Timeline for A normal distribution inequality
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2018 at 9:17 | vote | accept | Hans | ||
| Aug 18, 2019 at 19:59 | |||||
| Jul 8, 2013 at 13:09 | history | edited | cardinal | CC BY-SA 3.0 |
Minor cleanup.
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| Jul 8, 2013 at 13:00 | comment | added | cardinal | @Hans: $n$ is obviously exponentially bounded and so I was most concerned with matching the behavior near zero. Note that $n/N$ and $a e^{-ax}$ share the same derivative at zero as do their respective logs. | |
| Jul 8, 2013 at 12:44 | vote | accept | Hans | ||
| Jan 10, 2018 at 9:16 | |||||
| Jul 8, 2013 at 12:44 | vote | accept | Hans | ||
| Jul 8, 2013 at 12:44 | |||||
| Jul 8, 2013 at 4:46 | comment | added | Hans | I see your rationale, but can you describe what makes you think of the particular form of the exponential function $e^{-ax}$? Just a first lucky choice? And thanks for pointing out my typo. | |
| Jul 8, 2013 at 3:09 | comment | added | cardinal | @Hans: (Also, just a minor typo in your first comment: $u := \int_{-\infty}^x N(u)\,\mathrm du$. Cheers.) | |
| Jul 8, 2013 at 3:08 | comment | added | cardinal | Dear @Hans: Regarding motivation: Note that $n/N$ is decreasing and so I first tried the crudest thing possible, i.e, $x + n/N \leq x + n(0)/N(0) = x + a$. However, this doesn't work since it turns out by a similar argument to Lemma 2 that $(1-N)/n \geq (x+a)^{-1}$. So, I needed a function that decreased but stayed above $n/N$, while also decreasing fast enough that I'd still get an upper bound on $(1-N)/n$. Note that, actually, the same basic analysis as Lemma 2 will yield $(1-N)/n \leq (x+a e^{-bx})^{-1}$ where $b = \sqrt{\pi/2}-\sqrt{2/\pi}$, which is a little sharper, but unneeded here. | |
| Jul 8, 2013 at 2:40 | comment | added | Hans | It is unfortunate that there is only 1 point up vote allow for each person per answer. Otherwise, I would have put in ticked more. :-) | |
| Jul 8, 2013 at 2:36 | comment | added | Hans | Beautiful proof! I think the introduction of $u:=\int_{-\infty}^x n(t)dt$ is the key. Lemma 1 and Lemma 2 are useful result in their own rights. They connect the behavior of $N$ or $1-N$ near $0$ and $\infty$ smoothly. I will wait a while for others to check the computation, before I will check it as THE accepted answer, even though it is too pretty to be wrong. Meanwhile, could you please describe your motivation in coming up with the function $a e^{ax}$? | |
| Jul 7, 2013 at 20:40 | history | undeleted | cardinal | ||
| Jul 7, 2013 at 20:39 | history | edited | cardinal | CC BY-SA 3.0 |
Fixed up previous proof.
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| Jul 5, 2013 at 19:34 | history | deleted | cardinal | ||
| Jul 5, 2013 at 19:34 | history | edited | cardinal | CC BY-SA 3.0 |
deleted 923 characters in body
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| Jul 5, 2013 at 19:06 | history | answered | cardinal | CC BY-SA 3.0 |