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I was recently investigating the following question: Given $n$ independent $\text{Unif(0, 1)}$ variables $U_1,\ldots,U_n$, let $m$ be the number of elements in $[U_1,\ldots,U_n]$ that are greater than the mean $\frac{U_1+\ldots+U_n}{n}$. What is the distribution of $m$ given $n$? For example, if $U_1,U_2,U_3,U_4=0.2,0.7,0.8,0.9$, then the mean is $0.65$ and $U_2,U_3,U_4$ are greater than the mean, so $m=3$. Running lots of simulations in Python, it appears that the probability of having $m$ elements greater than the mean in $n$ independent $\text{Unif(0, 1)}$ variables is

$$P(m|n)=\frac{A(n-1,m-1)}{(n-1)!}$$

Where $A(n-1,m-1)$ is the Eulerian number equal to the number of permutations of $(1,\ldots,n-1)$ with $m-1$ ascents. More information about the Eulerian numbers can be found at https://en.wikipedia.org/wiki/Eulerian_number and https://oeis.org/A008292

Does anyone know of an existing proof relating the Eulerian numbers to number of uniform variables above the mean? The closest thing I could find is a proof that $P(k-1\leq\sum\limits_{i=1}^n U_i \leq k)=\frac{A(n,k)}{n!}$, in the following article: https://www.sciencedirect.com/science/article/pii/S0097316597928326

Emily Boyajian
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    I found a proof and posted it here: https://math.stackexchange.com/questions/4990849/p-x-k-overlinex-x-k1-is-given-by-the-eulerian-numbers/ – Bobjane Oct 28 '24 at 17:47

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Your question is fully answered by Theorem 1.1 in this pre-print on arXiv. Thanks to Bobjane for finding this paper.

S. Janson and W. Smith, Better-than-average uniform random variables and Eulerian numbers, or: How many candidates should a voter approve?, https://arxiv.org/abs/2403.02670, 2024.

Mike Earnest
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