Density Function – What Should the Integral of a CDF Be Called in Statistical Terminology?

cumulative distribution functiondensity functionpartial-momentsterminology

This is strictly a nomenclature question. I have no particular problem finding double integrals of the type $\int\int\text{pdf}(y) \, d y \,d x$, and I find them quite useful. Whereas we have a good name for $\int\text{pdf}(x) \, dx=\text{CDF}(\textit{x})$, where CDF is the cumulative distribution (credit: @NickCox, A.K.A., density) function, what I do not have is a good name for the integral of the CDF.

I suppose one could call it an accumulated cumulative distribution (ACD), DID (double integral of density) or CDF2, but I have not seen anything of the sort. For example, one would hesitate to use "ccdf" or "CCDF", as that is already taken as an abbreviation for complementary cumulative distribution function, which some prefer to saying "survival function," S$(t)$, as that latter is, strictly speaking, for an RV, whereas CCDF is not from an RV; it is a function equal to 1-CDF, which maybe a relate to probability, but does not have to. For example, PDF often refers to situations in which there are no probabilities, and a more general term for PDF is "density function". However, $df$ is already taken as "degrees of freedom", so the entire literature is stuck with "PDF". So what about DIPDF, "double integral of PFD, a bit long, that is. DIDF? ICDF for integral of cumulative distribution (density) function? How about ICD, integral of cumulative distribution? I like that one, it is short and says it all.

@whuber gave some examples of how these are used in his comment below and I quote "That's right. I establish a general formula for certain definite integrals of CDFs at stats.stackexchange.com/a/446404/919. Also closely related are stats.stackexchange.com/questions/413331, stats.stackexchange.com/questions/105509, stats.stackexchange.com/questions/222478, and stats.stackexchange.com/questions/18438 — and I know there are more."

Thanks to @whuber's contributions the text of this question is now more clear than prior versions. Regrets to @SextusEmpericus, we have both spent too much time on this.

And the accepted answer is "super-cumulative" distribution, because that name is catchy and has been used before, although frankly, without being told, I would not have known that, which is why, after all, I asked. Now, for the first time, we define SCD as its acronym. I wanted an acronym because unlike elsewhere, where $S(x)$ is used for SCD$(x)$ (not mentioning names), I wanted something that was unique enough to not cause confusion. Now granted, I may be using SCD outside of a purely statistical context in my own work, but as everyone uses PDF, even when there is no p to speak of, that is at most a venial sin.

Edit: Upon further consideration, I will call pdf as $f$ of whatever, e.g., $f(x)$, CDF as $F(x)$ and double integrals as $\mathcal{F}(x)$ just to make things simpler. The problem here is that it is unfortunately common practice to refer to all density functions, where a density function is a non-negative function, which when integrated over its support has a dimensionless area-under-the-curve of 1 as a pdf. Worse, pdf's are a small subset of density functions, such that just using $f$ and explaining that that refers to a "density function" is probably better practice. There is no end of confusion caused by even slight misidentification of concepts. For example, in pharmacokinetics there is a routine for maximum likelihood of residuals of concentration models, which rather than being called "maximum likelihood of residuals," where residuals can be interpreted as being random variates, is commonly referred to as "maximum likelihood," which makes no sense as concentration-time curves are 2D curves, and not a 1D collection of random realizations.

Best Answer

I am mentioning here one term for integral of CDF used by Prof. Avinash Dixit in his lecture note on Stochastic Dominance (which I happen to have very recently stumbled upon). Obviously, this is not a very generally accepted term otherwise it would have been discussed already on this thread.

He calls it super-cumulative distribution function and is used in an equivalent definition of Second Order Stochastic Dominance. Let $X$ and $Y$ be two r.v such that $E(X) = E(Y)$ and have same bounded support. Further, let $S_x(.), S_y(.)$ be the respective super cumulative distribution functions.

We say that $X$ is second order stochastic dominant over $Y$ iff $S_x(w) < S_y(w)$ for all values of $w$ in support of $X, Y$.

It may also be interesting to note that for First Order Stochastic Dominance, the condition gets simply replaced by CDF in place of super-cdf.

Related Solutions

Density Function – How to Understand Discrete Analog of CDF: Cumulative Mass Function?

The proper terminology is Cumulative Distribution Function, (CDF). The CDF is defined as $$F_X(x) = \mathrm{P}\{X \leq x\}.$$ With this definition, the nature of the random variable $X$ is irrelevant: continuous, discrete, or hybrids all have the same definition. As you note, for a discrete random variable the CDF has a very different appearance than for a continuous random variable. In the first case, it is a step function; in the second it is a continuous function.

CDF – How to Compute the Marginal CDF of a Joint Density

More generally, consider the distribution whose density is

$$f_p(u, \theta) = \theta C_p\int_1^\infty v^{-p} e^{-\theta u v}\,\mathrm{d}v = \theta C_pE_p(\theta u)$$

for $u \gt 0$ and parameters $p\gt 0,$ $\theta \gt 0.$ $E_p$ is the Exponential Integral function and $C_p$ is a normalizing constant (which we will find at the end). Notice, for future reference, that

$$E_{p+1}(0) = \int_1^\infty v^{-(p+1)}\,\mathrm{d}v = \frac{1}{p}.$$

Moreover, since for any $z\gt 0$

$$ 0\le E_p(z) = \int_1^\infty v^{-p} e^{-zv}\,\mathrm{d}v\le \int_1^\infty e^{-zv}\,\mathrm{d}v = \frac{1}{z},$$

it follows from the Squeeze Theorem (for limits) that

$$\lim_{z\to\infty} E_p(z) = 0.$$

The distribution function (cdf) is, by definition,

$$F_p(u,\theta) = \int_0^u f_p(z)\,\mathrm{d}z = C_p \int_0^u E_p(\theta z)\,\theta\mathrm{d}z = C_p \int_0^{\theta u} E_p(z)\,\mathrm{d}z.$$

The integrand defining $E_p$ is smooth and decreasing so quickly at its upper limit that we may interchange the operations of differentiating and integrating, showing

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}z}E_{p+1}(z) &= \frac{\mathrm{d}}{\mathrm{d}z}\int_1^\infty v^{-(p+1)} e^{-zv}\,\mathrm{d}v\\&=\int_1^\infty \frac{\mathrm{d}}{\mathrm{d}z}v^{-(p+1)} e^{-zv}\,\mathrm{d}v\\&=-\int_1^\infty v^{-p} e^{-zv}\,\mathrm{d}v\\&=-E_p(z). \end{aligned}$$

It is immediate (by the Fundamental Theorem of Calculus) that

$$F_p(u,\theta) = C_p \left[E_{p+1}(0) - E_{p+1}(\theta u)\right] = C_p \left[\frac{1}{p} - E_{p+1}(\theta u)\right].$$

Finally, since $\lim_{u\to\infty}F_p(u,\theta)=1$ (by the Law of Total Probability), we find

$$1 = \lim_{u\to\infty}C_p \left[\frac{1}{p} - E_{p+1}(\theta u)\right] = \frac{C_p}{p},$$

showing $C_p = p$ and giving

$$F_p(u,\theta) = 1 - pE_{p+1}(\theta u).$$

Set $p=5$ and $\theta=4$ for the answer to the question. Here are graphs of the density and its cdf:

These were drawn in gray using the analytical solution. As a check, over them are plotted, in red, the values obtained from numerically computing the single integral for $f$ and the double integral for $F.$ The R code that generated these graphs follows.

#
# Solution.
#
library(expint)
f <- function(u, p, theta) theta * p * expint_En(theta * u, p)
F <- function(u, p, theta) 1 - p * expint_En(theta * u, p+1)
#
# Brute force verification using numerical integration.
#
f. <- Vectorize(function(u, p, theta, ...) {
  theta * p * integrate(function(v) exp(-theta * u * v) / v^p, 1, Inf, ...)$value
}, "u")
F. <- Vectorize(function(u, p, theta, ...) {
  integrate(function(z) f.(z, p, theta), 0, u, ...)$value
}, "u")
#
# Example.
#
p <- 5
theta <- 4
par(mfrow=c(1,2))
curve(f(x, p, theta), 0, 1.5, n=501, lwd=2, col=gray(0.25),
      xlab="u", ylab="Density", 
      main=bquote(f[.(p)](u, .(theta))))
curve(f.(x, p, theta), add = TRUE, col="Red", lwd=2, lty=2)

curve(F(x, p, theta), 0, 1.5, n=501, lwd=2, col=gray(0.25),
      yaxp=c(0,1,1),
      xlab="u", ylab="Probability", 
      main=bquote(F[.(p)](u, .(theta))))
curve(F.(x, p, theta), add = TRUE, col="Red", lwd=2, lty=2)
par(mfrow=c(1,1))

Best Answer

Related Solutions

Density Function – How to Understand Discrete Analog of CDF: Cumulative Mass Function?

CDF – How to Compute the Marginal CDF of a Joint Density

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