# Probability distribution function that does not have a density function

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

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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: math.stackexchange.com/questions/4683/… – Byron Schmuland Jan 13 '12 at 16:41

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 $y=z=0$...

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+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.

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