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

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 $$\exists \text{ a random vector (X_0,X_1,X_2) such that }$$ $$(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)\sim P$$

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"

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?

• Random vectors with more than one dimension don't have CDFs. There are marginal CDFs for the coordinates, but no one CDF for the whole joint distribution. Dec 11, 2018 at 19:19
• If $X_1$ is not degenerate, then a necessary condition is that the correlation between $X_1$ and $X_2$ is $0.5$. In the special case where $(X_1, X_2)$ is a bivarate normal this, with the additional requirement that the mean is $(0,0)$, is both necessary and sufficient. It is easy to find counterexamples to sufficiency in the general case (e.g., if $X_1$ and $X_2$ are Rademacher variables then this is not sufficient). Additionally, note that it is necessary that $G_1 = G_2$.
– guy
Dec 11, 2018 at 19:23
• Since CDFs are right-continuous functions, your symmetry condition insists that the CDF be continuous at $0$ and have value $\frac 12$ at $0$. @Kodiologist: There exist joint CDFs for random vectors, for example, $F_{X.Y}(x,y) = P\{X \leq x, Y \leq y\}$ where the comma is commonly used to mean intersection. Easier to read than the more formal $$F_{X.Y}(x,y) = P\left(\{X \leq x\}\cap \{Y \leq y\}\right)$$ but YMMV.... Dec 11, 2018 at 21:13
• @guy: thanks, I had to modify slightly my question given the confusion in the comments. Could you explain why $G_1=G_2$? Thanks. I can see that my condition implies $X_1-X_0 \sim X_2-X_0\sim X_0-X_1$ and $X_1-X_2 \sim X_2-X_1\sim X_2-X_0$. Why this implies $X_1-X_0\sim X_1-X_2$?
– TEX
Dec 12, 2018 at 8:59
• @user Sorry, I’m not going to go back over your modified question :) I thought the original one was clear enough.
– guy
Dec 12, 2018 at 15:43

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}
• To avoid confusion I've slightly modify my question, adding $X_0$.
• The marginal distributions of $X$ and $Y$ are completely determined by the joint distribution of $X$ and $Y$. a.k.a. the distribution of $(X,Y)$. So, if you assume $(A,B) \sim (X.Y)$, you are implicitly assuming that $A\sim X$ and $B\sim Y$. Thus, your transitive assumption that $(X_1,X_1-X_2)\sim (-X_1, -X_2)$ is saying that $X_1 \sim -X_1$ which is your symmetry condition on $G_1$ but also that $X_1 - X_2 \sim -X_2$ which can only happen if $X_1$ is $0$ with probability $1$. Note that your $X_1-X_2 \sim -X_2$ also implies that.$X_1=0$. var($X_1-X_2)=2$, var$(-X_2)=1$ in your example. Dec 12, 2018 at 13:33