Timeline for A lower bound for the expectation of $\min\{X,n-X\}$ when $X$ follows a $\mathrm{Binomial}(n,p)$ distribution
Current License: CC BY-SA 4.0
15 events
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| Nov 30, 2021 at 1:16 | vote | accept | Xueyi Huang | ||
| Nov 29, 2021 at 2:09 | history | edited | Xueyi Huang | CC BY-SA 4.0 |
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| Nov 29, 2021 at 2:03 | comment | added | Xueyi Huang | @ClementC. I agree. I have revised the problem. | |
| Nov 29, 2021 at 2:02 | history | edited | Xueyi Huang | CC BY-SA 4.0 |
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| Nov 27, 2021 at 14:30 | comment | added | Brendan McKay | @ClementC. I agree. The best uniform estimate for $n\ge 2$ is $c=1/2$ however, and it occurs for $n=2,p=1/2$ and $n=3,p=1/2$. Experimentally, $1-1/\sqrt{2n}$ works and is sharp for $n=2$. | |
| Nov 27, 2021 at 9:00 | comment | added | Clement C. | @BrendanMcKay based on Aryeh's answer (see my comment below), one can get $c_n := 1−1/\sqrt{n}$ (and one must have $c\leq 1$). | |
| Nov 27, 2021 at 5:02 | comment | added | Brendan McKay | Indeed, my comment was about $p\le 1/2$. For this reason it would be more sensible to ask for $c$ such that $E(Y)\ge c\min\{p,1-p\}n$. | |
| Nov 27, 2021 at 2:06 | comment | added | Clement C. | Given that the distribution of $Y$ is invariant by replacing $p$ by $1-p$, shouldn't you aim for a lower bound symmetric that way as well? (In my linked question, I was only focusing on $p\leq 1/2$). | |
| Nov 26, 2021 at 13:45 | answer | added | Dmitry Krachun | timeline score: 3 | |
| Nov 26, 2021 at 11:33 | comment | added | Brendan McKay | I believe that for fixed $n$ the worst case is $p=1/2$ but I didn't prove it. That would mean you can take $c=1/2$ except if $n=1$. | |
| Nov 26, 2021 at 10:56 | history | edited | Nick Gill | CC BY-SA 4.0 |
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| Nov 26, 2021 at 10:00 | history | edited | Xueyi Huang | CC BY-SA 4.0 |
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| Nov 26, 2021 at 9:47 | answer | added | Aryeh Kontorovich | timeline score: 3 | |
| Nov 26, 2021 at 9:44 | history | edited | Xueyi Huang | CC BY-SA 4.0 |
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| Nov 26, 2021 at 9:03 | history | asked | Xueyi Huang | CC BY-SA 4.0 |