QUESTION:
Let $F$ be an absolutely continuous distribution function with density $f$, and $F_{n}$ be its nth empirical distribution. Suppose that $t\in (0,1)$ is constant. Is true the convergence $$nE\{F_{n}^{-}(t) - F^{-}(t)\}^{2}\stackrel{n}{\longrightarrow} \frac{t(1-t)}{f^{2}(F^{-}(t))}?$$ Here, $F_{n}^{-}(t)$ and $F^{-}(t)$ represents random variables, because $t$ is a constant.
CLARIFICATIONS:
The generalized inverse of distribution function $F$, or quantil function, is defined by $$F^{-}(t) = \inf\{x\in \mathbb{R}; F(x)\geq t\}.$$ Let $X_{1},\ldots, X_{n}$ be independent identically distributed random variables, defined in some $(\Omega, \mathcal{F},P)$, distributed with $F$. This collection can be treated like a sample random of $F$. The empirical distribution function of the random sample $X_{1},\ldots, X_{n}$ is defined by $$F_{n}(t,\omega) = \frac{\displaystyle{\sum_{i=1}^{n}I_{(X_{i}(\omega)\leq t)}}}{n}.$$ When the parameter $t$ is constant, the above expression represents a random variable. Indeed, we can use the symbol $F_{n}(t)$ to denote such random variable, and rewrite our definition this way: $$[F_{n}(t)](\omega) = \frac{\displaystyle{\sum_{i=1}^{n}I_{(X_{i}(\omega)\leq t)}}}{n}.$$ The generalized empirical inverse of the random variable $F_{n}(t)$, could be defined in a similar way to generalized inverse of $F$. The absolute continuity of $F$ lets correct the equality $$F_{n}^{-}(t) = F^{-}(U_{n}^{-}(t)),$$ where $U$ is the uniform distribution.
