Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is an example of the probability distribution function that does not have a density function?

share|cite|improve this question
Any discrete probability distribution, such as the one that picks an integer between 1 and 10 (inclusive) with equal probability. – Henning Makholm Jan 13 '12 at 16:32
Thank you. I should have said the probability distribution on a continuum set. – user12586 Jan 13 '12 at 16:33
Distribution of waiting time at a queue is an example: there is a non-zero probability that the queue is empty. So the $F_T(t)$ has a jump at $t=0$, i.e. it is not differentiable there, hence does not have density. – Sasha Jan 13 '12 at 16:35
Thank you. I should have said: do we have an example with atomless distributions? – user12586 Jan 13 '12 at 16:39
See the "devil's staircase" as in this answer:… – Byron Schmuland Jan 13 '12 at 16:41
up vote 6 down vote accepted

About genericity (see the comments), note that every probability distribution $\mu$ on the Borel line may be written uniquely as a sum $\mu=\mu_a+\mu_d+\mu_s$ of measures such that $\mu_a$ is absolutely continuous with respect to Lebesgue measure, $\mu_d$ is discrete and $\mu_s$ is... well, the remaining part.

Thus, for every Borel set $B$, $\mu_a(B)=\displaystyle\int_Bf(x)\mathrm dx$ for some nonnegative integrable density $f$, $\mu_d(B)=\displaystyle\sum\limits_{n}p_n\cdot[x_n\in B]$ for some finite or infinite sequence $(x_n)_n$ of points of the real line and some sequence $(p_n)_n$ of nonnegative weights. The measure $\mu_a$ is called the densitable part of $\mu$. The measure $\mu_d$ is called the discrete part of $\mu$. The third measure $\mu_s$ is called the singular part of $\mu$ and is somewhat the most mysterious part since $\mu_s$ is atomless AND has no density.

The measures $\mu_a$, $\mu_d$ and $\mu_s$ are mutually singular, in the sense that there exists some disjoint Borel sets $A$, $D$ and $S$ such that $\mu_a(\mathbb R\setminus A)=\mu_d(\mathbb R\setminus D)=\mu_s(\mathbb R\setminus S)=0$. The set $D$ is always discrete, hence at most countable. The set $S$ might be a Cantor set with Lebesgue measure zero.

One sees that, in a sense, probability distribution functions with a density are the opposite of generic, since they correspond to measures $\mu$ such that $\mu_d=\mu_s=0$. And asking that $\mu=\mu_a$ is a bit like asking that a point $(x,y,z)$ in $\mathbb R^3$ is in fact located on the first coordinate axis $y=z=0$...

share|cite|improve this answer
+1 Actually, my questions there were sparkled from your reply here: 1. Can singular continuous measures be generalized to a more general measure space than Lebesgue measure space R? 2. The purpose of knowing it is that to what extent the decomposition of a singular measure into a discrete measure and a singular continuous measure exist, wrt some reference measure? – Tim Jan 14 '12 at 20:08

Take $f$ to be the Cantor function, then it has no density, but is continuous.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.