Given a probability space $\Gamma:=\langle\Omega,\Sigma,P\rangle$ and a random variable $\mathit{X}:\Omega\rightarrow\mathbb{R}^n$, I do not fully understand the definition of absolute continuity of $\mathit{X}$. I'm aware that a sound understanding of absolute continuity requires an equally sound knowledge of measure theory and Lebesgue integration, but I've been trying to get some sense out of it without delving too deep into the weeds.

My current picture is the following:

1) Absolute continuity implies uniform continuity

2) $\mathit{X}$ is continuous iff the range of $\mathit{X}$ is uncountably infinite

3) If $\mathit{X}$ is absolutely continuous, then there exists a function $f_X:\mathbb{R^n}\rightarrow[0,1]\subseteq\mathbb{R}$ such that the cumulative distribution function $F_X(u_1,\dots,u_n)=\displaystyle\idotsint_Uf_X(u_1,\dots,u_n)du_1 \dots du_n$

Are (1), (2), (3) correct? Is there a proper definition for an absolutely continuous random variable?

Thanks in advance!

  • $\begingroup$ There is no reason to have a topology on $\Omega $. The measure on $\mathbb R$ induce by $X$ is absolutely continuous wrt the Lebesgue measure if there is $f$ s.t. $\int_{A}\mathbb P_X(dx)=\int_Af(x)dx$ $\endgroup$
    – Pierre
    Mar 20 '19 at 8:48
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    $\begingroup$ (2) is not correct, and (3) is also an if and only if. $\endgroup$
    – user647486
    Mar 20 '19 at 8:49
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    $\begingroup$ 1) and 3) are true but 2) is false. Definition of absolute continuity can be found in nay book on measure theoretic probability like the one by Chung. $\endgroup$ Mar 20 '19 at 8:50
  • $\begingroup$ Thanks. As far as 2) is concerned, why isn't it correct? As far as I know, $\mathit{X}$ is discrete if its range is either finite or countably infinite? $\endgroup$ Mar 20 '19 at 8:57
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    $\begingroup$ @NTAuthority $X$ is continuous iff CDF $F_X$ is a continuous function. It is possible that the image of $X$ is not countable but nevertheless the CDF of $X$ is not continuous at some point. $\endgroup$
    – drhab
    Mar 20 '19 at 9:30

Let $X$ be a random variable and let $\mathcal B$ denote the $\sigma$-algebra of Borel subsets of $\mathbb R$.

Then $X$ is absolutely continuous iff for every $B\in\mathcal B$ with $\lambda(B)=0$ we also have $P(X\in B)=0$.

Notation for this: $P_X\ll\lambda$.

Here $\lambda$ denotes the Lebesgue meausure on $(\mathbb R,\mathcal B)$ and $P_X$ denotes the measure on $(\mathbb R,\mathcal B)$ that is prescribed by $B\mapsto P(X\in B)$.

As remarked already in the comments on your question: 2) in your question is not correct.

The mentioned condition is necessary but not sufficient.

$X$ is continuous iff the CDF of $X$ is continuous.

Here an example of a random variable having uncountable image that is not continuous:$$X=B+(1-B)Y$$ where $B,Y$ are independent, $Y$ has standard normal distribution and $B$ has Bernoulli distribution with parameter $p\in(0,1)$.

Observe that $P(X=1)=P(B=1)=p>0$, showing that the CDF of $X$ is not continuous at $1$.


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