Timeline for Maximum of two normal random variables
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 24, 2019 at 9:20 | comment | added | Alexey Ustinov | Poincare's Bread: ceadserv1.nku.edu/longa//confs/biomath2008/Poincare.html | |
| S Sep 24, 2019 at 1:54 | history | suggested | Thomas Dybdahl Ahle | CC BY-SA 4.0 |
Added precise expression
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| Sep 23, 2019 at 19:35 | review | Suggested edits | |||
| S Sep 24, 2019 at 1:54 | |||||
| Jun 21, 2014 at 15:51 | answer | added | wolfies | timeline score: 4 | |
| Jun 20, 2014 at 21:13 | comment | added | fedja | Unfortunately, the whole point I'm trying to make (perhaps too aggressively, I apologize again for that) is that the second method is pretty much as illogical as the first and it works better just because to get a bound worse than that of the first method without immediately seeing a way to improve it is next to impossible. So, any comparison of these two will illuminate next to nothing. What should be done instead is to think for a few minutes of what the worst possible case is (I posted it in my answer) and then you'll see everything there is to see here yourself :-) | |
| Jun 20, 2014 at 20:48 | comment | added | user0820 | I didn't claim that the first method is anything close to smart, it was just something that can be compared with the second method. The whole point of asking the question was to get some intuitive understanding about what's happening in the latter one and why it is logical to come up with it, i.e. remove the word "trick" for the description. | |
| Jun 20, 2014 at 20:37 | answer | added | fedja | timeline score: 6 | |
| Jun 20, 2014 at 20:29 | answer | added | ofer zeitouni | timeline score: 17 | |
| Jun 20, 2014 at 20:03 | comment | added | fedja | The ultimate reason in the case presented is that doing something even minimally smart is always better than doing something stupid in every possible respect. It is not the Laplace trick that is better here, but the estimate $\max(X,Y)\le|X|+|Y|$ that is worse. Even if you improve it to something as obvious as $\max(X,Y)\le \max(X,0)+\max(Y,0)$, you get a two-fold gain immediately resulting in $1.59/2\approx 0.8$, which beats the Laplace estimate hands down. | |
| Jun 20, 2014 at 19:46 | comment | added | Qiaochu Yuan | The fact that you know $\mathbb{E}(e^{\lambda X})$ is a reflection of the fact that you have control over all of the moments of $X$; the first argument uses weaker assumptions, and in particular doesn't assume that even the second moments exist (but on the other hand applies in much greater generality). It shouldn't be surprising that assuming more about moments gets you a stronger bound. (I picked up this idea from Terence Tao: see, for example, terrytao.wordpress.com/2010/01/03/… ). | |
| Jun 20, 2014 at 19:32 | history | edited | user0820 | CC BY-SA 3.0 |
deleted 1 character in body
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| Jun 20, 2014 at 18:47 | history | edited | user0820 | CC BY-SA 3.0 |
edited body
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| Jun 20, 2014 at 18:33 | comment | added | user0820 | I am not sure this is satisfactory enough. It's still not clear to me why the second method should be better on a conceptual level. In other words, I would hope for something that wouldn't rely on the fact that we know the exact distribution of $\max(X,Y)$. For example, let's consider the general situation when X and Y are not necessarily independent. | |
| Jun 20, 2014 at 17:11 | history | edited | user0820 | CC BY-SA 3.0 |
deleted 54 characters in body
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| Jun 20, 2014 at 16:55 | comment | added | Steve Huntsman | en.wikipedia.org/wiki/… | |
| Jun 20, 2014 at 16:31 | review | First posts | |||
| Jun 20, 2014 at 16:53 | |||||
| Jun 20, 2014 at 16:15 | history | asked | user0820 | CC BY-SA 3.0 |