Confidence Intervals for ECDF

Question

The Dvoretzky–Kiefer–Wolfowitz inequality is the following:

$Pr(\text{sup}|\hat{F}_n(x)-F(x)|>\epsilon)\leq 2\exp(-2n\epsilon^2)$,

and it predicts how close an empirically determined distribution function will be to the distribution function from which the empirical samples are drawn. Using this inequality we are able to draw confidence intervals (CI's) around $\hat{F}_n(x)$ (ECDF). But these CI's will be equal in distance around every point of the ECDF .

What I wonder, is there another way to construct a CI around the ECDF?

Reading about ordered statistics we find that the asymptotic distribution of the ordered statistic is the following:

Now, first off, what does the $np$-index with those symbols mean?

Main question: are we able to use this result, together with the delta method (see below), to provide CI's for the ECDF. I mean, the ECDF is a function of the ordered statistic, right? But at the same time the ECDF is a non-parametric function, so is this a dead end?

We know that $E(\hat{F}_n(x))=F(x)$ and $\text{Var}(\hat{F}_n(x))=\frac{F(x)(1-F(x))}{n}$

I hope I'm clear as to what I'm getting at here, and appreciate any help.

EDIT:

Delta method: If you have a sequence of random variables $X_n$ satisfying

,

and $\theta$ and $\sigma^2$ are finite, then the following is satisfied:

,

for any function g satisfying the property that $g′(\theta)$ exists, is non-zero valued, and is polynomially bounded with the random variable (quote wikipedia)

Answer 1

I see no way of using the delta method, but...

Reading about the convergence of the empirical distribution function we read that the central limit theorem gives us:

$\sqrt{n}(\hat{F}_n(x)-F(x)) \rightarrow \mathcal{N}(0,F(x)(1-F(x)))$

We can use this to create varying CI's around each $\hat{F}_n(x)$:

$\hat{F}_n(x) \pm 1.96\frac{\hat{F}_n(x)(1-\hat{F}_n(x))}{n}$,

since $E(\hat{F}_n(x))=F(x)$, $\hat{F}_n(x)$ is our best estimate of $F(x)$.

Using the following R-code:

#confidenc ebands calculation:
sim_norm<-rnorm(100)
plot(sim_norm)
hist(sim_norm)
sim_norm_sort<-sort(sim_norm)
n = sum(!is.na(sim_norm_sort))
plot(sim_norm_sort, (1:n)/n, type = 's', ylim = c(0, 1), 
     xlab = 'sample', ylab = '', main = 'Empirical Cumluative Distribution')

# Dvoretzky–Kiefer–Wolfowitz inequality:
# P ( sup|F_n - F| > epsilon  ) leq 2*exp(-2n*epsilon^2)
# set alpha to 0.05 and alpha=2*exp(-2n*epsilon^2):
# --> epsilon_n = sqrt(-log(0.5*0.05)/(2*n))
#
#lower and upper bands:
L<-1:n
U<-1:n


  epsilon_i = sqrt(log(2/0.05)/(2*n))

  L=pmax(1:n/n-epsilon_i, 0)
  U=pmin(1:n/n+epsilon_i, 1)
  lines(sim_norm_sort, U, col="blue")
  lines(sim_norm_sort, L, col="blue")

#using clt:
U2=(1:n/n)+1.96*sqrt( (1:n/n)*(1-1:n/n)/n )
L2=(1:n/n)-1.96*sqrt( (1:n/n)*(1-1:n/n)/n )
lines(sim_norm_sort, L2, col="red")
lines(sim_norm_sort, U2, col="red")

We get:

We see that the red bands (from the CLT method) gives us more narrow confidence bands.

EDIT: As @Kjetil B Halvorsen pointed out - these two types of bands are different types. I had @Glen_b explain exactly what he meant:

Very different kinds of confidence bands. With a pointwise confidence band you'd expect a number of points outside the band even if it was the distribution from which the data were drawn. With simultaneous bands you wouldn't. If you have a 95% pointwise band, on average 5% of the points for the correct distribution would be outside the bands. With simultaneous bands, there's a 5% chance that the point with the biggest deviation is outside.

Many thanks to both!