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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?

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  • $\begingroup$ 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. $\endgroup$
    – Kodiologist
    Commented Dec 11, 2018 at 19:19
  • $\begingroup$ 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 $G_1 = G_2$. $\endgroup$
    – guy
    Commented Dec 11, 2018 at 19:23
  • $\begingroup$ 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.... $\endgroup$
    – Dilip Sarwate
    Commented Dec 11, 2018 at 21:13
  • $\begingroup$ @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 $X_1-X_0\sim X_1-X_2$? $\endgroup$
    – Star
    Commented Dec 12, 2018 at 8:59
  • $\begingroup$ @user Sorry, I’m not going to go back over your modified question :) I thought the original one was clear enough. $\endgroup$
    – guy
    Commented Dec 12, 2018 at 15:43

1 Answer 1

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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}

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  • $\begingroup$ To avoid confusion I've slightly modify my question, adding $X_0$. $\endgroup$
    – Star
    Commented Dec 12, 2018 at 8:55
  • $\begingroup$ 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. $\endgroup$
    – Dilip Sarwate
    Commented Dec 12, 2018 at 13:33

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