# Example of singularly continuous random variable

I am looking for a useful example of a continuous but not absolutely continuous random variable, that is for which the cumulative distribution function is not differentiable.

By useful example, I mean that it has a practical application.

• @Glen_b Yes, I meant continuous but not absolutely continuous random variable. I have edited the question. Jun 21, 2015 at 20:17
• All funtions $F$ that are singular continuos distributions are differentiable with $F'=0$ almost everywhere with respect to Lebesgue measure . Jul 2, 2022 at 2:10

I am not aware of any "useful" examples of a singularly continuous random variable which must be defined as such (but see last 3 paragraphs below). As far as I know, the only practical application of one dimensional singularly continuous distributions is to getting a Ph.D. or habilitation (European system), getting papers published, getting promoted, getting tenure, etc. For many people, these are the most important applications there are. But other than that (for one dimensional distributions), no.

Do see however, http://www.math.northwestern.edu/~ehsu/Class%20of%20Singular%20Contnuous%20Functions.pdf , which gets a little closer to "practical application" of a one dimensional singularly continuous distribution than the prototypical Cantor distribution https://en.wikipedia.org/wiki/Cantor_distribution , which is a special case of the class considered in the paper.

Actually though, there are many singularly continuous multivariate distributions, mainly arising when "support" for the random variable is concentrated in a lower dimensional space. For example, consider a Bivariate Normal distribution with correlation = 1 between the components, which threfore has a singular covariance matrix. That is singularly continuous when viewed as a 2 dimensional random variable, but would be absolutely continuous if viewed as a 1 dimensional random variable. There are numerous similar examples, for instance a 2 dimensional random variable whose support is entirely on a line or circle, or in general concentrated in a lower dimensional space than that in which the random variable is defined. You can find numerous examples in Feller, 'An Introduction to Probability Theory and Its Applications, Vol. II'. Aside from scattered mentions throughout the book, the 2 page long Section V.3a is devoted entirely to singularly continuous distributions.

As to whether any of these are practical applications beyond the type of applications I listed in the first paragraph? You'll have to judge for yourself. But it can be convenient to define a multivariate random variable in its "original" space in which it is singularly continuous, rather than in a lower dimensional space. Is that absolutely necessary to do? Perhaps not, but it can be convenient. A Multivariate Normal having a singular covariance matrix is perhaps the most common "real life" example of a singularly continuous distribution.

Singularly continuous distributions frequently manifest themselves in probability and statistics "under the hood", but not necessarily "above the hood", mainly on account of people going out of their way to avoid singularity. Who wants to deal with a singularly continuous Multivariate Normal for which you can not even write out a density? Therefore, people go out of their way to avoid them, by doing such things as finding a lower dimensional space in which they are not singularly continuous. What happens when you estimate a covariance matrix as the sample covariance from a number of vector data points which is less than the dimension of the vector? You get a singular estimated covariance matrix, which if taken as the covariance matrix of a Multivariate Normal, corresponds to a singularly continuous random variable - this is an extremely common situation as several recent threads in this forum, such as such as When the sample covariance matrix becomes singular demonstrate. Now what happens when you try to compute the Cholesky factorization of this singular covariance so that you can generate random variables from the distribution? Well, not too good, is it?

Edit: Adding definition of singular distribution from Feller Vol. II p. 141 in response to comment by whuber.

*3a. Singular Distributions The condition (3.6) of the Radon-Nikodym theorem leads one to the study of the extreme counterpart of absolutely continuous distributions.

Definition. The probability distribution F is singular with respect to U if it is concentrated on a set N such that U{N} = O.

The Lebesgue measure U{dx} = dx plays a special role and the word "singular" without further qualification refers to it. Every atomic distribution is singular with respect to dx, but the Cantor distribution of example I, II (d) shows that there exist continuous distributions in R^1 that are singular with respect to dx. Such distributions are not tractable by the methods of calculus and explicit representations are in practice impossible. For analytic purposes one is therefore forced to choose a framework which leads to absolutely continuous or atomic distributions. Conceptually, however, singular distributions play an important role and many statistical tests depend on their existence. This situation is obscured by the cliche that "in practice" singular distributions do not occur.

Edit 2: Per my comment below, I am defining continuous distribution in r dimensions per Feller Vol. II. In agreement with 2nd comment from whuber, per Billingsley "Probability and Measure" 3rd edition, my examples of multivariate singularly continuous distributions, which are all from Feller Vol. II, would not be considered to be continuous in the r dimensional space in which they are defined.

• Your uses of "singularly continuous" seem very much at odds with the sense intended in the question. Could you provide us a definition or description of what you mean by this phrase? The question concerns distributions that are continuous but not absolutely continuous in the sense of measures.
– whuber
Jun 20, 2015 at 14:20
• See my edit. If you don't like the definition, then you can take it up with Feller, but he died 45 1/2 years ago. In my opinion, Feller Vol. II still sets the standard for matters of taste in probability. Let me ask you, is a Bivariate Normal which has.correlation coefficient between the components equal to 1 continuous (answer: yes), and does it have a probability density in the 2D space in which the (cumulative) distribution function is defined (answer: no)? Therefore, it is singularly continuous. QED. Everything you need to know about probability is in Feller Vol. II. Jun 20, 2015 at 14:42
• I think our understandings of "continuous" might differ: a bivariate normal with correlation $1$ is singular but I believe it is not continuous. Regardless, being wrong or right is not the principal issue, because first we must try to communicate with people. This post remains ambiguous. Your inclusion of Feller's definition helps, but it still does not seem entirely germane to the question. The question is phrased in terms of a CDF, suggesting a concern with univariate distributions. Are there any practical applications of continuous but not absolutely continuous univariate distributions?
– whuber
Jun 20, 2015 at 14:44
• whuber, you are right per p. 260 Billingsley amazon.com/Probability-Measure-Patrick-Billingsley/dp/… "If k > 1 , F can have discontinuity points even if J.L has no point masses: mu corresponds to a uniform distribution of mass over the segment B = [(x, 0): 0 < x < 1] in the plane (.u {A) = lambda A[x: 0 < x < 1 , (x,0) € A]), then F is discontinuous at each point of B. This also shows that F can be discontinuous at uncountably many points." I am right per Feller Vol. II p. 138 in chapter on r dimensional distributions, "A distribution without atoms is called continuous". Jun 20, 2015 at 19:40
• Thank you for the research, Mark. I apologize for the earlier version of my comment, which had not been well considered. I modified that comment when I independently realized there might be differences in the literature concerning the meaning of "continuous" in more than one dimension. That is one reason I am suggesting focusing on one dimension, where there should be less room for conflicting definitions. Once more I would like to emphasize that I am not trying to challenge the correctness of your post but am only concerned about whether it will be interpreted as you intended.
– whuber
Jun 20, 2015 at 19:51