The following question, which arose during my research, seems deceivingly simple to me, but I could not find any elegant and formal argument.


For a binomially distributed variable $X \sim \text{Bin} \left( n, \frac{1}{\sqrt[4]{n}} \right)$, I am looking for a preferably slick and short formal argument that: $$\text{Pr}[ X \geq \mathbb{E}[X]] = \Omega(1)$$ asymptotically, i.e. that the probability of $X$ lying above its expectation is lower bounded by a constant as $n$ grows large.


Any help is well appreciated. Thank you!



Simple anticoncentration bound for binomially distributed variable
Let $Z\sim N(0,1)$, $p\_n:=n^{-1/4}$, $q\_n:=1-p\_n$. By the [Berry--Esseen inequality](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Identically_distributed_summands),
$$P(X\ge EX)\ge P(Z\ge0)-0.5\frac{n(p\_nq\_n^3+q\_np\_n^3)}{(np\_nq\_n)^{3/2}}=\frac12-o(1)$$
as $n\to\infty$. $\quad\Box$


More explicitly, (i) for $p\in(0,0.68]$ and $q=1-p$ we have $\dfrac{pq^3+qp^3}{pq^{3/2}}\le1$ and (ii) $p\_n<0.68$ for
$n\ge5$, so that
$$P(X\ge EX)\ge \frac12-0.4748\, n^{-1/8}$$
for $n\ge5$.


