# Convergence in distribution with empirical distribution function (EDF)

I'm struggling with the following question:

Let $F_n(x)$ denote the EDF of a random sample. Show that

$\sqrt{n}(F_n(x)-F(x))\xrightarrow[]{d}N(0,F(x)(1-F(x)))$.

I think that the course of action should be:

1. prove that $F_n(x) \xrightarrow[]{p} F(x)$ using the WLLN and
2. prove the convergence in the question using the CLT.

This is my take on the first step: $F_n(x)=\frac{1}{n} \sum_{i=1}^n \mathbf{1}(x_i \leq x)$ where $\mathbf{1}(\cdot)$ is the indicator function. Then, for all $x$, $\mathbf{1}(x_i \leq x)$ is an iid random variable with expectation $F(x)$. Thus, by the WLLN, $F_n(x)$ is a consistent estimator of $F(x)$: $F_n(x) \xrightarrow[]{p} F(x)$.

So far, so good?

My real troubles begin with the second step. I have set up the expression

$\sqrt{n}(F_n(x)-F(x)) = \sqrt{n}\left(\frac{1}{n} \sum_{i=1}^n \mathbf{1}(x_i \leq x) - F(x)\right)$

but I don't really know where to go from there. Any pointers would be greatly appreciated!

• What do you know? what did you try? – kjetil b halvorsen Feb 10 '15 at 15:37
• Thus far, very little. I did not really know where to start. Apparently, F_n(x) converge in distribution to F(x) by WLLN and the expression in the question should then be derived from the CLT. – Fredrik P Feb 10 '15 at 16:49
• Fix $x$. Then think about variables $Z_i=1(x_i\le x)$. Are they independent? Are they identically distributed? What is their mean and variance? – mpiktas Feb 11 '15 at 16:54
• Note if $x$ is not fixed, then the question becomes much harder. – mpiktas Feb 11 '15 at 16:55
• Thanks, @mpiktas ! If you post those two comments as an answer, then I can mark the question as answered. And, a final question, does this mean that my first step is not really needed to answer the question? – Fredrik P Feb 12 '15 at 8:36

Fix $x$. Then think about variables $Z_i=1(x_i\le x)$. Are they independent? Are they identically distributed? What is their mean and variance?
Note that if $x$ is not fixed, then the question becomes much harder.