# Empirical distribution function

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!

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

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: $$F^{-1}_n\left(\frac{i}{n}\right)=\xi_{(i)}$$

2)

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 $$(0,1)$$:$$P(\Phi(\eta)\leq x)=P(\eta\leq F(x))=F(x)$$