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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?

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If I recall correctly, there is a restriction to at most countable discontinuities. –  Drew Christianson May 26 '12 at 16:58
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At least one discontinuity is common in densities of practical importance. –  André Nicolas May 26 '12 at 17:31
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@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

3 Answers 3

up vote 5 down vote accepted

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

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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.

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(+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?

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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

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