Timeline for inequality with exponents
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2016 at 23:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 20, 2016 at 22:22 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 21, 2016 at 21:23 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 25, 2015 at 23:50 | comment | added | Marek Adamczyk | Ok, that will not work. When we take a lot lot of isolated vertices with $a_v=1$ with small $\gamma_v$, then the LHS essentially becomes $e^\lambda$ | |
| Aug 25, 2015 at 23:27 | comment | added | Marek Adamczyk | Now, if $a_v = 0$, then the exponent with decrease in the system plus the gain is clearly less than 1. Therefore, one can maybe argue that the expected value of the $E \exp( -(e^\lambda-1)decrease + \lambda \cdot gain)$ will be bigger if we would condition on choosing only $v$-s for which $a_v=1$. This expectation with conditioning may be easier to handle. | |
| Aug 25, 2015 at 23:27 | comment | added | Marek Adamczyk | Thank you Fedor for your help. From the case of $k=1$ it seems like we can drop the assumption of $\sum \gamma_v = 1$, and we can substitute it with $\sum \gamma_v \leq 1$. Also in place of integer $k$ we can just take the sum $\sum \gamma_v$. With this maybe the following idea from probabilistic interpretation will be helpful plus your remark that only 0-1 $a$-s are important: if we pick $\gamma_v$ with probability $\gamma_v/\sum \gamma_v$, then $a_v$ is the gain we collect, and $\sum_{u\in \delta(f)} a_u\gamma_u$ is the decrease in the system. | |
| Aug 25, 2015 at 17:59 | comment | added | Fedor Petrov | Alas, it becomes false. But maybe this could be modified. | |
| Aug 25, 2015 at 17:24 | comment | added | Fedor Petrov | Maybe, it makes sense to estimate $\gamma e^{\lambda a}\leq \gamma-1+e^{\gamma a (e^{\lambda}-1)}$ in each summand. At least it becomes more pretty, though maybe false. | |
| Aug 25, 2015 at 15:11 | comment | added | Marek Adamczyk | Good point, can it be combined with the argument below? | |
| Aug 25, 2015 at 15:06 | answer | added | Marek Adamczyk | timeline score: 1 | |
| Aug 25, 2015 at 15:06 | comment | added | Fedor Petrov | LHS is convex in any of variables $a_v$, hence it suffices to take $a_v\in \{0,1\}$ | |
| S Aug 25, 2015 at 14:50 | history | suggested | Tadashi |
Added relevant tag
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| Aug 25, 2015 at 14:41 | review | Suggested edits | |||
| S Aug 25, 2015 at 14:50 | |||||
| Aug 25, 2015 at 14:22 | comment | added | Marek Adamczyk | Yes, the neighbors of $v$. | |
| Aug 25, 2015 at 14:17 | comment | added | Steve Huntsman | What is $\delta(v)$? Is it the set of vertices adjacent to $v$? | |
| Aug 25, 2015 at 13:47 | history | edited | Marek Adamczyk | CC BY-SA 3.0 |
added 40 characters in body
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| Aug 25, 2015 at 13:32 | history | asked | Marek Adamczyk | CC BY-SA 3.0 |