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.


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.


1 Answer 1


Asymptotics for L2 functionals of the empirical quantile process, with applications to tests of fit based on weighted Wasserstein distances by EUSTASIO DEL BARRIO, EVARIST GINE´ and FREDERIC UTZET in Bernoulli 11(1), 2005, 131–189, Section 1.1 discusses L2 convergence and should answer your question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.