The empirical cdf is an estimate of the cdf. What kind of estimation method (such as method of moments, MLE, ...) constructs the empirical cdf?

Is the empirical cdf a nonparametric estimate? Do nonparametric estimates have construction methods different from MOM, MLE, and others?

For example, the empirical pmf of a finite-valued discrete distribution is constructed by MLE. But that is a special case, and not the whole story. Moreover, it is not an empirical cdf but an empirical pmf.

  • $\begingroup$ Not asking the definition of empirical cdf, but by what kind of estimation method is it constructed. $\endgroup$
    – Tim
    Mar 31, 2014 at 2:34
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    $\begingroup$ You don't seem to realise that estimates can be produced by any method you like. I could have an estimator that said always guess 42; it would be a lousy estimator in all but very special circumstances, but it would be an estimator. In other words, the class of estimators is not restricted to good or sometimes good named methods such as moments or MLE. The method used for calculating CDF is often just empirical fractions; it doesn't have to be thought of as formal estimation. $\endgroup$
    – Nick Cox
    Mar 31, 2014 at 7:46
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    $\begingroup$ @NickCox But it would answer the ultimate question about life, the universe, and everything quite nicely. ;) $\endgroup$
    – Alexis
    Sep 8, 2014 at 19:00
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    $\begingroup$ @Alexis Glad someone caught the allusion. $\endgroup$
    – Nick Cox
    Sep 8, 2014 at 19:05
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    $\begingroup$ @NickCox I remember the precise moment I first heard about those books in a radio interview with Douglas Adams I was listening to as an 11-year old in the car on 580 in the Bay Area. Worth catching, and they stand rereading well. :) $\endgroup$
    – Alexis
    Sep 8, 2014 at 19:12

2 Answers 2


In An Introduction to the Bootstrap, Efron and Tibshirani find it useful to characterize the empirical cumulative distribution function (ecdf) as the nonparametric maximum likelihood estimate of the "underlying population" $F$.

Given data $x_1, x_2, \ldots, x_n$, the likelihood function (by definition) is the product of the probabilities

$$L(F) = \prod_{i=1}^n {\Pr}_F(x_i).$$

E&T claim this is maximized by the ecdf. Since they leave it as an exercise, let's work out the solution here. It's not completely trivial, because we have to account for the possibility of duplicates among the data. Let's take care with the notation, then. Let $x_1, \ldots, x_m$ be the distinct data values, with $x_i$ appearing $k_i \ge 1$ times in the dataset. (Thus, $x_{m+1}, \ldots, x_n$ are all duplicates of the first $m$ values.) The ecdf is the discrete distribution that assigns probability $k_i/n$ to $x_i$ for $1 \le i \le m$.

For any distribution $F$, the likelihood $L(F)$ has $k_i$ terms equal to $p_i = {\Pr}_F(x_i)$ for each $i$. It therefore is completely determined by the vector $p=(p_1, p_2, \ldots, p_m)$ and can be computed as

$$L(F) = L(p) = \prod_{i=1}^m p_i^{k_i}.$$

Since the likelihood for the ecdf is nonzero, the maximum likelihood will be nonzero. Therefore, for any distribution $\hat F$ that maximizes the likelihood, $p_i = {\Pr}_{\hat F}(x_i)$ must be nonzero for all the data. The Axiom of Total Probability asserts the sum of the $p_i$ is at most $1$. This reduces the problem to a constrained optimization:

$$\text{Maximize } L(p) = \prod_{i=1}^m p_i^{k_i}$$

subject to

$$p_i \gt 0, i=1, 2, \ldots m;\quad \sum_{i=1}^m p_i \le 1.$$

This can be solved in many ways. Perhaps the most direct is to use a Lagrange multiplier $\lambda$ to optimize $\log L$, which produces the critical equations

$$\left(\frac{p_1}{k_1}, \frac{p_2}{k_2}, \ldots, \frac{p_m}{k_m}\right) = \lambda\left(1, 1, \ldots, 1\right)$$

with unique solution $$\hat p_i = \frac{k_i}{k_1+\cdots+k_m} = \frac{k_i}{n},$$

precisely the ecdf, QED.

Why is this point of view important? Here are E&T:

As a result, [any] functional statistic $t(\hat F)$ is the nonparametric maximum likelihood estimate of the parameter $t(F)$. In this sense, the nonparametric bootstrap carries out nonparametric maximum likelihood inference.

[Section 21.7, p. 310]

Some words of explanation: "as a result" follows from the (easily proven) fact that the MLE (maximum likelihood estimate) of any function of a parameter is that function of the MLE of the parameter. A "functional statistic" (or "plug-in" statistic) is one that depends only on the distribution function. As an example of this distinction, E&T point out that the usual unbiased variance estimator $s^2 = \sum (x_i-\bar x)^2/(n-1) $ is not a functional statistic because if you were to double all the data, the ecdf would not change, but the $s^2$ would be multiplied by $2(n-1)/(2n-1)$, which does change (albeit only slightly). Functional statistics are crucial to understanding and analyzing the Bootstrap.


Bradley Efron and Robert J. Tibshirani, An Introduction to the Bootstrap. Chapman & Hall, 1993.

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    $\begingroup$ And it turns out this is not just a fun exercise: with right censored data, the Kaplan Myer curves can be shown to be an MLE in a similar fashion, although they also have a MOM justification. But for interval censored data, maximizing the non-parametric likelihood is, as far as I know, the only justification for the basic CDF estimator. $\endgroup$
    – Cliff AB
    Jan 14, 2020 at 18:53
  • $\begingroup$ It seems this only makes sense for discrete problems. For continuous distributions with a density, this makes no sense. $\endgroup$
    – gondolier
    Jul 16, 2020 at 5:31
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    $\begingroup$ @gondolier In what way does it make no sense? A great many (very successful) applications of bootstrapping use models with continuous distributions. $\endgroup$
    – whuber
    Jul 16, 2020 at 13:26
  • $\begingroup$ @whuber I'm a bit confused by this definition of the likelihood function. For any continuous distribution $F$, the likelihood would be $L(F) = 0$, right? (Since point mass is $0$ for any given point.) $\endgroup$
    – zxmkn
    Jan 29 at 7:41
  • $\begingroup$ @zxmkn That's correct. It follows that the (usual) bootstrap always estimates a discrete $\hat F.$ In retrospect this is kind of obvious because no matter how large the dataset is, there are only finitely many possible bootstrap samples. $\endgroup$
    – whuber
    Jan 29 at 14:53

For a discrete random variable, the standard definition of the empirical cumulative distribution function (cdf) can be seen as a Method-of-Moments estimator. Consider the discrete random variable $X$ taking values $\{k_1 <k_2 <...\}$. Then its cdf is defined as

$$F_X(k_m) =\Pr(X\le k_m)= \sum_{i=1}^m\Pr(X=k_i)$$

We have that $\Pr(X=k_i) = E[I_{\{X=k_i\}}]$, where $I_{\{X=k_i\}}$ is the indicator function taking values $1$ if $X=k_i$, $0$ otherwise. Substituting we have

$$F_X(k_m) = \sum_{i=1}^mE[I_{\{X=k_i\}}]$$

If we have available a sample of size $n$, $\{x_j,\, j=1,...,n\}$, of realizations of $X$, the sample analogue of the RHS is

$$\sum_{i=1}^m\left(\frac 1n\sum_{j=1}^nI_{\{x_j=k_i\}}\right) = \frac 1n\sum_{j=1}^nI_{\{x_j\le k_m\}} = \hat F_X(k_m)$$

i.e it is the standard expression for the empirical cumulative distribution function. So, since it uses the sample analogue of expected values (which here are moments of the indicator functions which in turn are Bernoulli r.v.'s), it can be seen as a Method-of-Moments estimator.


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