What should the integral of a CDF be called?

Question

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.

Answer 1

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.

Answer 2

Disclaimer

What should the integral of a CDF be called

I suggest the following name "integral of a CDF". Unless there is something intuitive about this integral I do not see why we should aim for a different name. The following answer will only show that the current status is that there is no intuitive idea behind the double integral of a PDF or integral of a CDF (and that the examples are not examples of integrals of a CDF). It is not a direct answer to the question (instead it is an answer to why we can not answer the question).

This is not an answer suggesting a name. It is a summary of several comments that may be helpful to achieve an answer.

At the moment it is, to me, not very clear what the double integral of the probability density function is supposed to mean. The two examples have some problems: 1 Your examples are physics and not probability. Is there use for the double integral of a probability density? 2 In addition, the examples are not examples of a double integration.

In this answer I will argue why the double integral of a pdf is problematic* **, and possibly this may lead to clarifications of the examples, and eventually inspiration for a name for this integral.

* There are several notions of the integral of $1-CDF$ like in the questions:

• Expected value of a random variable by integrating $1-CDF$ when lower limit $a\neq 0$? where the integral is $$\int_a^\infty 1-CDF(x) dx$$

• What is the expected partial value function really called? where the integral is $$\int_{-\infty}^a 1-CDF(x) dx$$

but I do not know of anything that integrates the $CDF$

** By problematic I mean that it is an integral of an extensive property but not in an additive way with disjoint sets. Or, the integrand $dx$ a measure of space is the quantity which we add up and weighed by 1-CDF(x), so we must see it intuitively as a sum over $dx$.

The integral over $1-F(x)$ can be converted into a sum over the quantile function $\int_0^b (1-F(x)) dx = \int_{F(0)}^{F(b)} Q(p) dp$ and these are related by the integral of inverse functions making the integral over $1-F(x)$ equivalent to an integral over the quantile function. For the integral over $F(x)$ you do not have the same equivalence. Without this equivalence I do not see any intuition for the use of such integrals and it becomes difficult to come up with a name.

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Densities

The meaning of density has been a subject in this question: What do we exactly mean by "density" in Probability Density function (PDF)?

In my answer to that question I relate densities to the Radon-Nikodym derivative

• Densities as the ratio of two measures on the same space. $$\rho = \frac{d \nu}{d \mu}$$
• These two quantities/measures are extensive properties. The ratio is an intensive property
• By integration of this density you get an extensive property. $$\nu(S) = \int_S \rho d \mu$$

So the integral of a probability density (or a normalized density as used in your examples) will give 'probability' as outcome. However an integral of the extensive property 'probability' gives a value with no clear use.

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Example 2

In your second example, decay of some amount of radiactive material, your double integral is not resulting from a double integral of an intensive propery.

The amount of material $M(t)$ follows a differential equation (with $\dot{}$ referring to differentiation in time):

$$\dot{M}(t)= -\frac{ln(2)}{\tau} \cdot M(t) = -\lambda \cdot M(t)$$

where $\tau$ is the half time, and $\lambda$ is the rate of decay. The solution is:

$$\begin{array}{rlcrcl} \text{amount of mass} &[mass] &:& & M(t) &=& 1-e^{-\lambda t} \\ \text{loss rate} &[mass/time]&:& & \dot{M}(t) &=& \lambda e^{-\lambda t} \\ \end{array}$$

Because of that differential equation we can write $\dot{M}(t)$ or $M(t)$ as an integral of itselve by using $M(t) - M(r) = - \int_t^r \dot{M}(s) ds$ and if $M(\infty) = 0$ then

$$M(t) = M(t) - M(\infty) = - \int_t^\infty \dot{M}(s) ds = \lambda \int_t^\infty {M}(s) ds $$

In your example you compute the total loss $Q(a,b)$ (and related the average loss is $Q(a,b)/(b-a)$) in some time period from $a$ to $b$ as a function of the mass. It is in that way that you get the double integral

$$\begin{array}{rrcl} \text{total loss between $a$ and $b$} :& Q(a,b) &=& \int_a^b \dot M(t) dt = M(b) - M(a)\\ &&=& \int_a^b -\lambda M(t) dt \\ &&=& \int_{a}^b - \lambda \left(\lambda \int_t^\infty {M}(s) ds \right) dt \\ && =& - \lambda^2 \int_{a}^b \int_t^\infty {M}(s) ds dt \end{array}$$

BTW. In this example the integral $\int_t^\infty {M}(s) ds$ does actually not relate to an integral of the CDF but instead it is an integral of the survival function.

So, in this example the double integral arrises from the relationship $\dot{M}(t) \propto M(t)$ and it is not so much a double integral of the intensive property 'density'. There is a factor $\lambda$ with units $[1/time]$ which changes the extensive property 'amount of mass' into a intensive property 'loss rate'.

Plainly integrating two times the pdf has no meaning, and it gets only a meaning through the differential equation.

This indicates that for those examples where this double integral occurs we can use the actual physical meaning of the integral to 'give a name' to the double integral.

BTW, in your example the average radiation exposure (as a fraction) is

$$\dfrac{\text{CDF}(t_2) - \text{CDF}(t_1)}{t_2-t_1} \quad \text{with units} \frac{1}{[time]}$$

instead of

$$\dfrac{\int_{0}^{t_2}\text{CDF}(t)\,d t-\int_{0}^{t_1}\text{CDF}(t)\,d t}{t_2-t_1} \quad \text{with units} \frac{[time]}{[time]}$$

You can see this based on the units. The total fraction of radiation exposure is unit less. The average fraction of radiation exposure must have units $[1/time]$. The coefficient $\lambda$ is missing to give the expression the right dimensions.

Example 1

You can shift up and down one integral because the quantity is an integral of itself. This is also clear from the article that you link from the comments 'Comparison of the gamma-Pareto convolution with conventional methods of characterising metformin pharmacokinetics in dogs' Journal of Pharmacokinetics and Pharmacodynamics volume 47, pages19–45(2020).

In that article it is written

the average mass over the dose interval, which written from the survival function equals $\Delta S(t)/\tau$, i.e., $S \tau(i) = \frac{1}{\tau} \lbrace S[\tau(i-1)] - S(\tau i) \rbrace$, for $i=1,2,3, \dots$.

In the question you write

Then to find the average drug mass during a dosing interval, we need an integral average of the summed CCDF during that interval

which relates to the integral $\dfrac{\int_{0}^{t_2}\text{CDF}(t)\,d t-\int_{0}^{t_1}\text{CDF}(t)\,d t}{t_2-t_1}$

If you are looking for a name of this integral, then why not just use the name for the equivalent $\Delta S(t)/\tau$?