The behavior of a uniform order statistic near zero

Question

Let $X_{(k)}$ be the $k$th order statistic out of $n$ uniform $[0,1]$ random variables. Let $q$ be the location of the $p$ quantile of $X_{(k)}$, i.e. $\Pr[X_{(k)}\leq q] = p$. For small $p$, Is it true that $q = O(p^{1/k} \frac{k}{n})$?

Answer 1

Your conjecture is true. More specifically, \begin{equation*} q \lesssim 2C p^{1/k} \frac kn \tag{1} \end{equation*} uniformly as \begin{equation*} p\to0,\quad C_1k\ge\ln n, \quad n-k\to\infty, \tag{1a} \end{equation*} where $C_1$ is any positive real constant and $C$ is any positive real constant such that \begin{equation} C>C_2:=e^{C_1/2}. \tag{2} \end{equation}

Indeed, $X_k:=X_{(k)}$ has the beta distribution with parameters $k,n-k+1$. So, letting
\begin{equation*} c:=k/n, \end{equation*} for the mean and the variance of $X_k$ we have \begin{equation*} EX_k=\frac k{n+1}\le c,\quad Var\,X_k\le\frac{c(1-c)}n\le\frac cn. \end{equation*} So, letting \begin{equation*} F(q):=P(X_k\le q) \end{equation*} and using Chebyshev's inequality, we have \begin{equation*} F(2c)\ge1-\frac{c/n}{c^2}=1-\frac1k\to1. \end{equation*} So, eventually $q\le2c$, which implies (1) unless \begin{equation*} Cp^{1/k}\le1, \tag{3} \end{equation*} which may and will be henceforth assumed.

Let \begin{equation*} q_*:=Ccp^{1/k}, \end{equation*} where $C$ is as in (2), so that $q_*\le c$. So, in view of Stirling's formula, (1a), and (2), \begin{align*} F(q_*)&=k\binom nk\int_0^{q_*} x^{k-1}(1-x)^{n-k}\,dx \\ &\ge k\binom nk\frac{q_*^k}k(1-q_*)^{n-k} \\ &\gtrsim\frac1{\sqrt n}\, \frac{n^n}{k^k(n-k)^{n-k}}\,q_*^k(1-q_*)^{n-k} \\ &=\frac1{\sqrt n}\,\Big(\frac{q_*}c\Big)^k\Big(\frac{1-q_*}{1-c}\Big)^{n-k} \\ &\ge\frac1{\sqrt n}\, \Big(\frac{q_*}c\Big)^k \ge \Big(\frac{q_*}{C_2c}\Big)^k=\Big(\frac C{C_2}\Big)^k p, \end{align*} so that $F(q_*)>p$ eventually, whence (1) follows as well in the case when (3) holds.