Timeline for Bounding integral arising from expectation of a random variable satisfying Bernstein's inequality
Current License: CC BY-SA 4.0
6 events
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| Mar 17, 2019 at 14:02 | comment | added | Iosif Pinelis | Since any probability is $\le1$, the tail bound actually available to us is of the form $1\wedge(Cq(t))$, which is nonlinear in $C$, and, in contrast with $Cq(t)$, does not grow with $C$ at all as soon as $C$ becomes large enough. After integration, the resulting bound does grow with $C$, but only logarithmically in $C$, much slower than $C$ itself. | |
| Mar 17, 2019 at 13:46 | comment | added | B Merlot | Thank you! On a more conceptual note, how is it that there is no linear dependence on $C$? I almost convinced myself this is impossible, since the LHS can be written as C*[some constant integral], whereas the right hand side only grows as $log(C)$. | |
| Mar 17, 2019 at 13:43 | vote | accept | B Merlot | ||
| Mar 17, 2019 at 3:58 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 17, 2019 at 2:55 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Mar 17, 2019 at 2:49 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |