# Questions on a new definition of continuous multivariate distribution

For a univariate distribution or a univariate random variable, we call it continuous/absolutely continuous if its cumulative distribution function (CDF) is continuous/absolutely continuous. Now I am trying to extend this concept to the multivariate case, and want the following holds: The continuity of a random vector does not change under homeomorphisms.

If we still define the continuity of a distribution only according to the continuity of its CDF, then the invariance property above does not hold. This can be seen from the following example.

For any random vector $(X,X)$ where $X$ is continuous, it is easy to check that this vector is continuous (according to the definition above). We transform $(X,X)$ into $(2X,0)$ by the homeomorphism $(x,y)\mapsto (x+y,x-y)$. Obviously, $(2X,0)$ is not continuous.

Now I give two new definitions of the continuity of a random vector.

(Definition 1) For random vector $(X,Y)$, we call it continuous if $F_{X'Y'}(x,y)$ is jointly continuous in $(x,y)$ for any homeomorphism $(X,Y)\mapsto (X',Y')$. Here $F_{X'Y'}$ is the CDF of $(X',Y')$.

(Definition 2) For random vector $(X,Y)$, we call it continuous if $F_X(x)$ is continuous, and $F_{Y|X}(y|x)$ is continuous in $y$ for $P_X$-almost every $x$. Here $F_X$ and $F_{Y|X}$ are the CDF of $X$ and the conditional CDF of $Y$ given $X$, and $P_X$ is the distribution of $X$.

My question is: For Definition 2, does the continuity of a random vector change under homeomorphisms? If yes, for any continuous random vector $(X,Y)$, is $(Y,X)$ also continuous?

1. I believe the term continuous distribution (or a continuous measure) is often used to refer to a distribution without atoms.

2. Homeomorphisms of $\mathbb{R}^2$ can be very singular. For example, there is a homeomorphism $\Phi$ that maps the vertical interval $\{(0, y) : y \in [0, 1]\}$ into the Osgood curve. If $(X,Y)$ has a uniform distribution on a square that contains this Osgood curve and $$(X',Y') = \Phi^{-1}(X,Y),$$ then $X'$ has an atom at zero.

3. For the above reason, any definition of a continuous distribution in $\mathbb{R}^2$ which is invariant under homeomorphisms must include distributions with discontinuous marginals. In particular, your Definition 1 is void, and the answer to your first question about Definition 2 is: yes, it may change.

4. If $(X,Y)$ satisfies your Definition 2, then clearly $Y$ is a continuous random variable: $$\mathbb{P}(Y = y) = \int \mathbb{P}(Y = y | X = x) P_X(dx) = 0.$$ However, $(Y,X)$ may fail to satisfy Definition 2, a counterexample is given below. In other words, the answer to your second question about Definition 2 is: no.

5. Take $X,Z$ to be two independent random variables uniformly concentrated on a Cantor set $$C = \left\{\sum_{n = 1}^\infty \frac{a_n}{5^n} : a_n \in \{0,1\}\right\},$$ and define $Y = X + 2 Z$. Then $X$ has a continuous distribution, and for every $x \in C$, $Y$ has a continuous conditional distribution $x + 2 Z$ given $X = x$. On the other hand, for every $y \in [0, 1)$ there is at most one pair $(x, z) \in C \times C$ such $y = x + 2 z$, and therefore the conditional distribution of $X$ given $Y = y$ is always a Dirac delta.

• Thank you for your answer! Could you provide a reference that involves "continuous measures"? I only found that a distribution/measure without atoms is called a atomless or non-atomic distribution/measure. It is called a continuous distribution only for univariate cases.
– Leon
Jul 22, 2018 at 14:46
• BTW, I was going to construct a distribution like yours in 5. Thank you for providing it to me first!
– Leon
Jul 22, 2018 at 14:51
• @LeiYu: You're welcome. I had the feeling that atomless', non-atomic' and `continuous' are synonyms here (a quick Google search yields several examples, like this, this, this or this, but indeed this may be too strong a statement; I'll edit the answer in a minute. Jul 22, 2018 at 15:41