I am stuck on a problem which seems to simple, meaning there is probably more to it than I am thinking.

Let ${\xi_1,...,\xi_n}$ be independent and identically distributed over ${(\Omega, \mathcal{A}, \mathbb{P})}$, with the continuous distribution function ${F}$. Then $${F_n:= [0,1] \ni t \rightarrow F_n(t) := \frac{1}{n} \sum_{i=1}^n \chi(\xi_i \leq t)}$$ is the empirical distribution function, ${\chi}$ being the indicator function.

1) Find ${nF_n(\xi_i)}$ and ${F_{n}^{-1}(\frac{i}{n})}$ for ${1 \leq i \leq n}$

I doubt that I am right about this, but i simply wrote $${n*F_n(\xi_i) = n*\frac{1}{n} * \sum_{j=1}^n \chi(\xi_j \leq \xi_i) = \sum_{j=1}^n \chi(\xi_j \leq \xi_i)}$$

And my approach for ${F_{n}^{-1}(\frac{i}{n})}$ would be the same, just using $${F_n^{-1}(u):= inf\{ x \in X: F_n(x) \geq u\}}$$

Is my approach correct? Because I did not use the continuous function ${F}$ at all and does not really seem like a solution.

2) Determine the distribution of ${F_n^{-1}(\eta)}$, with ${\eta}$ being equally distributed over ${[0,1]}$

I think here it is a direct conclusion of Skorochod's theorem that ${F_n^{-1}}$ in this case must have the distribution ${F_n}$, although I do not know how to show this other than just to state "implied by Skorochod".

Any help will be appreciated!

  • $\begingroup$ what does $n*$ mean? $\endgroup$
    – user587377
    Jan 30, 2019 at 1:47

1 Answer 1



For every $i\in\{1,\dots,n\}$ there is a $k\in\{1,\dots,n\}$ such that $\xi_i=\xi_{(k)}$ where $\xi_{(k)}$ denotes the $k$-th order statistic.

Observe that - if $i\neq j$ - we have $\chi(\xi_i\leq\xi_j)\stackrel{a.s.}{=}\chi(\xi_i<\xi_j)$ because $F$ is a continuous distribution.

Actually the continuity of $F$ allows you to assume that: $$\xi_{(1)}<\xi_{(2)}<\cdots<\xi_{(n)}\tag1$$where $<$ replaces $\leq$.

Based on this it is not difficult to find that $nF_n(\xi_i)=k$ where $k$ is the integer that satisfies $\xi_i=\xi_{(k)}$.

Your expression for $F_n^{-1}(u)$ is okay and again applying $(1)$ we find: $$F^{-1}_n\left(\frac{i}{n}\right)=\xi_{(i)}$$


You are correct.

In general if $F$ is a CDF and $\Phi:(0,1)\to\mathbb R$ is defined by: $$\Phi(u)=\inf(\{x\in\mathbb R\mid F(x)\geq u\}$$then it can be deduced that: $$u\leq F(x)\iff \Phi(u)\leq x$$

Based on that we find for $\eta$ uniformly distributed on $(0,1)$:$$P(\Phi(\eta)\leq x)=P(\eta\leq F(x))=F(x)$$


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