# What should the integral of a CDF be called?

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.

• Comments are not for extended discussion; this conversation has been moved to chat. – whuber Dec 16 '20 at 14:07
• This question seems not to be about integrating CDFs, but rather about integrating over the indexing space ("time") of a stochastic process. This makes it rather vague and confusing and also suggests there should not be any general term for such a broad, ill-defined procedure. – whuber Dec 16 '20 at 15:02
• @whuber The question is about integrating CDF's from the lower limit of the CDF support to some independent variable value. Whether this is a stochastic process or not is not relevant. Whether the independent variable is time or not is irrelevant. Due to irrelevant concerns and a few relevant ones, this question has gained the dubious title of 'most commented.' Perhaps in the future, I will not state that I have a use for something as that invites the reader to think that they can judiciously second guess my intentions without having a full deck of cards. – Carl Dec 20 '20 at 16:36
• Your comments apply to the completely revised version of the question. There was no second guessing when I wrote my previous comments: they were based on evidence of confusion between the indefinite integral of the CDF and time-integrals of time-varying CDFs. – whuber Dec 20 '20 at 18:34
• @whuber I accept blame for confusing the reader. However, it is also important that you now state that the confusion no longer pertains. I leave you with the following, I may have been confusing, but I was not confused. – Carl Dec 21 '20 at 14:31

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

• The lecture note link is not available to me when I follow the link. It may be that you have privileges for viewing it that are not shared. Perhaps you could obtain permission to share these notes. – Carl Dec 19 '20 at 2:55
• @Carl: i think it should be accessible. You can try right clicking and click on 'save link as' as this link is directly to pdf file. Alternatively try this link and go to lecture note 4. – Dayne Dec 19 '20 at 3:03
• Indeed, these notes call the integral of the CDF to be a "super-cumulative" distribution function on page 3, and propose using it to compare Second Order Stochastic Dominance. – Carl Dec 19 '20 at 15:15
• @Carl: this second link says "sometimes called ...". So it appears that more than one writer has used this terminology. – Dayne Dec 19 '20 at 15:45
• @Carl: Sure if it becomes more commonly used then SCD will surely come in fashion (maybe not as much as cdf - as that will remain more widely used). – Dayne Dec 19 '20 at 18:26

### 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:

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.

### 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.

### 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

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

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

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 $$\Delta S(t)/\tau$$?

• Re "problematic:" integrals of the CDF directly give what one might call "partial expectations." They can be used (with proper normalization) to give succinct formulas for expectations of truncated versions of the variable. – whuber Dec 16 '20 at 14:23
• 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. – whuber Dec 16 '20 at 14:55
• My comments are intended only to point out what appears to be a fundamental ambiguity in this question. IMHO, it won't have a good answer and likely can't be answered adequately unless the respondent makes the effort to refine and narrow the question. – whuber Dec 17 '20 at 15:09
• @Carl, what you do not know is how others do not know (or interpret) your question. If my answer shows a lack of understanding, and as whuber says there is a lot of ambiguity, then you could try to improve the question-answer on your side as well (even if the question is not an answer it is a feed-back about your concept). If it is so different to introduce a new concept that desires a new name does it deserve a new name? First the concept needs to be explained clearly such that other's can understand it without too much trouble. – Sextus Empiricus Dec 17 '20 at 15:12
• @Carl the matter is not so difficult to understand (it's just double integrals). That is not the problem. The problem is that it is being sort of decorated with confusing terms, and there is no double integral of the CDF. Not in the examples, not in the linked article, neither in Whuber's links. – Sextus Empiricus Dec 17 '20 at 18:50