Is anything known about the behavior of the function $$f(x)=\prod_{i=1}^n x_i^{x_i}$$ on the standard simplex, i.e. the set $\{x\in\mathbb{R}^n:\sum_{i=1}^n x_i=1, x_i\geq0\}$? I ask because I have done a lot of numerical experiments that suggest that $$1\leq\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\left(\sum_{i=1}^{n}x_{i}^{1/2}\right)^{2}\leq 2$$ for all $n$ and all $x$ on the standard simplex, and I have no clue how to prove this. Of course, one could also re-write this more compactly as $$1\leq\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\|x\|_{1/2}\leq 2$$but this still doesn't seem particularly helpful to me.
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1$\begingroup$ 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.) $\endgroup$– Andrej BauerCommented Apr 12, 2016 at 8:27
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$\begingroup$ @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. $\endgroup$– Jennifer GaoCommented Apr 12, 2016 at 8:32
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$\begingroup$ You assume $0^0=1$, right? $\endgroup$– joroCommented Apr 12, 2016 at 9:22
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1$\begingroup$ 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 $\prod_i x_i^{x_i}<1$. $\endgroup$– martin crippsCommented Apr 12, 2016 at 9:27
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$\begingroup$ Sorry too many square roots in the last inequality - it should read $\sum_i\sqrt{x_i}\geq \prod_ix_i^{-2x_i}$! $\endgroup$– martin crippsCommented Apr 12, 2016 at 9:46
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The upper bound is not correct. E.g., let $n=101$, $x_1=\dots=x_{100}=1/1000$, $x_{101}=9/10$. Then $$\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\left(\sum_{i=1}^{n}x_{i}^{1/2}\right)^{2}>7>2.$$
More generally, take any $a\in(0,1)$ and any natural $k$, and then let $n=k+1$, $x_1=\dots=x_k=a/k$, $x_{k+1}=1-a$. Then $$\ln\left[\left(\prod_{i=1}^{n}x_{i}^{x_{i}}\right)\left(\sum_{i=1}^{n}x_{i}^{1/2}\right)^{2}\right]=a\ln\frac ak+(1-a)\ln(1-a)+2\ln(\sqrt{ak}+\sqrt{1-a})\to\infty$$ as $k\to\infty$.
answered Apr 12, 2016 at 14:24
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2$\begingroup$ Thanks, you taught me a valuable lesson about sampling from the boundary of a simplex :) $\endgroup$ Commented Apr 12, 2016 at 16:37
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$\begingroup$ I am glad this was useful. $\endgroup$ Commented Apr 12, 2016 at 16:58