Timeline for Maximizing the expectation of a polynomial function of iid random variables
Current License: CC BY-SA 4.0
9 events
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Feb 25, 2019 at 8:08 | comment | added | user44143 | @JairoBochi, I tried that, but the integrals are long enough that Mathematica stops without an answer, so I stopped too. | |
Feb 25, 2019 at 7:51 | comment | added | Jairo Bochi | So far I haven't been able to prove that the optimizer must contain a discrete (ie atomic) component. Maybe this is false, and what is going on is that the density tends to infinity at 0 and 1... I ask you the following: What if we combine two Beta distributions, one with parameters $>1$, another with parameters $<1$? Does this lead to an improvement? | |
Feb 23, 2019 at 14:57 | history | edited | user44143 | CC BY-SA 4.0 |
added 3 characters in body
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Feb 23, 2019 at 7:21 | comment | added | user44143 | The symmetric Beta distribution is one of the simplest distributions on the unit interval. Perhaps it maximizes a function of the order statistics like $E[X_{(2)}-X_{(1)}]/\sigma$; I'd be surprised if it's not the solution to some optimization. | |
Feb 23, 2019 at 7:11 | comment | added | Jairo Bochi | Since it seems out of reach to find an exact solution for the baby problem (unless a miracle happens), I think that a more modest but still very nice problem would be to prove that any (the unique?) optimizing distribution is indeed a non-trivial mixture of discrete and continuous distributions. | |
Feb 23, 2019 at 7:09 | comment | added | Jairo Bochi | I see. But is there any reason to believe that the true optimizer will have a Beta component? | |
Feb 23, 2019 at 7:07 | comment | added | user44143 | @JairoBochi, I figure what it lacks in optimality, it makes up for with round parameters and being exactly calculable. | |
Feb 23, 2019 at 7:05 | comment | added | Jairo Bochi | Nice! This agrees much more with my experiments. As you say, your distribution seems almost optimal, but not exactly optimal. Indeed, the optimal distribution supported on a uniform mesh of 20 points yields an integral >.0615. | |
Feb 23, 2019 at 6:50 | history | answered | user44143 | CC BY-SA 4.0 |