I am working on a research paper where I need to investigate conditions for the existence of probability distributions satisfying certain characteristics. I have already asked a related question ([here](https://mathoverflow.net/questions/401905/conditions-for-existence-of-a-distribution-with-full-support)), whose answers allowed me to frame better in my mind the problems I'm facing. In what follows, I report a close, although different, question. I will highlight the key differences below.


Consider a $6\times 1$ random vector
$$
\eta\equiv (\eta\_1,\eta\_2,..., \eta\_6)
$$
satisfying the following property (hereafter, called Property 1):


**Property 1:** $$
 \begin{pmatrix}
\eta\_1\\
\eta\_2\\
\eta\_3
\end{pmatrix} \sim \begin{pmatrix}
\eta\_4\\
-\eta\_2\\
\eta\_5
\end{pmatrix} \sim \begin{pmatrix}
\eta\_6\\
-\eta\_3\\
-\eta\_5
\end{pmatrix} \sim \begin{pmatrix}
-\eta\_1\\
-\eta\_4\\
-\eta\_6
\end{pmatrix} \sim G
$$
where "$\sim$" denotes "distributed as" and $G$ is an absolutely continuous distribution with full support on $\mathbb{R}^3$.




---


**Question A:** Let $\mathcal{G}$ denote the family of absolutely continuous distribution with full support on $\mathbb{R}^3$ and whose marginals are *symmetric around zero and identical*. For each $G\in \mathcal{G}$, does there exists a vector $\eta$ satisfying Property 1?




---


**Question B:** Let $\epsilon$ be a $4\times 1$ random vector
$$
\epsilon\equiv \begin{pmatrix}
\epsilon\_1\\
\epsilon\_2\\
\epsilon\_3\\
\epsilon\_0\\
\end{pmatrix} $$


For each $(G,\eta)$ satisfying Property 1, does there exist $\epsilon$ satisfying Property 2 described below?


**Property 2:**
$$ 
\begin{pmatrix}
1 & 0 & 0 & -1\\
1 & -1 & 0 & 0\\
1 & 0 & -1 & 0\\
0 & 1 & 0 & -1\\
0 & 1 & -1 & 0\\
0 & 0 & 1 & -1\\
\end{pmatrix}\*\epsilon=\begin{pmatrix}
\eta\_1\\
\eta\_2\\
\eta\_3\\
\eta\_4\\
\eta\_5\\
\eta\_6
\end{pmatrix}
$$
and the distribution $F$ of $\epsilon$ is absolutely continuous with full support on $\mathbb{R}^4$?




---


**My thoughts:**


I believe that the answer to Question A is "Yes": any distribution in $\mathcal{G}$ satisfies Property 1. Certainly, there exist distributions outside $\mathcal{G}$ that can also satisfy Property 1.


I believe that the answer to Question B is "Yes" as well. However, I'm not 100% sure and I would appreciate your help. The answers [here](https://mathoverflow.net/questions/401905/conditions-for-existence-of-a-distribution-with-full-support) suggest that: if $F$ is absolutely continuous with full support on $\mathbb{R}^4$, then $G$ is absolutely continuous with full support on $\mathbb{R}^3$. Here, however, I'm asking something different: if $G$ is absolutely continuous with full support on $\mathbb{R}^3$, can we always find a distribution $F$ for $\epsilon$ that is absolutely continuous with full support on $\mathbb{R}^4$?



From probability distribution in $\mathbb{R}^3$ to probability distribution in $\mathbb{R}^4$
I think the answer to both questions is negative.


**Question A.**
You ask whether for all $G\in\mathcal{G}$, there exist a random vector $\eta=(\eta\_i)\_{1\le i\le 6}$ such that they satisfy Condition 1 together.


If $G$ were not assumed to be fully supported, the answer would be easily seen to be negative: taking $G$ supported on $\{(x,y,z)\in\mathbb{R}^3 \mid x=y\}$ would force $\eta\_1=\eta\_2$, $\eta\_4=-\eta\_2$ and $-\eta\_1=-\eta\_4$ almost everywhere, which are incompatible.


But you can start from there and change $G$ slightly to be fully supported: simply start with any distribution supported on $\{(x,y,z)\in\mathbb{R}^3 \mid x=y\}$ with equal and centrally symmetric marginals; assume further that the marginals give a mass less than $1/5$ to $[-1,1]$ (**added in edit:** previously was $1/4$, but we need some room because of the perturbation). Define $G$ as a convolution of that distribution with a Gaussian $\sim\mathcal{N}(0,\varepsilon)$ for some small positive $\varepsilon$. Taking $\varepsilon$ small enough, you can ensure $G(\{(x,y,z)\in\mathbb{R}^3 \mid \lvert x-y\rvert>1/4\})<1/4$ and that any marginal of $G$ give a mass less than $1/5$ to $[-1,1]$. If a random vector $\eta$ were to satisfy condition 1 with $G$, you would have
\begin{align\*}
\mathbb{P}(\lvert \eta\_4\rvert\le 1) &<1/4 \\
\mathbb{P}(\lvert \eta\_1-\eta\_2\rvert>1/4) &<1/4 \\
\mathbb{P}(\lvert \eta\_4+\eta\_2\rvert>1/4) &<1/4 \\
\mathbb{P}(\lvert \eta\_1-\eta\_4\rvert>1/4) &<1/4 \\
\end{align\*}


With positive probability, we would thus have $\lvert \eta\_1-\eta\_2\rvert\le 1/4$, $\lvert \eta\_4+\eta\_2\rvert\le 1/4$, $\lvert \eta\_1-\eta\_4\rvert \le1/4$ and $\lvert \eta\_4\rvert > 1$. Now, this is impossible since
$$\lvert \eta\_4+\eta\_4\rvert \le \lvert \eta\_4-\eta\_1\rvert + \lvert\eta\_1-\eta\_2\rvert+\lvert\eta\_2+\eta\_4\rvert. $$
(I took more room than needed, but that does the trick.)


**Question B. (added in edit)**


Simply take $\eta$ a normal vector $\sim\mathcal{N}(0,I\_6)$. Then condition 1 holds with $G$ a normal distribution, but for any $\epsilon$ its image under the matrix is contained in its image vector space, which has dimension $4$ at most since it is a $6\times 4$ matrix. This cannot have a fully supported law, hence cannot equal $\eta$.


