Continuity of a random variable and of its cumulative distribution function

Question

Could you help me to understand if the following statements are correct?

Let $X$ be a random variable with cdf $F$ (so $F$ is, by definition, non-decreasing, right continuous, $\lim_{x\rightarrow-\infty}F(x)=0$, $\lim_{x\rightarrow\infty}F(x)=1$)

1) $X$ is a continuous random variable if and only if $F$ is a continuous functions

2) $F$ is a continuous function strictly monotone (increasing) if and only if $F^{-1}$ exists

3) $X$ is a continuous random variable and $F$ strictly monotone function (increasing) if and only if $F^{-1}$ exists

Answer 1

The first statement: $\implies$: $X$ being a continuous random variable means that it has a density. Just write $F$ as an integral of this density to see that $F$ is continuous. $\impliedby$: If $F$ is continuous, then it is absolutely continuous with respect to the Lebesgue measure so then by the Radon-Nikodyn theorem $X$ has a density.

The second statement: This has nothing to do with probability. $\implies$ is obvious I hope. For $\impliedby$ note that $F$ has an inverse if and only if it is bijective. Since $F$ is nondecreasing, it must be strictly increasing. A strictly increasing (or decreasing) bijective function has to be continuous. See this for a proof.

The third statement is just the first two combined.

Answer 2

I saw the accepted answer has an error in it (and I don't have enough reputation to comment on that), so I'll note the following.

Statement 1) is false -- usually "$X$ is a continuous random variable" means "there is an integrable function $f$ (called the density of $X$) such that $\mathbb{P}(X \leq t) = \int_{-\infty}^t f(x)\,dx$". Since $f$ is integrable, this expression (which gives the cdf) is continuous. But the converse need not be true. The Cantor function is the classic example of a continuous cdf which is not absolutely continuous (i.e. cannot be expressed as a cumulative integral of some other function).

Statement 3) is false for basically the same reason as statement 1) is; there are continuous and strictly increasing (i.e. invertible) cdfs which are not absolutely continuous -- for an explicit example, take the sum of any continuous and strictly increasing cdf (say, that of the normal distribution, for example) and the Cantor function, and divide by 2 (this is just a trick to get a variant of the Cantor function which is strictly increasing).