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

user12586
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    Any discrete probability distribution, such as the one that picks an integer between 1 and 10 (inclusive) with equal probability. – hmakholm left over Monica Jan 13 '12 at 16:32
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    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
  • @user12586: It is still a distribution on $\mathbb R$ -- it just happens never to pick a non-integer. You could also take something like "pick the number 0 with probability 1/2, otherwise pick from a normal distribution with mean 0 and standard deviation 1". – hmakholm left over Monica Jan 13 '12 at 16:36
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    Thank you. I should have said: do we have an example with atomless distributions? – user12586 Jan 13 '12 at 16:39
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    See the "devil's staircase" as in this answer: http://math.stackexchange.com/questions/4683/continuous-and-bounded-variation-does-not-imply-absolutely-continuous –  Jan 13 '12 at 16:41
  • Thank you very much. So there is a monotone, increasing, but not AC function. – user12586 Jan 13 '12 at 16:44
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    @user12586 Yes! Weird isn't it? Actually "non-decreasing" is a bit more accurate than "increasing" since the function is virtually always flat. –  Jan 13 '12 at 16:46
  • But can it be said that these functions are non-generic in some sense? – user12586 Jan 13 '12 at 17:06

2 Answers2

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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 $y=z=0$...

Did
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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
  • This is a wonderful answer. Where could one look for some examples of this decomposition $\mu=\mu_a+\mu_d+\mu_s$? I remember seeing that this decomposition always exists in a measure theory course, but at the time I did not have time to build intuition for what kinds of measures induce nonzero components $\mu_a, \mu_d, \mu_s$. – kdbanman Sep 17 '20 at 19:00
  • For those wondering, this decomposition is called Lebesgue's decomposition theorem : https://en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem – Voidt Apr 01 '23 at 11:39
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Take $f$ to be the Cantor function, then it has no density, but is continuous.

Amber Xue
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