Timeline for Ordering preference for two zero mean Gaussian outcomes
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 19, 2021 at 5:25 | answer | added | fedja | timeline score: 3 | |
| S Apr 9, 2021 at 20:01 | history | bounty ended | CommunityBot | ||
| S Apr 9, 2021 at 20:01 | history | notice removed | CommunityBot | ||
| S Apr 1, 2021 at 18:20 | history | bounty started | Sam Zbarsky | ||
| S Apr 1, 2021 at 18:20 | history | notice added | Sam Zbarsky | Draw attention | |
| Mar 10, 2021 at 18:30 | comment | added | Sam Zbarsky | Do you have an idea for what probability distributions this seems to be true? If you care instrumentally about getting the result for the Gaussian, it seems like a computer-assisted proof could work (divide into high $a$, low $a$, intermediate $a$ regions, similarly for $b$, in divide the intermediate region into small patches, bound the derivative when $a$ is close to $b$). | |
| Mar 9, 2021 at 12:52 | comment | added | Suman Chakraborty | @Pierre I understand. Since one can write $f_a(x)$ in terms of the CDF of Gaussian explicitly (en.wikipedia.org/wiki/Folded_normal_distribution), I thought that might be a possible approach. If I can work out the details, I will post it as an answer. | |
| Mar 9, 2021 at 7:00 | comment | added | Pierre | I have tried many ways before comming to you, including explicit computation. If you replace « Gaussian » by « Exponential », then an explicit computation works, but is actually already hard to lead. For a uniform distribution on [x,y], there is a nite argument involving mixture decomposition and convexity of $f_a$. I am wondering if such a neat argument can be in a way extented to the Gaussian case. | |
| Mar 8, 2021 at 13:32 | comment | added | fedja | which you can compute explicitly Really? That depends on your definition of "explicit", of course, but even then how do you prove the resulting "explicit" inequality? | |
| Mar 7, 2021 at 22:47 | comment | added | Suman Chakraborty | Have you tried computing the expectation directly and compare? More precisely, $f_a(x) = \frac{1}{2}(x+\mathbb{E}|aX-x|)$ which you can compute explicitly. | |
| Mar 7, 2021 at 17:40 | review | First posts | |||
| Mar 7, 2021 at 19:58 | |||||
| Mar 7, 2021 at 17:34 | history | asked | Pierre | CC BY-SA 4.0 |