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?



Conditions for existence of a distribution with full support
$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$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{equation}
 \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}
\end{equation}
we have
\begin{equation}
 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}
\end{equation}


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.


