# If $X$ is a continuous random variable with support $A$, does this imply that the cdf of $X$ is strictly increasing on $A$?

If $$X$$ is a continuous random variable with support $$A$$, does this imply that the cdf of $$X$$ is strictly increasing on $$A$$?

My guess is yes. But just in case, let me know if you can think of any counterexamples.

A counter example is made by taking the cdf as the Cantor distribution $$\mathfrak c(\cdot)$$ on $$[0,1]$$, whose support is the Cantor set $$\mathfrak C$$:

1. it is a continuous (if not absolutely continuous) distribution
2. its support $$\mathfrak C$$ is closed and of Lebesgue measure zero
3. the points 1⁄3 and 2⁄3 are adjacent in $$\mathfrak C$$, i.e., $$\mathfrak C\cap(1/3,2/3)=\emptyset$$ while $$1/3,2/3\in\mathfrak C$$
4. hence $$\mathfrak c(1/3)=\mathfrak c(2/3)$$ and $$\mathfrak c(\cdot)$$ is not strictly increasing on $$\mathfrak C$$
• Brilliant counter-example. What happens if the density exists as discussed in the other answer? – cross-entropy Dec 21 '20 at 10:57
• @Cross Wherever the density might be zero, the cdf is not strictly increasing. It is possible for densities to equal zero on the support of the random variable--they just cannot be zero throughout any open interval. – whuber Dec 21 '20 at 12:35
• Thanks, Xi'an! This is very helpful. – T34driver Dec 21 '20 at 22:57
• @Xi'an@whuber Then requiring the density to be everywhere positive would suffice, right? – T34driver Dec 21 '20 at 22:58

Assuming the density exists, and ignoring the boundary points, a function strictly increases if its derivative is strictly positive. CDF’s derivative is PDF and it is strictly positive in the support. So, CDF must strictly increase in the support.

• A continuous random variable being a random variable whose cumulative distribution function (cdf) is continuous everywhere, this cdf may be differentiable nowhere, meaning there is no density (pdf). – Xi'an Dec 21 '20 at 8:01
• @gunes Thank you very much! – T34driver Dec 23 '20 at 20:23