3
$\begingroup$

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!

Improve this question
$\endgroup$
3
  • $\begingroup$ Did you try the central limit theorem (Lindelöf) to show that $\lim_{n \to \infty} \mathbb{P}(X_n \geq \mathbb{E}X_n) = 1/2$? Where is the problem? $\endgroup$
    – Dieter Kadelka
    Commented May 19, 2023 at 12:32
  • 1
    $\begingroup$ First, I could not find on the web the Lindelöf variant. I guess you mean the Lindeberg (i.e. classical) variant as cited here: wikipedia? If so, I am not sure I can apply it, as my mean $\mu$ is a function of the number of samples $n$. $\endgroup$
    – reservoir
    Commented May 19, 2023 at 12:45
  • $\begingroup$ Yes, thanks a lot! And sorry, I was afk during the weekend. $\endgroup$
    – reservoir
    Commented May 22, 2023 at 8:55

1 Answer 1

Reset to default
10
$\begingroup$

Let $Z\sim N(0,1)$, $p_n:=n^{-1/4}$, $q_n:=1-p_n$. By the Berry--Esseen inequality, $$P(X\ge EX)\ge P(Z\ge0)-0.5\frac{n(p_nq_n^3+q_np_n^3)}{(np_nq_n)^{3/2}}=\frac12-o(1)$$ 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 $n\ge5$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .