We follow the proof in [1]}.
By definition of the empirical quantile, we have
\(\alpha \le F_{n+1}(Q_{n+1}(\alpha )) = \frac{1}{n+1}\sum _{i=1}^{n+1} \mathbb {1}_{E_i \le Q_{n+1}(\alpha )} \hspace{5.0pt}.\) 

Мы следуем доказательству в [1].

По определению эмпирического квантиля, у нас имеем

\(\alpha \le F_{n+1}(Q_{n+1}(\alpha )) = \frac{1}{n+1}\sum _{i=1}^{n+1} \mathbb {1}_{E_i \le Q_{n+1}(\alpha )} \hspace{5.0pt}.\)