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Empirical CDF functions are usually estimated by a step function. Is there a reason why this is done in such a way and not by using a linear interpolation? Does the step function has any interesting theoretical properties which make us prefer it?

Here is an example of the two:

ecdf2 <- function (x) {
  x <- sort(x)
  n <- length(x)
  if (n < 1) 
    stop("'x' must have 1 or more non-missing values")
  vals <- unique(x)
  rval <- approxfun(vals, cumsum(tabulate(match(x, vals)))/n, 
                    method = "linear", yleft = 0, yright = 1, f = 0, ties = "ordered")
  class(rval) <- c("ecdf", class(rval))
  assign("nobs", n, envir = environment(rval))
  attr(rval, "call") <- sys.call()
  rval
}


set.seed(2016-08-18)
a <- rnorm(10)
a2 <- ecdf(a)
a3 <- ecdf2(a)

par(mfrow = c(1,2))
curve(a2, -2,2, main = "step function ecdf")
curve(a3, -2,2, main = "linear interpolation function ecdf")

enter image description here

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  • $\begingroup$ Related................................... $\endgroup$
    – user128129
    Aug 18, 2016 at 12:48
  • 8
    $\begingroup$ "...estimated by a step function" belies a subtle misconception: the ECDF is not merely estimated by a step function; it is such a function by definition. It is identical to the CDF of a random variable. Specifically, given any finite sequence of numbers $x_1, x_2, \ldots, x_n$, define a probability space $(\Omega,\mathfrak{S},\mathbb{P})$ with $\Omega=\{1,2,\ldots, n\}$, $\mathfrak{S}$ discrete, and $\mathbb{P}$ uniform. Let $X$ be the random variable assigning $x_i$ to $i$. The ECDF is the CDF of $X$. This enormous conceptual simplification is a compelling argument for the definition. $\endgroup$
    – whuber
    Aug 18, 2016 at 13:52

1 Answer 1

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It's by definition.

The empirical distribution function of a set of observations $(X_n)$ is defined by

$$F_e(t) = \frac{\#\{X_n \mid X_n \le t\}}n$$

Where $\#$ is the set cardinality. This is, by nature, a step function. It converges to the actual CDF almost surely.

Also note that for any distribution with $P(X = x) \ne 0$ for at least two $x$ (especially nondegenerate discrete distributions), your variant of ECDF does not converge to the actual CDF. For example consider a Bernoulli distribution with CDF

$$F_X(x) = p \chi_{x \ge 0} + (1-p) \chi_{x \ge 1}$$ this is a step function whereas ecdf2 will converge to $\chi_{x\ge 0} \cdot (p + (1-p)\min(x, 1))$ (a piecewise linear function connecting $(0,p)$ and $(1,1)$.

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  • $\begingroup$ Thanks Alex. So is there another name for the function I wrote? (because I would guess it also converges to the actual CDF) $\endgroup$
    – Tal Galili
    Aug 18, 2016 at 10:19
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    $\begingroup$ @TalGalili It doesn't. Consider a Bernoulli distribution. Your ecdf2 will not converge in this case. You could call it a smoothed ecdf. I suspect it will converge to the actual CDF iff the actual CDF has no points with nonzero probability except for extreme points (where you don't smooth) $\endgroup$
    – AlexR
    Aug 18, 2016 at 11:02
  • $\begingroup$ @AlexR you could edit your answer to add this comment since discrete distributions are the reason for such definite -- so it answers the "why" question. $\endgroup$
    – Tim
    Aug 18, 2016 at 13:02
  • 1
    $\begingroup$ @Tim Done. ${}{}$ $\endgroup$
    – AlexR
    Aug 18, 2016 at 14:33
  • $\begingroup$ Thanks. Is there a way to define a continuous empirical function that would converge to the step function but would be fully monotone (i.e.: without any sharp "jumps")? $\endgroup$
    – Tal Galili
    Aug 19, 2016 at 10:02

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