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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?

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  • $\begingroup$ dont you mean "there exists a random vector $(X_0, X_1, X_2)$..."? otherwise what is $X_0$? $\endgroup$
    – antkam
    Dec 13, 2018 at 12:40
  • $\begingroup$ the requirement is also not what i expected, because it doesnt have cyclic symmetry. are you sure? $\endgroup$
    – antkam
    Dec 13, 2018 at 12:41
  • $\begingroup$ @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? $\endgroup$
    – TEX
    Dec 13, 2018 at 12:45
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    $\begingroup$ 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. :) $\endgroup$
    – antkam
    Dec 13, 2018 at 13:30
  • $\begingroup$ 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. $\endgroup$
    – TEX
    Dec 13, 2018 at 13:36

1 Answer 1

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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: $$\mathrm{Var}(V-U)=\mathrm{Var}+\mathrm{Var}(U) = 2>1=\mathrm{Var}(V)$$

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  • $\begingroup$ 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? $\endgroup$
    – TEX
    Dec 23, 2018 at 9:38
  • $\begingroup$ Regarding (1), I'm stuck here: $(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)$ $\Rightarrow$ $\begin{cases}(X_1-X_2)\sim( X_2-X_1)\sim (X_0-X_2)\\ (X_1-X_0)\sim (X_2-X_0) \sim (X_0-X_1) \end{cases}$ $\Rightarrow$ ...? $\endgroup$
    – TEX
    Dec 23, 2018 at 9:42
  • $\begingroup$ 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 $(X_1-X_0, X_1-X_2)\sim (X_2-X_0, X_2-X_1)\sim (X_0-X_1, X_0-X_2)$. $\endgroup$
    – TEX
    Dec 23, 2018 at 9:47
  • $\begingroup$ @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 $V\sim Y$. $\endgroup$
    – Bjørn Kjos-Hanssen
    Dec 23, 2018 at 11:00
  • $\begingroup$ Thanks. (2) Can you make a parametric example of $V-U\sim V$? $\endgroup$
    – TEX
    Dec 23, 2018 at 11:08

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