Timeline for How large can $\mathbf{P}[X_1 + X_2 + X_3 < 2 X_4]$ get?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 21 at 5:28 | comment | added | Tobias Fritz | I've updated my answer with an improved upper bound. | |
| Jul 14 at 10:55 | answer | added | Tobias Fritz | timeline score: 4 | |
| Jul 13 at 7:14 | comment | added | Tobias Fritz | @DrewBrady: thanks for asking, as this made me realize that what I wrote wasn't quite right; I believe that $\sup_\mu \mathbf{P}[X_1 + X_2 + X_3 < 3 X_4]$ is actually $3/4$, but can't quite prove it yet. I've corrected the argument now. It may be a bit terse, so let me know in case that you'd like me to expand. | |
| Jul 13 at 7:12 | history | edited | Tobias Fritz | CC BY-SA 4.0 |
correction
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| Jul 12 at 17:30 | comment | added | Drew Brady | What do you mean by "a symmetrization argument shows that..."? | |
| Jul 12 at 15:44 | answer | added | jlewk | timeline score: 10 | |
| Jul 12 at 15:23 | history | edited | Tobias Fritz | CC BY-SA 4.0 |
typo
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| Jul 12 at 14:58 | history | edited | Tobias Fritz | CC BY-SA 4.0 |
clearer language concerning sup
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| Jul 12 at 13:43 | comment | added | Will Sawin | For a distribution to be globally optimal a necessary condition is that it be locally optimal, which in particular implies that the function sending $Y$ to $ 3 \mathbb P[X_1 + X_2 + Y < 3 X_4] + \mathbb P [X_1+X_2+X_3< Y]$ is maximized for $Y$ in the support of the distribution. I didn't check rigorously but I'm pretty sure this is not true for the Gamma distribution, so it can't literally be the optimum, although it could be close. | |
| Jul 12 at 11:10 | history | asked | Tobias Fritz | CC BY-SA 4.0 |