Timeline for Existence of a Lyapunov function for a log-concave measure
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 24, 2022 at 17:57 | comment | added | 0xbadf00d | @NawafBou-Rabee Thank you. | |
| Jul 24, 2022 at 14:20 | comment | added | Nawaf Bou-Rabee | Sorry to hear about your bereavement. | |
| Jul 24, 2022 at 8:19 | comment | added | 0xbadf00d | @NawafBou-Rabee I'm sorry. I've immediately started a boundty (worth 100 rep) after I wrote to you. But I couldn't award the bounty before a period of 24 hours. Then I had a bereavement in my family and I forgot the bounty. Now I've started another bounty (worth 200 rep), but I need to wait 24 hours again. | |
| Jul 23, 2022 at 13:25 | comment | added | Nawaf Bou-Rabee | I thought you said "Could you undelete your answer? I will accept it and reward you the bounty". The bounty has now expired without a reward made. | |
| Jul 14, 2022 at 18:39 | comment | added | 0xbadf00d | @NawafBou-Rabee Thank you very much in advance. | |
| Jul 14, 2022 at 18:18 | comment | added | Nawaf Bou-Rabee | Done. I will provide a short explanation for why E[III] = O(t^2) separately and within the next two weeks (really busy right now). | |
| Jul 14, 2022 at 17:12 | comment | added | 0xbadf00d | @NawafBou-Rabee Could you undelete your answer? I will accept it and reward you the bounty. It would be very kind of you if you could do that and additionally explain how you intend to show that $\operatorname E[\text{III}]$ is really $\mathcal O(t^2)$ (you can wait for the bounty, if you like). I'm failing to prove that. | |
| Jun 29, 2022 at 15:33 | comment | added | 0xbadf00d | @NawafBou-Rabee I don't like the fact that the time I can give you the bounty is limited. However, once the issues have been solved, it would be no problem to give you a new bounty- | |
| Jun 29, 2022 at 14:56 | comment | added | Nawaf Bou-Rabee | as you know, the bounty on that question was allowed to expire even though I went out of my way to answer the (many) auxiliary questions that came up days before the expiration. meanwhile you seemed to have vanished. I'm sorry things couldn't work out, next time I'll be more selective. | |
| Jun 29, 2022 at 14:50 | comment | added | 0xbadf00d | @NawafBou-Rabee Unfortunately, you've deleted your answer to this question before I was able to accept it. There were still some issues which prevented me to do so. The essential issue was that I wasn't sure how you justified $\operatorname E[\text{II}\mid\tau_1=s]=\operatorname E[f(Y^1_0)-f(Y^0_s)]1_{\{\:t\:\ge\:s\:\}}$. I've asked for that separateley here: math.stackexchange.com/q/4482416/47771. And the isssue itself is described here: math.stackexchange.com/q/4482321/47771. | |
| S Feb 9, 2019 at 18:02 | history | bounty ended | CommunityBot | ||
| S Feb 9, 2019 at 18:02 | history | notice removed | CommunityBot | ||
| Feb 5, 2019 at 10:12 | vote | accept | 0xbadf00d | ||
| Feb 1, 2019 at 18:48 | answer | added | 0xbadf00d | timeline score: 2 | |
| S Feb 1, 2019 at 16:48 | history | bounty started | 0xbadf00d | ||
| S Feb 1, 2019 at 16:48 | history | notice added | 0xbadf00d | Canonical answer required | |
| Feb 1, 2019 at 14:03 | comment | added | 0xbadf00d | @NawafBou-Rabee You're right. Don't know what I thought. Please take note of my edit. | |
| Feb 1, 2019 at 14:03 | history | edited | 0xbadf00d | CC BY-SA 4.0 |
added 294 characters in body
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| Jan 30, 2019 at 17:27 | comment | added | Nawaf Bou-Rabee | How does (8) follow from (7)? The sign of the RHS of (8) looks off. Seems like (7) only implies that $A J / (\lambda J) \le 1$. | |
| Jan 27, 2019 at 19:34 | comment | added | 0xbadf00d | @NawafBou-Rabee By definition of $\lambda$ (assuming $\lambda<\infty$), $$\forall\varepsilon>0:\exists R>0:\forall r\ge R:\left|\lambda-\inf_{|x|\ge r}\frac{\langle\nabla f(x),x\rangle}{|x|}\right|<\varepsilon.$$ And $\lambda\ge\inf_{|x|\ge r}\frac{\langle\nabla f(x),x\rangle}{|x|}$ is trivial for all $r>0$, since $\lambda$ is the supremum of the values on the right-hand side over all $r>0$. As you noted, the right-hand side is increasing in $r$. | |
| Jan 27, 2019 at 19:03 | comment | added | Nawaf Bou-Rabee | Still don't see how (5) follows from the definition of $\lambda$. Suppose $R \ge r$. Then $\inf_{|x| \ge R} \nabla f(x) \cdot x / |x| \ge \inf_{|x| \ge r} \nabla f(x) \cdot x / |x|$, since the infimum in the RHS of the inequality is taken over a larger set because $\{ |x| \ge R \} \subseteq \{ |x| \ge r \}$, and so $\lambda \ge \inf_{|x| \ge r} \nabla f(x) \cdot x / |x|$. This conclusion seems different from (5). | |
| Jan 27, 2019 at 18:23 | comment | added | 0xbadf00d | @NawafBou-Rabee By definition, $\lambda=\lim_{r\to\infty}\inf_{|x|\ge r}\frac{\langle\nabla f(x),x\rangle}{|x|}$. The only problematic thing could be if $\lambda=\infty$ (for example, when $f(x)=\frac12 x^2+\text{constant}$). Am I missing something? | |
| Jan 27, 2019 at 16:24 | comment | added | Nawaf Bou-Rabee | Why does (5) follow from (4)? | |
| Jan 27, 2019 at 13:49 | history | asked | 0xbadf00d | CC BY-SA 4.0 |