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Let $X$ be a continuous random variable and let $g$ be a non-constant real-valued continuous function. Prove or disprove that $g(X)$ is a continuous random variable.

Note : Here it is assumed that $g(X)$ is a random variable.

Michael Hardy
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karfai
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1 Answers1

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Let $$ g(x)=\begin{cases} 0 & \text{if }x<0, \\ x & \text{if }0\le x\le 1, \\ 1 & \text{if } x>1. \end{cases} $$ Suppose $X$ is normally distributed with expectation $0$ and variance $1$.

Then $g(X)$ puts probability $1/2$ at $0$ and probability about $0.16$ at $1$, so $g(X)$ is not a continuous random variable.

Michael Hardy
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  • The usual definition of a continuous random variable $Y$ (as opposed to a discrete one) is that the image of $Y$ is uncountably infinite. Here, the image of your $g(X)$ is the whole of $[0,1]$ which is uncountably infinite. So your $g(X)$ is in fact continuous. – Malkin Aug 09 '18 at 22:16
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    @Malkin : No, that's not the usual definition at all. One definition sometimes seen is that the c.d.f. is continuous. For this random variable, the c.d.f. has two discontinuities. It is not the case that every probability distribution that is not discrete is continuous. This distribution is not discrete, but it has a discrete part. Nor is it continuous. Another definition of continuous random variable is one that is "absolutely continuous", i.e. it assigns probability $0$ to every set with Lebesgue measure $0.$ A theorem tells us that happens if and only if there is a density function. – Michael Hardy Aug 09 '18 at 22:32
  • @Malkin : Many books define "discrete random variable" as one whose image is finite or countably infinite. But a random variable that is not discrete by that definition certainly need not be continuous, and the one I give here is not, since its c.d.f. has discontinuities at two points. In my opinion, a better definition of "discrete", for probability distributions, is that a discrete distribution is one for which point masses account for all of the probability. – Michael Hardy Aug 09 '18 at 22:34
  • @Malkin : You should go back and review the definitions. – Michael Hardy Aug 09 '18 at 22:36
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    My original comment seems somewhat blunt now: I apologise! I should have posed a question to the OP about their definition "continuous random variable". Indeed, with your definition "that the c.d.f. is continuous", $g(X)$ is obviously not continuous. – Malkin Aug 09 '18 at 22:41
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    Maybe I should add that the two definitions I mentioned do not agree. Some distributions are continuous in the sense of having a continuous c.d.f. but not in the sense of having a density. The Cantor distribution is usually cited as an example. – Michael Hardy Aug 11 '18 at 11:20