Timeline for Is there a lower bound for variance in terms of curvature?
Current License: CC BY-SA 3.0
13 events
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| Apr 18, 2013 at 14:40 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
added 629 characters in body
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| Apr 15, 2013 at 16:41 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
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| Apr 15, 2013 at 16:26 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
clarified hypothesis
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| Apr 15, 2013 at 13:38 | answer | added | Benoît Kloeckner | timeline score: 2 | |
| Apr 15, 2013 at 12:31 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
clarified curvature
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| Apr 15, 2013 at 12:26 | comment | added | Mikhail Katz | The "unit area" is with respect to the standard area form $dxdy$. | |
| Apr 15, 2013 at 12:24 | comment | added | Mikhail Katz | The variance is the minimum of $\int(f-m)^2$ over constant $m$. The minimum is attained for $m=E(f)$, the expected value of $f$. Moreover, $E(f^2)-(E(f))^2=Var(f)$. From this combined with uniformisation one immediately deduces the strengthened form of Loewner's torus inequality. | |
| Apr 15, 2013 at 12:19 | comment | added | Robert Bryant | @katz: I'm not familiar with your notion of '$Var(f)$'. Could you write down an explicit definition or formula? Without it, I don't see how you expect to 'quantify' any conjectured relationship. Also, when you write 'unit area domain', do you mean with respect to the standard measure in the $xy$-plane or with respect to the $g$-measure? | |
| Apr 15, 2013 at 9:31 | comment | added | Mikhail Katz | Say, $f$ is defined in a unit-area domain in the $x,y$ plane. In Loewner's torus inequality, this is a fundamental domain for the torus. | |
| Apr 15, 2013 at 9:27 | comment | added | Liviu Nicolaescu | Variance with respect to what probability measure? | |
| Apr 15, 2013 at 8:01 | history | edited | Mikhail Katz |
tag
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| Apr 15, 2013 at 7:50 | history | edited | Mikhail Katz | CC BY-SA 3.0 |
add tag; added 2 characters in body
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| Apr 15, 2013 at 7:33 | history | asked | Mikhail Katz | CC BY-SA 3.0 |