# Conditions for existence of a distribution with full support

Consider a $$6\times 1$$ continuous random vector $$\eta\equiv (\eta_1,\eta_2,..., \eta_6)$$ satisfying the following property: $$\underbrace{\begin{pmatrix} \eta_1\\ \eta_2\\ \eta_3 \end{pmatrix}}_{\equiv x_1} \sim \underbrace{\begin{pmatrix} \eta_1\\ \eta_4\\ \eta_5 \end{pmatrix}}_{\equiv x_2} \sim \underbrace{\begin{pmatrix} \eta_2\\ \eta_4\\ \eta_6 \end{pmatrix}}_{\equiv x_3} \sim \underbrace{\begin{pmatrix} \eta_3\\ \eta_5\\ \eta_6 \end{pmatrix}}_{\equiv x_4} \sim G$$ where $$\sim$$'' denotes "distributed as" and $$G$$ is some distribution with support $$\subseteq \mathbb{R}^3$$.

Question: I am looking for necessary and sufficient conditions on the support of $$G$$ such that there exists a $$4\times 1$$ random vector $$\epsilon \equiv (\epsilon_0, \epsilon_1,\epsilon_2,\epsilon_3)$$ having an absolutely continuous distribution with full support on $$\mathbb{R}^4$$ and such that (*) \quad \quad \begin{aligned} \eta_1= \epsilon_1-\epsilon_0\\ \eta_2= \epsilon_2-\epsilon_0\\ \eta_3= \epsilon_3-\epsilon_0\\ \eta_4= \epsilon_1-\epsilon_2\\ \eta_5= \epsilon_1-\epsilon_3\\ \eta_6= \epsilon_2-\epsilon_3\\ \end{aligned}

In particular, I would like to understand the following:

A) if the distribution of $$\epsilon$$ has full support on $$\mathbb{R}^4$$, then $$G$$ CANNOT have full support on $$\mathbb{R}^3$$. Is this correct? Why?

B) if A is correct, then there should be certain boxes in $$\mathbb{R}^3$$ where $$G$$ is zero. Can we characterise those boxes?

• Do you have any examples of a distribution that works, albeit with holes in G ? It seems to me that full support should not be an issue. If you produced any G whose support contained an interval containing 0, then randomly scaling the $\epsilon$ would produce a different G with full support.. I would guess, that if G did not live on a linear subspace, then you could do something like this to get full support.
– mike
Aug 18, 2021 at 8:22
• I don't t have any example unfortunately. My doubts stem from the fact that $\epsilon$ is a linear combination of $\eta$, which makes me worry that if the distribution of $\epsilon$ has full support on $\mathbb{R}^4$, then $G$ has to be somehow degenerate in $\mathbb{R}^3$. Can we exclude that?
– Star
Aug 18, 2021 at 11:30
• Unless I misunderstood the question the opposite of A) is true. If $\epsilon$ has full support, then so does $G$ since $\epsilon \mapsto x_1$ is surjective. Aug 20, 2021 at 20:28

$$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$$There is no necessary and sufficient condition in terms of the support of $$G$$ for the following: there exists a $$4\times 1$$ random vector $$\ep:=(\epsilon_0, \ep_1,\ep_2,\ep_3)$$ having an absolutely continuous distribution with full support on $$\R^4$$ and such that for \begin{aligned} \eta_1&:= \ep_1-\ep_0, \\ \eta_2&:= \ep_2-\ep_0, \\ \eta_3&:= \ep_3-\ep_0, \\ \eta_4&:= \ep_1-\ep_2, \\ \eta_5&:= \ep_1-\ep_3, \\ \eta_6&:= \ep_2-\ep_3 \end{aligned} \tag{1} we have $$$$x_1:=\begin{pmatrix} \eta_1\\ \eta_2\\ \eta_3 \end{pmatrix} \sim \begin{pmatrix} \eta_1\\ \eta_4\\ \eta_5 \end{pmatrix} \sim \begin{pmatrix} \eta_2\\ \eta_4\\ \eta_6 \end{pmatrix} \sim \begin{pmatrix} \eta_3\\ \eta_5\\ \eta_6 \end{pmatrix} \sim G. \tag{2}$$$$

Indeed, as was noted in Martin Hairer's comment, the map given by the first three definitions in (1) that maps the random vector $$\ep$$ to the random vector $$x_1$$ is surjective and continuous. So, it is necessary that (the distribution $$G$$ of) $$x_1$$ have full support on $$\R^3$$, because $$\ep$$ has full support on $$\R^4$$. (Indeed, take any nonempty open ball $$B_3\subset\R^3$$. The preimage of $$B_3$$ under the mentioned continuous map contains a nonempty open ball $$B_4\subset\R^4$$. So, $$P(x_1\in B_3)\ge P(\ep\in B_4)>0$$.)

On the other hand, (2) implies that $$\eta_1\sim\eta_2$$. Letting now $$G$$ be any probability distribution on $$\R^3$$ with full support whose first two one-dimensional marginals are not the same, we see that the condition $$x_1\sim G$$ in (2) cannot be satisfied.

So, there is no necessary condition on the support of $$G$$ that would also be sufficient.

• Thank you. In the second to last paragraph: "letting now $G$ being any probability distribution on $\mathbb{R}^4$". You meant $\mathbb{R}^3$?
– Star
Aug 22, 2021 at 11:07
• @TEX : Yes, thank you for your comment. This is now fixed. Aug 22, 2021 at 14:42

I will argue that (A) is incorrect: $$G$$ always has full support.

The proof is simple. Define $$a_1=\eta_1$$, $$a_2=\eta_2$$, $$a_3=\eta_3$$, $$a_4=\epsilon_0$$. Clearly $$a$$ is just a change of basis of $$\epsilon$$, and therefore is absolutely continuous with full support on $$\mathbb R^4$$. Therefore there exists a positive measurable function $$f:\mathbb R^4\to\mathbb R$$ which integrates to one, and which is the Radon-Nikodym derivative of the law of $$a$$ with respect to $$\lambda^4$$.

By Fubini's theorem, the Radon-Nikodym derivative of $$x_1$$ with respect to $$\lambda^3$$ is given by

$$g:\mathbb R^3\to\mathbb R,\,y\mapsto \int f(y_1,y_2,y_3,u)d\lambda(u),$$

which is well-defined almost everywhere. Since $$f$$ is positive, $$g$$ is positive almost everywhere. Since $$G$$ has the law of $$x_1$$, we are done.

You introduced a lot of other notation as well; I think they are not needed for providing an answer to your question.