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Let $X_1,...,X_n$ denote independent and identically distributed random variables, with $X_i \sim F$, $1 \leq i \leq n$. Assume $F$ is continuous. Then we know that its generalized inverse (quantile function) $F^{\leftharpoonup}(u):= \inf\{x: \, F(x)>u\}$ exists. If $F_n$ denotes the empirical cumulative distribution function, i.e. $F_n(x)=\frac{1}{n} \sum_{i=1}^n \textbf{1}(X_i \leq x)$, by Glivenko-Cantelli we know $\Vert F_n - F \Vert_\infty \overset{a.s.}{\to}0$. Now, what can we say about $ \Vert {F_n}^{\leftharpoonup}- F^{\leftharpoonup} \Vert_\infty $? Are there conditions under which a.s. uniform convergence can be obtained for the empirical quantile function ${F_n}^{\leftharpoonup}$?

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  • $\begingroup$ Do we need uniform continuity of $F^{\leftharpoonup}$ ? $\endgroup$ – Jack London Nov 15 '16 at 18:43
  • $\begingroup$ Look at the final comment under my answer here: math.stackexchange.com/questions/93983/… $\endgroup$ – user940 Nov 16 '16 at 16:27
  • $\begingroup$ @Byron Schmuland thank you for your comment. My only concern is: in your link the point is: do the theoretical quantile functions of the sequence of random variables $(X_i)$ converge to the theoretical quantile function of $X$, if $X_i \to_d X$? Here the point is different, I think. I am asking: do the EMPIRICAL quantile function of the sample $X_1,..., X_n$ defined via $ {F_n}^{\leftharpoonup}(u)= \inf\{x: \, \frac{1}{n} \sum_{i=1}^n \textbf{1}(X_n \leq x) > u\}$ converges a.s. and uniformly to the theoretical quantile function, identical for all the random variables $X_1,...,X_n$? $\endgroup$ – Jack London Nov 17 '16 at 7:30
  • $\begingroup$ The empirical quantile function is a special case of the general result mentioned by commenter Egor in the linked answer. Egor forgot to say that the support of $F$ must be a finite interval as well, for the result to be true. In other words, you are quite correct: it is necessary to assume that the limit quantile function is uniformly continuous on $(0,1).$ $\endgroup$ – user940 Nov 19 '16 at 15:23

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