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

From other materials that I've read, the probability density of a continuous random variable must itself be continuous. Is this correct? If it is, I don't understand why that would be so, why can't the probability change abruptly?

share|cite|improve this question
If I recall correctly, there is a restriction to at most countable discontinuities. – Drew Christianson May 26 '12 at 16:58
At least one discontinuity is common in densities of practical importance. – André Nicolas May 26 '12 at 17:31
@DrewChristianson: Perhaps you are thinking of probability distribution functions. A density function can even be everywhere discontinuous. – Nate Eldredge May 26 '12 at 18:48
"Continuous" distribution means the cdf (cumulative distribution function) is continuous. This does not mean the density is continuous, or even that a density exists. – GEdgar May 27 '12 at 18:34
up vote 7 down vote accepted

Take $f(x) = 2x$, $0\le x \le 1$, and 0 otherwise. This is a density function which is not continuous.

share|cite|improve this answer
I thought so. Thanks! – Paul Manta May 26 '12 at 16:51

Michael Chernick asks for an example of a probability distribution with a density that is everywhere discontinuous.

As discussed in this question, there exists a measurable set $A \subset \mathbb{R}$ such that for every interval $I$, we have $0 < m(A \cap I) < m(I)$, and moreover $m(A) < \infty$. Then $f(x) = \frac{1}{m(A)} 1_A(x)$ is a nonnegative measurable function with $\int_\mathbb{R} f(x)\,dx = 1$, so it can be taken as the density of a continuous probability distribution. $f$ is nowhere continuous because every interval contains points of $A$ and $A^C$. Moreover, any function $g$ with $f=g$ a.e. is also nowhere continuous.

share|cite|improve this answer
(+1) More generally, a construction similar to the following should work: Let $f$ be a probability density function continuous on $\mathbb R$. Take $g = f \cdot 1_{(\mathbb R \setminus \mathbb Q)}$. – cardinal May 27 '12 at 18:43

Although valid I don't think the triangular density given by ncmathsadist is a good example. The U[0.1] density is discontinuous too because of the abrupt rise at x=0 and drop at x=1.But both these densities are continuous within their domain. I think a better example would be one with a discontinuity in its domain. Consider the density f(x)=2x for 0<=x<=1/2

and f(x)=(5-2x)/16 for 1/2

Certainly a density can have many such discontinuities. But can anyone give an example of a true probability distribution with a density that is everywhere discontinuous?

share|cite|improve this answer
In some sense, all probability distributions that are absolutely continuous have (a version of) a density that is everywhere discontinuous on the closure of its support. – cardinal May 27 '12 at 16:26
@cardinal: Could you please supply a reference or post an answer to explicate your assertion? Thank you. – Hans Dec 14 '15 at 20:10

No, need not be. However, the cumulative density function (CDF), is always continuous (mayn't be differentiable though) for a continuous random variable. For discrete random variables, CDF is discontinuous.

Taken from MIT6_041F10 Lecture slides

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.