Timeline for Uniform Convergence for Vectors
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 28, 2017 at 13:21 | comment | added | Aryeh Kontorovich | @omerbp I think I have a fix. | |
| Nov 28, 2017 at 13:20 | history | edited | Aryeh Kontorovich | CC BY-SA 3.0 |
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| Nov 28, 2017 at 10:07 | comment | added | AvidLearner | Not sure it helps, but $g$ is defined over the set of binary vectors only. | |
| Nov 28, 2017 at 9:57 | comment | added | Aryeh Kontorovich | You're right -- see my edit. | |
| Nov 28, 2017 at 9:56 | history | edited | Aryeh Kontorovich | CC BY-SA 3.0 |
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| Nov 27, 2017 at 23:44 | comment | added | AvidLearner | I probably miss something. How can one, using the first display and the empirical distribution analog, bound $||L_D(f)-\hat L_S(f)||$ using the $\tilde f$ equivalents ? I would appreciate if you could provide the claims for bounding $||L_D(f)-\hat L_S(f)||$ using $L_D(\tilde f),\hat L_S(\tilde f)$. Sorry for being blind here.... | |
| Nov 27, 2017 at 11:50 | history | edited | Aryeh Kontorovich | CC BY-SA 3.0 |
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| Nov 27, 2017 at 10:41 | comment | added | Aryeh Kontorovich | ... as in that first display. | |
| Nov 27, 2017 at 10:40 | comment | added | Aryeh Kontorovich | Because f is sandwiched between 1/2 tilde f and tilde f. | |
| Nov 27, 2017 at 10:37 | comment | added | AvidLearner | Could you please elaborate on why the last inequality holds for $f$? I can only see why it holds for $\tilde f$... | |
| Nov 27, 2017 at 10:35 | comment | added | Aryeh Kontorovich | It was used to establish the first large-display inequality. | |
| Nov 27, 2017 at 9:08 | comment | added | AvidLearner | Thanks for the response. I can see why the last inequality holds if we replace $f$ with $\tilde f$ (as the text just above the inequality explains), but I fail to see why the inequality holds as you wrote it . Where did you use $\frac12\tilde g\le g\le\tilde g$? | |
| Nov 22, 2017 at 19:44 | history | answered | Aryeh Kontorovich | CC BY-SA 3.0 |