# From probability distribution in $\mathbb{R}^3$ to probability distribution in $\mathbb{R}^4$

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), 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 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 $$\mathbb{R}^4$$?

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.)

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 $$\eta$$.
• Thanks. I think I'm not understanding the conclusion "[...] this is impossible since [...]". Is there any "tractable" way to restrict $\mathcal{G}$ to $\mathcal{G}_{\text{restricted}}$ such that property 1 holds for each $G\in \mathcal{G}_{\text{restricted}}$? How about question B)?
• I think this is a clear counterexample. Perhaps another way to argue is that the restriction of full support is quite useless, since Property 1 remains when taking limits in distribution (i.e., if $\eta^\varepsilon$ satisfy Property 1 and $\eta^\varepsilon$ converges weakly to $\eta$, then also $\eta$ satisfies Property 1; where we simply define $G$ as the distribution of $(\eta_1, \eta_2, \eta_3)$ and same for $G^\varepsilon$). Sep 3, 2021 at 7:05
• @TEX: the last, implicit step is that with positive probability $2<2\lvert\eta_4\rvert\le 3/4$. Question B is not formulated precisely enough for me to answer. Maybe you ask whether for all $G,\eta$ such that condition 1 holds, there is an $\epsilon$ such that the second in-line equation holds, but you should rephrase things more precisely. Sep 3, 2021 at 8:56
• @TEX given the absolute values, I don't see the difference between the last statement and your proposal. The first statement comes from the assumption on the initial marginals (in fact you get something slightly larger than $1/4$ because of the perturbation I will correct). Sep 3, 2021 at 14:23