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Aug 29, 2021 at 20:56 comment added Iosif Pinelis @MattF. : To have a closure here, are you satisfied with this answer?
Jun 14, 2021 at 13:38 comment added Iosif Pinelis A typo in Lemma 1 is fixed: $g''<0$.
Jun 14, 2021 at 13:29 comment added Iosif Pinelis @AnthonyQuas : Thank you for your comment.
Jun 14, 2021 at 13:29 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 14, 2021 at 5:06 comment added Anthony Quas Nice! To make this self-contained, Lemma 1 can be proved by noting that if $g$ is convex and $g(u)=0$, then $k(x)=\frac 12g(x)/(x-u)$ is increasing. The function $h$ is $\int_u^x k(x)(x-u)\big / \int_u^x (x-u)$, which is an average of an increasing function, and hence increasing.
Jun 13, 2021 at 21:17 comment added Iosif Pinelis Now Mathematica does not have to be used at all.
Jun 13, 2021 at 21:17 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 17:07 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 16:51 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 16:15 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 16:09 comment added Iosif Pinelis @MattF. : Such an explicit example is now given.
Jun 13, 2021 at 16:09 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 15:20 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 15:19 comment added user44143 Thanks, Iosif. I deleted my previous comments. Can you give an explicit example of a distribution which attains this factor, or some factor better than $1/4$?
Jun 13, 2021 at 15:14 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 15:04 history edited user44143 CC BY-SA 4.0
fixed typos, changed $l_1\to r$, $l_2\to s$.
Jun 13, 2021 at 14:27 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 14:22 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 13:07 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 7:05 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 6:58 history edited Iosif Pinelis CC BY-SA 4.0
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Jun 13, 2021 at 6:53 history answered Iosif Pinelis CC BY-SA 4.0