Timeline for Prove $x\mapsto\frac{E[f(X)1(f(X)\geq x X)]}{1+E[X1(f(X)\geq x X)]}$ is Lipschitz-continuous
Current License: CC BY-SA 4.0
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| Apr 29, 2020 at 21:17 | comment | added | Nik Weaver | @Iques: to echo that, you are a new contributer and wouldn't know this, but it is considered bad form on Mathoverflow to modify a question after the original version has been answered. Nothing wrong with posting the modified question separately. | |
| Apr 29, 2020 at 17:21 | comment | added | Iosif Pinelis | @Iques : As Nik Weaver noted, even after your change of the question the function $g$ can still be discontinuous, as is now detailed in my answer. More importantly, if your posted question is not what you actually meant, then you (and not the answerer) should take the responsibility for your mistake. That is, you should not change your question so as to invalidate a valid answer. Rather, you may want to post the amended question separately. | |
| Apr 29, 2020 at 17:11 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 32 characters in body
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| Apr 29, 2020 at 14:53 | comment | added | Nik Weaver | @Iques I think you can take Iosif's example with larger $A$ to make $E[f(x)]$ larger. | |
| Apr 29, 2020 at 13:32 | comment | added | Iques | Thank you for your answer, however I forgot to explicit the domain of $g$, which is $[0,E[f(X)]]$. Hence in you example $g$ is constant. | |
| Apr 29, 2020 at 12:37 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |