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There is a lemma in Jaeckel (19171) paper "Some flexible estimates", which basically states that under some conditions $X_(i) - F^{-1}(i^*)$ is $O(n^{\frac{1}{2}})$ or that $\left[X_{(i)} - F^{-1}(i^*)\right]$ tends to $0$ in probability uniformly, when $i^* = \frac{i}{n+1}$. I cannot understand it, because $F^{-1}(i^*)$ in my mind gives out probabilities, but $X_{(i)}$ is order statistic, that's equal to $i-th$ value of ordered sample. So, let's say that $X \sim N(0,1)$, sample size $n=30$, and $i=15$. Then $X_{15}$ could be equal to 0.2. Then $F^{-1}\left(\frac{0.2}{31}\right) = F^{-1}(0) = 0.5$. Doesn't seem that they could be equal. What am I missing?

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  • $\begingroup$ $F^{-1}(i^*)$ should take a probability and give a value in the support of $X$. For example, for a standard normal distribution $\Phi^{-1}(0.975) \approx 1.96$. $\endgroup$
    – Henry
    May 9 at 18:40
  • $\begingroup$ @Henry But in my given example then $i^* \approx 0.006 \Rightarrow i = 0.006\cdot 31 = 0.186$. Then what's the $X_{0.186}$? $\endgroup$
    – user
    May 9 at 19:07

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