Timeline for Dynamics of a random stretch map
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Jul 26 at 11:39 | vote | accept | Nate River | ||
| Jul 25 at 5:33 | answer | added | Anthony Quas | timeline score: 2 | |
| Jul 24 at 13:44 | comment | added | Nate River | @MartinHairer You're right, its trivially true if the stretch factor is uniform between $0$ and $2$. I have changed it to be uniform between $0$ and $e$ instead, I think this is the nontrivial case. | |
| Jul 24 at 13:43 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 24 at 13:33 | comment | added | Nate River | @MartinHairer The notation has been corrected, though I think what you said might still work... | |
| Jul 24 at 13:30 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 24 at 13:27 | comment | added | Nate River | @MartinHairer My apologies, I think the additive notation might be confusing people a bit. I will switch to using exponentials instead, one second. | |
| Jul 24 at 13:26 | comment | added | Martin Hairer | @NateRiver By Jensen, $E \log(2\epsilon_i) < 0$, so $D_n(x) \to 0$ by the LLN and the statement is trivially true, no? | |
| Jul 24 at 13:22 | comment | added | Karl Fabian | But then better identify $S^1$ with $e^{i\,\phi}$ and $2\,\epsilon_k$ with $e^{i\,\alpha_k}$ with $\alpha_k \in [0, 4\,\pi]$. It then looks like there is no reason to believe that $D_n(x)$ and $D_n(y)$ should converge a.e. because $\sum \alpha_k$ should still be equally distributed. | |
| Jul 24 at 13:13 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 24 at 13:00 | comment | added | Nate River | @Karl Fabian Sorry, that distance was a typo. It is a little bit ugly to write out explicitly anyhow, so I will just say it is the length metric. | |
| Jul 24 at 12:58 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 24 at 12:58 | comment | added | Karl Fabian | That also related to your original $\min(|x-y|,|y-x|)$ metric which lives on$[0,1]$. | |
| Jul 24 at 12:55 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 24 at 12:54 | comment | added | Nate River | @KarlFabian I don't quite follow your sketch. $D_n (x)$ is an element of $S^1$. | |
| Jul 24 at 12:53 | comment | added | Karl Fabian | Doesn't follow this directly from $D_n(x)\leq\,2^n\,\prod \epsilon_i \to 0$ almost surely? (One has $a^n\,\prod \epsilon_i \to 0$ for $a<e$.) | |
| Jul 24 at 12:52 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 24 at 12:46 | comment | added | user479223 | @mathworker21 I assume the probability space the $\epsilon_i$ are defined on. | |
| Jul 24 at 12:44 | comment | added | mathworker21 | what's $\Omega$? | |
| Jul 24 at 12:20 | history | edited | Nate River | CC BY-SA 4.0 |
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| Jul 24 at 12:05 | history | asked | Nate River | CC BY-SA 4.0 |