# Conditions on a bivariate distribution to be the distribution of $(X_1-X_0, X_1-X_2)$, $(X_2-X_0, X_2-X_1)$ and $(X_0-X_1, X_0-X_2)$

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

Are there necessary and sufficient conditions on the cumulative distribution function (CDF) associated with $$P$$ (joint or marginal) ensuring that $$\exists \text{ a random vector (X_0,X_1,X_2) such that }$$ $$(X_1-X_0, X_1-X_2), (X_2-X_0, X_2-X_1), (X_0-X_1, X_0-X_2)$$ $$\text{ have all probability distribution P? }$$

Notice:

$$(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)$$ 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.

My thoughts: among the necessary conditions, I would list the following: let $$G$$ be the CDF associated with $$P$$ and let $$G_1,G_2$$ be the two marginal CDFs. Then it should be that $$\begin{cases} G_1 \text{ is symmetric around zero, i.e., G_1(a)=1-G_1(-a) \forall a \in \mathbb{R}}\\ G_2 \text{ is symmetric around zero, i.e., G_2(a)=1-G_2(-a) \forall a \in \mathbb{R}}\\ \end{cases}$$

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

• dont you mean "there exists a random vector $(X_0, X_1, X_2)$..."? otherwise what is $X_0$? Dec 13, 2018 at 12:40
• the requirement is also not what i expected, because it doesnt have cyclic symmetry. are you sure? Dec 13, 2018 at 12:41
• @antkam (1) Yes, edited. (2) What did you expect instead? What do you mean by "cyclic symmetry"? Is your "cyclic symmetry" an implication of exchangeability?
– TEX
Dec 13, 2018 at 12:45
• i mean if you cyclically substitute $X_0 \rightarrow X_1 \rightarrow X_2 \rightarrow X_0$ then you would generate the similarity requirements like this: $(X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)\sim (X_1- X_2, X_1-X_0)$. note the difference in the 3rd term. however, you can of course require your version. i am not sure it made much difference to be honest. :) Dec 13, 2018 at 13:30
• I see. You are just flipping the components of my first argument. I think also your relation is an implication of exchangeability. I don't know if considering your relation rather than mine can make things easier, though.
– TEX
Dec 13, 2018 at 13:36

When you have a vector of random variables, or equivalently a random variable taking values in $$\mathbb R^2$$, we can write it as $$(U,V)$$ where $$U$$ is the $$x$$-coordinate and $$V$$ is the $$y$$-coordinate of the random vector. So $$G_1(u)=\mathbb P(U\le u),$$ $$G_2(v)=\mathbb P(V\le v).$$

Now, in general if $$X$$ and $$Y$$ are random variables and $$F_X(x)=\mathbb P(X\le x)$$, $$F_Y(y)=\mathbb P(Y\le y)$$, then we write $$X\sim Y$$ if $$F_X=F_Y$$.

Besides the conditions you give,

namely: if $$(U,V)$$ is a random variable on $$R^2$$ as desired then $$U\sim -U$$ and $$V\sim -V$$, where $$\sim$$ denotes "has the same distribution as",

there's also

$$V-U\sim (X_0-X_2)-(X_0-X_1)= X_1-X_2\sim V$$

And note that $$V\sim -V$$, $$U\sim -U$$ does not imply $$V-U\sim V$$, e.g., take $$U$$, $$V$$ to be independent standard normal $$N(0,1)$$ random variables: $$\mathrm{Var}(V-U)=\mathrm{Var}+\mathrm{Var}(U) = 2>1=\mathrm{Var}(V)$$

• Thanks: 2 questions on that (1) How do I show that your condition is necessary? (2) Is your condition implying that one variable between $X_1, X_2$ should be zero almost surely?
– TEX
Dec 23, 2018 at 9:38
• Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)$ $\Rightarrow$ $\begin{cases}(X_1-X_2)\sim( X_2-X_1)\sim (X_0-X_2)\\ (X_1-X_0)\sim (X_2-X_0) \sim (X_0-X_1) \end{cases}$ $\Rightarrow$ ...?
– TEX
Dec 23, 2018 at 9:42
• Regarding (2), I don't think that $(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)$ should imply that one variable between $X_1, X_2$ should be zero almost surely. Indeed, if $(X_0, X_1, X_2)$ exchangeable then $(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)$.
– TEX
Dec 23, 2018 at 9:47
• @STF I agree with you regarding (2). $V-U\sim V$ doesn't imply $U=0$. And regarding (1), the rule I use is that if $(U,V)\sim (X,Y)$ then $U\sim X$ and $V\sim Y$. Dec 23, 2018 at 11:00
• Thanks. (2) Can you make a parametric example of $V-U\sim V$?
– TEX
Dec 23, 2018 at 11:08