# What is $B=\prod_{i=1}^{n}s_{i}^{-s_{i}}$ and its maximum?

I'm having trouble with this one. Doing some research I find that some variable of interest depends on $$B=\left[\prod_{i=1}^{n}s_{i}^{s_{i}}\right]^{-1}$$ where $$0 for each $$i$$, $$i=1\ldots n$$ and $$\sum_{i=1}^{n}s_{i}=1$$.

To situate: imagine a stick of length 1 which is divided into $$n$$ shares denoted $$s_i$$. There is no relationship among the $$s_i$$ except $$\sum_{i=1}^{n}s_{i}=1$$

I want to study and say everything there is to say about $$B$$. Namely I have two questions at the moment:

1. is $$B$$ of any recognizable form? It looks like (the inverse of) a weighted geometric average of the shares $$s_i$$, but with weights equal to the shares themselves, which makes it strange to me.
2. Find the maximum of $$B$$. By numerical simulation it seems that $$B$$ reaches a maximum for “equitable shares” as in $$s_i=(\frac{1}{2},\frac{1}{2})$$ for $$n=2$$ shares ... $$s_i=(\frac{1}{n},\dots\frac{1}{n})$$ for $$n$$ shares. The maximum is then reached at $$B=n$$. (equivalently: it seems that the maximum of $$B$$ is reached for the arithmetic mean shares).

Any insights greatly appreciated. I'm more than rusty.

• Notice that $\log B$ is the entropy of the probability distribution $s = (s_1, \cdots, s_n)$. It is well-known that it achieves the maximum at the uniform distribution $s=(\frac{1}{n},\cdots,\frac{1}{n})$, and this can be easily shown from the method of Lagrange multiplier together with strict concavity of the map $s\mapsto\log B$. Jun 28 '19 at 23:34
• I suggest you look at the min of $B^{-1}$ which is achieved at the same point as the min of $\sum s_i\log s_i$ since $\log$ is increasing. Observe that $r\to r\log r$ is convex on $[0,1]$ (taking it equal zero at zero for continuity) and thus the sum $\sum s_i\log s_i$ is convex as well. Jun 28 '19 at 23:41