Timeline for An inequality on the simplex involving $x^x$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2016 at 20:06 | comment | added | VSJ | Another way to pose the question would be to take logarithms. The function $-\sum_{i = 1}^n x_i \log x_i$ is the popular entropy function of the distribution. The lower bound (which might still be true) then looks as follows: $H(\mathbf{x}) \geq 2\log \sum_{i=1}^n \sqrt x_i - \log 2.$ Perhaps some inequality like Pinker's might help to establish this. | |
| Apr 12, 2016 at 16:37 | vote | accept | Jennifer Gao | ||
| Apr 12, 2016 at 14:24 | answer | added | Iosif Pinelis | timeline score: 12 | |
| Apr 12, 2016 at 9:46 | comment | added | martin cripps | Sorry too many square roots in the last inequality - it should read $\sum_i\sqrt{x_i}\geq \prod_ix_i^{-2x_i}$! | |
| Apr 12, 2016 at 9:27 | comment | added | martin cripps | If you use $\log\sum_i \sqrt{x_i} = \log\sum_i x_i \frac{1}{\sqrt{x_i}}\geq \sum_i x_i \log(\frac{1}{\sqrt{x_i}})=-2\sum_ix_i\log x_i$ (where the inequality is Jensen's) you get $\sum_i \sqrt{x_i}\geq \sqrt{\prod_i x_i^{-2x_i}} $. Which is enough for the lower bound, because $\prod_i x_i^{x_i}<1$. | |
| Apr 12, 2016 at 9:22 | comment | added | joro | You assume $0^0=1$, right? | |
| Apr 12, 2016 at 8:32 | comment | added | Jennifer Gao | @AndrejBauer Definitely a real problem, but it's a long story. I am hoping it is true because it allows me to describe the probability distribution of a sum of random variables whose cdfs have a certain form. | |
| Apr 12, 2016 at 8:27 | comment | added | Andrej Bauer | Is this the sort of idel curiosity that lead to the discovery of so many things, or a question motivated by a "real problem"? (I am asking out of idle curiosity because I can't image how and why anyone would think of these inequalities.) | |
| Apr 12, 2016 at 8:23 | review | First posts | |||
| Apr 12, 2016 at 8:39 | |||||
| Apr 12, 2016 at 8:22 | history | asked | Jennifer Gao | CC BY-SA 3.0 |