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Jun 30, 2021 at 18:52 comment added Iosif Pinelis @FeiCao : I have added the description of these $a,b,c$.
Jun 30, 2021 at 18:52 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 30, 2021 at 18:31 comment added Fei Cao In you illustration, may I know what are the specific values of $a$ and $b$ and $c$?
Jun 30, 2021 at 18:25 history edited Iosif Pinelis CC BY-SA 4.0
added 180 characters in body
Jun 30, 2021 at 18:12 comment added Iosif Pinelis @FeiCao : Concerning $G$, take $U=X-Y$ and $V_n=Z_n-W_n$, where $X,Y$ are iid random variables each with density $p$, and $Z_n,W_n$ are iid centered normal random variables each with variance $1/n$, say, such that $Z_n,W_n$ are independent of $X,Y$. Similarly, for $H$.
Jun 30, 2021 at 17:48 vote accept Fei Cao
Jun 30, 2021 at 17:43 comment added Fei Cao sorry, I am a bit confused, what is your $V_n$ and $U$ in the context of this problem/answer?
Jun 30, 2021 at 17:36 comment added Iosif Pinelis @FeiCao : Yes, of course, these limits have to be taken. These limits are immediate from the triangle inequality: If $E|V_n|\to0$, then $E|U+V_n|\to E|U|$.
Jun 30, 2021 at 17:32 comment added Fei Cao Thanks! But then what you showed at the end is $H[\rho*g] \geq G[\rho*g] / 14434$ right? I believe certain limit should be taken to reach the conclusion that $H[\rho] \geq G[\rho]/14434$
Jun 30, 2021 at 17:26 comment added Iosif Pinelis @FeiCao : Oops! I guess I did not save the latest edit with the detail on the approximation. This is now done.
Jun 30, 2021 at 17:24 history edited Iosif Pinelis CC BY-SA 4.0
added 393 characters in body
Jun 30, 2021 at 17:09 comment added Fei Cao Thanks! But your sentence "By approximation, without loss of generality (wlog), $\rho(x) >0$ for all real $x$" is a bit obscure. As I mentioned earlier, I really need to impose that the support of $\rho$ is contained in $[0,\infty)$. How are you going to approximate $\rho(x) = \mathrm{e}^{-x}\mathbf{1}_{x\geq 0}$ by some density $p$ with $p > 0$ for all $x \in \mathbb R$?
Jun 30, 2021 at 13:02 comment added Iosif Pinelis @FeiCao : As was written, "By approximation, without loss of generality (wlog), $p(x)>0$ for all real $x$". I have now added a detail on that. Please let me know if anything else seems unclear here.
Jun 30, 2021 at 6:23 comment added Fei Cao Also, I have another question if you don't mind. The exponential density $\rho(x) = \mathrm{e}^{-x}$ is clearly log-concave (it is even log-linear!), so your $c = 0$ and $p_* = 1$. But there do not exist two numbers $a < 0 < b$ such that $\rho(a) = \rho(b) = 1/\mathrm{e}$...
Jun 30, 2021 at 6:17 comment added Fei Cao Thanks! Actually I checked your profile and found that you are a very good researcher in the area of probability, I might use (not very sure) this answer for a publishable research project, may I know whether you will be interested in co-author a paper? (I am merely a Ph.D student in applied math at ASU who just finished my fourth year) If you are interested, you can email me at [email protected], and I am very happy to share the potential research project behind it....Best regards.
Jun 30, 2021 at 6:16 vote accept Fei Cao
Jun 30, 2021 at 6:24
Jun 30, 2021 at 5:51 comment added Iosif Pinelis @FeiCao : Your case is just a particular case of the general one dealt with in this answer. So, I do not see any difficulty here. Also, you can do the same calculations with an arbitrary real $a$, without any shifting -- only the calculations will be a bit more complicated.
Jun 30, 2021 at 5:47 comment added Fei Cao Remarkable effort in writing this answer. Although I have one concern, in my problem, I really need to impose that the support of $\rho$ is contained in $[0,\infty)$ (i.e., no mass outside $\mathbb{R}_+$), so in that case, your sentence "By shifting, wlog, $a =0 $" causes me some trouble...
Jun 30, 2021 at 5:42 history edited Iosif Pinelis CC BY-SA 4.0
added 167 characters in body
Jun 30, 2021 at 5:31 history answered Iosif Pinelis CC BY-SA 4.0