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Jun 27, 2021 at 4:12 comment added Fei Cao Thank you very much!
Jun 27, 2021 at 2:47 comment added Iosif Pinelis @FeiCao : The assumption that the mean $\mu$ was fixed was not stated in your OP. It was only said that $\mu>0$. However, even assuming $\mu$ is fixed, Remark 1 in my answer gives essentially the same conclusion, that $f=0$. I have added details on this.
Jun 27, 2021 at 2:44 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 26, 2021 at 2:39 vote accept Fei Cao
Jun 25, 2021 at 23:24 comment added Fei Cao In fact, I thought your answer is still missing something. In my OP I indeed said that $\rho$ has some fixed mean $\mu > 0$, so your example will only work when $a = \mu$ (you are forbidden to take any real $a>0$). With your argument, we can only say that $f(\mu) = 0$, am I right?
Jun 25, 2021 at 20:06 comment added Fei Cao I agree with you. Thank you!
Jun 25, 2021 at 20:06 vote accept Fei Cao
Jun 25, 2021 at 23:25
Jun 25, 2021 at 19:59 comment added Iosif Pinelis @FeiCao : As briefly noted in my answer, no degree of smoothness can possibly help here, because any distribution can be appropriately approximated by an arbitrarily smooth one. If you want to impose additional, "structural" conditions on the distribution (such as, say, the log-concavity of the pdf), that would be quite a different question, to be posted separately. For now, your question, as posted, has been fully answered.
Jun 25, 2021 at 19:48 comment added Fei Cao Thanks for your answer! But I don't expect that the relation $H\geq f(G)$ to hold for all p.d.f. $\rho$ whose support is contained in $[0,\infty)$. For instance, is it possible to show something like this for "smooth" $\rho$ (instead of being a combination of Dirac masses)?
Jun 25, 2021 at 19:34 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 25, 2021 at 19:27 history answered Iosif Pinelis CC BY-SA 4.0