/n$ (asymptotically). | |
| Mar 27, 2012 at 15:19 | history | answered | Andreas Thom | CC BY-SA 3.0 |