# How can one understand the expression $X_{(i)} - F^{-1}(i)$, where $X_{(i)}$ is order statistics and $F$ is distribution function.

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?

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