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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.

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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 $\mathfrak C$
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    $\begingroup$ Brilliant counter-example. What happens if the density exists as discussed in the other answer? $\endgroup$ – cross-entropy Dec 21 '20 at 10:57
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    $\begingroup$ @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. $\endgroup$ – whuber Dec 21 '20 at 12:35
  • $\begingroup$ Thanks, Xi'an! This is very helpful. $\endgroup$ – T34driver Dec 21 '20 at 22:57
  • $\begingroup$ @Xi'an@whuber Then requiring the density to be everywhere positive would suffice, right? $\endgroup$ – T34driver Dec 21 '20 at 22:58
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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.

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    $\begingroup$ 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). $\endgroup$ – Xi'an Dec 21 '20 at 8:01
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    $\begingroup$ @gunes Thank you very much! $\endgroup$ – T34driver Dec 23 '20 at 20:23

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