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Let $X:\Omega\to\mathbb{R}$ be a continuous random variable in the sense that its range is an uncountable set. Is it possible that its cdf $F_X:\mathbb{R}\to\mathbb{R}$ is not a continuous function? I did learn that if $X$ is absolutely continuous (i.e., has a density function), then $F_X$ is continuous.

Some authors define a continuous random variable to be those whose CDF is continuous, while others define it to be those whose range is an uncountable set. So I guess my question is whether these two definitions are equivalent.

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  • $\begingroup$ May be monotone continuity from below and monotone continuity from above of the underlying probability measure will establish continuity of $F_X$? $\endgroup$
    – Dinesh
    Aug 12, 2020 at 8:38

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The definitions are not equivalent.

Let $\Omega=\mathbb R$ be equipped with the $\sigma$-algebra of Borel subsets and a probability measure $P$ that satisfies $P([0,1])=p>0$.

Now prescribe $X:\Omega\to\mathbb R$ by $\omega\mapsto0$ if $\omega\in[0,1]$ and $\omega\mapsto\omega$ otherwise.

Then clearly the range of $X$ is uncountable, but $P(X=0)=p>0$.

That means that $F_X$ is not continuous at $0$.

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Let $X$ have standard normal distribution. Let $Y=X$ if $X <1$ and $2$ if $X \geq 1$. Then $Y$ takes uncountably many values. But $P( Y \leq 2)=1$ and $P(Y\leq t) =P(X<1) $ for $t <2$. Hence the CDF of $Y$ has jump discontinuity at $2$.

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Let $X=0$ with probability $1/2$, and otherwise take values in $[1/2,1]$ uniformly randomly. Its law is $$ \mathcal L(A) = \frac12\delta_0(A) + |A\cap [1/2,1]|. $$ Its CDF $P(X<t) = \mathcal L((-\infty,t])$ is not continuous at zero.

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