1
$\begingroup$

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 $\mathbb{R}^4$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $\eta$.

$\endgroup$
11
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.