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?

  • $\begingroup$ 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. $\endgroup$
    – mike
    Aug 18, 2021 at 8:22
  • $\begingroup$ 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? $\endgroup$
    – Star
    Aug 18, 2021 at 11:30
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 20, 2021 at 20:28

2 Answers 2


$\newcommand{\ep}{\epsilon}\newcommand{\R}{\mathbb R}$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.

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


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