When $(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim(X_0-X_1, X_0-X_2)$?

Question

Consider a bivariate distribution function $P: \mathbb{R}^2\rightarrow [0,1]$. I have the following question:

Are there necessary and sufficient conditions on $P$ (or on its marginals) ensuring that $$ \exists \text{ a random vector $(X_0,X_1,X_2)$ such that } $$ $$ (X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)\sim P $$

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Remarks:

(I) $(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)$ does not imply that some of the random variables among $X_1, X_2, X_0$ are degenerate.

For example, $(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)$ is implied by $(X_0, X_1, X_2)$ exchangeable.

(II) The symbol "$\sim$" denotes "DISTRIBUTED AS"

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My thoughts: among the necessary conditions, I would list the following: let $P_1,P_2$ be the two marginals of $P$. Then it should be that $$ \begin{cases} P_1 \text{ is symmetric around zero, i.e., $P_1(a)=1-P_1(-a)$ $\forall a \in \mathbb{R}$}\\ P_2 \text{ is symmetric around zero, i.e., $P_2(a)=1-P_2(-a)$ $\forall a \in \mathbb{R}$}\\ \end{cases} $$

Should $P$ be as well symmetric at zero?

Are these conditions also sufficient? If not, what else should be added to get an exhaustive set of sufficient and necessary conditions?

Answer 1

Answer to the original version of the question:

The answer to your question is No except trivially when $(X_1,X_2) = (0,0)$ with probability $1$. The $(X_1, X_1-X_2) \sim (X_2, X_2-X_1)$ part is easy enough to satisfy (e.g. $X_1, X_2$ are iid normal) but transitivity of $\sim$ implies that $(X_1, X_1-X_2) \sim (-X_1, -X_2)$ which says that $X_1-X_2$ has the same distribution as $-X_2$. So, $X_1$ must be $0$ with probability $1$, no? And then \begin{align}(X_1, X_1-X_2) &\sim (0, -X_2) &\text{as just proven}\\ (X_1, X_1-X_2)&\sim (X_2, X_2-X_1) &\text{as given}\\ &\implies X_2 \sim 0 ~\text{ wp } 1\end{align}