Order statistics in limit

Question

Let $X_1,X_2,\cdots$ denote a sequence of $iid$ continuous random variables with some distribution $F(\cdot)$. Condition on $X_1 = x$ and fix $\rho\in (0,1)$. Define $$ A_n = \{X_1 \text{ is among the least $\rho n$ random variables in $\{X_i\}_{i=1}^n$} \}. $$ I am interested in $\lim_{n\to\infty} P(A_n)$. My guess is that it converges to one if $x \leq c$ where $F(c) = \rho$, and zero otherwise.

Answer 1

For simplicity, consider the case where $X_k$'s are i.i.d. and uniformly distributed between $[0, 1]$. For each $n \geq 1$, let $X_{n,1} \leq \cdots \leq X_{n,n}$ denote the $n$-th order statistic. Then it is not hard to show that, for each $\rho \in (0, 1)$,

$$ \sqrt{n} ( X_{n,\lfloor n\rho\rfloor} - \rho ) \xrightarrow[n\to\infty]{\text{law}} \mathcal{N}(0, \sigma_{\rho}^2) $$

for some $\sigma_{\rho} > 0$ which we can explicitly figure out. Then

$$ P(A_n) = P(X_{n,\lfloor n\rho\rfloor} > x) = P\big(\sqrt{n}(X_{n,\lfloor n\rho\rfloor} - \rho) > \sqrt{n}(x - \rho)\big), $$

and so, it follows that

$$ \lim_{n\to\infty} P(A_n) = \begin{cases} 0, & \text{if $x > \rho$}; \\ \frac{1}{2}, & \text{if $x = \rho$}; \\ 1, & \text{if $x < \rho$}; \end{cases} $$