Let $X_1,X_2,\ldots,X_n$ be i.i.d. random variables in $\mathbb{R}$ with common cumulative distribution function (CDF) $F(x)$. The empirical approximation to $F(x)$ is defined as follows:

$$\hat{F}_n(x)=\frac{1}{n}\sum_{i=1}^n\mathbf{1}_{\{X_i\leq x\}}(x)\tag{1}$$

where $\mathbf{1}_A(x)$ is an indicator function: unity when $x\in A$ and zero otherwise.

Per an answer to my previous question, by Donsker's Theorem the scaled error process $G_n(x)\equiv\sqrt{n}\left(\hat{F}_n(x)-F(x)\right)$ converges to the Brownian bridge process $B(F(x))$.

I am interested in the error of a point estimate $\hat{F}_n(x_0)$. The first sentence of the "History" section of the Wikipedia article on the Donsker's Theorem states the following:

By the classical central limit theorem, for fixed $x$, the random variable $G_n(x)$ converges in distribution to a Gaussian (normal) random variable $G(x)$ with zero mean and variance $F(x)(1 − F(x))$ as the sample size $n$ grows.

I have two questions about this statement (in order of importance):

  1. What is the standard approach when $F(x)$ is unknown? Can one use the estimate $\hat{F}_n(x)$? (similarly to how we use the sample variance estimator to approximate the confidence interval for unknown mean given unknown variance)

  2. While I am familiar with the standard proofs and the use of the classical CLT, I don't see how one proves the statement above using standard CLT. Can someone provide a hint?


1 Answer 1


If you're only interested in estimating $F$ at a single point $x_0$, then you are really just estimating the probability $p_0 = \mathbb{P}\left(X\in (-\infty,x_0]\right)$. The indicator variables $B_i = [X_i\leq x_0]$ are i.i.d. Bernoulli with (potentially unknown) probability $p_0$, so you can apply the usual CLT machinery to get the stated result and from there a confidence interval for $p_0$.

Similar methods work if you care about $F$ at finitely many points. You only really need Donsker's Theorem to understand the whole process $G$ at once.

  • $\begingroup$ Aha -- the key word is "Bernoulli"! I think I understand how this works now. All this is new to me, and I appreciate your help. $\endgroup$
    – Bullmoose
    Commented Mar 24, 2014 at 21:19

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