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 $\mathbb{R}^4$?


1 Answer 1


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 $\eta$.

  • $\begingroup$ 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)? $\endgroup$
    – Star
    Sep 2, 2021 at 20:27
  • 2
    $\begingroup$ 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 $G^\varepsilon$). $\endgroup$
    – Steve
    Sep 3, 2021 at 7:05
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    $\begingroup$ @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. $\endgroup$ Sep 3, 2021 at 8:56
  • $\begingroup$ @BenoîtKloeckner Thank you. I have rephrased also Question B following your suggestion. Does it sound better? $\endgroup$
    – Star
    Sep 3, 2021 at 10:26
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    $\begingroup$ @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). $\endgroup$ Sep 3, 2021 at 14:23

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