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Consider a random vector $X\equiv (X_1,...,X_L)$. Assume that each $X_l$ is continuously distributed with support $\mathbb{R}$, for $l=1,...,L$. Does this imply that also $X$ should be continuously distributed (although, not necessarily with support $\mathbb{R}^L$)?

I clarify that, by support of $X_l$, I intend the smallest closed set $\mathcal{X}$ such that $Pr(X_l\in \mathcal{X})=1$.

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    $\begingroup$ Wikipedia says that "A continuous probability distribution is a probability distribution whose support is an uncountable set". Under this definition, the answer is trivially yes. Are you working with some other definition of a continuous distribution? $\endgroup$ – Stephan Kolassa Oct 5 '20 at 20:06
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    $\begingroup$ @Stephan Although the Wikipedia article is vague, it is evident the context of the quotation is a univariate probability distribution. The quotation is obviously incorrect in more than one dimension; take e.g. the random variable $(Z,Z)$ where $Z$ has a standard Normal distribution. Incidentally, the term "support" in the quotation is misleading. That is usually taken to be the smallest closed set having unit probability; but there exist discrete distributions whose support is the entire real line. $\endgroup$ – whuber Oct 5 '20 at 20:20
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    $\begingroup$ @Dave A binary $Y$ can hardly be said to be continuously distributed with support $\mathbb R$ as the OP's hypotheses aver. $\endgroup$ – Dilip Sarwate Oct 5 '20 at 20:38
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    $\begingroup$ @whuber: Thank you. I think what you are saying is that the closure of the rational numbers is the real numbers (not something I had thought about) and that it is possible to have a discrete distribution on the rational numbers (something I had previously thought about) $\endgroup$ – Henry Oct 5 '20 at 21:32
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    $\begingroup$ The Wikipedia definition seems incorrect. A mixed distribution will also have uncountable support. $\endgroup$ – John Coleman Oct 6 '20 at 9:38
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No, if the individual random variables are continuous and thus their marginal distributions can be described using pdfs, it is not necessarily the case that they enjoy a joint pdf. A standard counterexample to the OP's "claim" is when $X \sim N(0,1)$ and $Z$ is an independent discrete random variable taking on values $\pm 1$ with equal probability. Then, $Y = ZX \sim N(0,1)$ also, but $(X,Y)$ does not have a joint pdf (measured in units of probability mass per unit area). All the probability mass lies on the lines $y=x$ and $y=-x$ and since lines have zero area, $(X,Y)$ does not enjoy a joint pdf (measured in units of probability mass per unit area). See, for example, this answer by Macro.

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  • $\begingroup$ Thanks. Your example seems close to the first example of the other answer. However, the other answers seems to say that there can be a proper pdf on the line $y=x$. Which one is correct? $\endgroup$ – TEX Oct 5 '20 at 23:59
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    $\begingroup$ There is a pdf on the straight line but it is not a joint pdf which has units probability mass per unit area but rather a univariate pdf measured in units of probability mass _per unit length as measured along the straight line. Looked at another way, the joint CDF $F_{X,Y}(x,y)=P(X\leq x,Y\leq y)$ is not continuous everywhere in the plane, and so is not differentiable everywhere in the plane either, and so we cannot write $f_{X,Y}(x,y)=\frac{\partial^2}{\partial x \partial y}F_{X,Y}(x,y)$ as we can for jointly continuous random variables. $\endgroup$ – Dilip Sarwate Oct 6 '20 at 16:56
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No, a very simple counterexample is $(Z, Z)$ where $Z \sim \mathcal{Norm}(0,1)$ where the marginals are standard normal but the joint distribution is concentrated on the diagonal line $y=x$. So the joint distribution do not have a density with respect to the Lebesgue measure on the plane, but it does indeed have a density with respect to the Lebesgue density on that line $y=x$.

Another simple example, but here the marginals have a density which is positive on the segment $[-1, 1]$. Let $(X,Y)$ have the uniform distribution on the unit circle, that is, we can represent it as $X=\cos(\theta), Y=\sin(\theta)$ where $\theta \sim \mathcal{Uniform}(0, 2\pi)$. I did a simulation in R:

enter image description here

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  • $\begingroup$ Thanks, but in both examples the vector $X$ is continuosly distributed (although not with support equal to the plane). Hence it does not contradict my statement. $\endgroup$ – TEX Oct 5 '20 at 23:54
  • $\begingroup$ Also the first example seems to contradict with the other answer which says that there can't be a proper pdf on the line $y=x$ $\endgroup$ – TEX Oct 5 '20 at 23:55
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    $\begingroup$ A density is always with respect to a base measure There can be no density on $y=x$ with respect to Leb Measure on the plane (as then the line has measure 0), but it can be *with respect to Leb measure on the line itself. $\endgroup$ – kjetil b halvorsen Oct 5 '20 at 23:58
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    $\begingroup$ In both examples the vector $X$ is concentrated on a set with measure 0 with respect to Leb measure on the plane. So it is not absolutely continuous with respect to that measure. This makes clear that in these examples we need to be more precise with the terminology! $\endgroup$ – kjetil b halvorsen Oct 6 '20 at 0:01
  • $\begingroup$ See stats.stackexchange.com/questions/298293/… for details about absolute continuity! $\endgroup$ – kjetil b halvorsen Oct 6 '20 at 17:23

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