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Consider a trivariate cumulative distribution function (cdf) $G$.

  • Is there a collection of necessary conditions on $G$ ensuring that $$ \exists \text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has cdf $G$} $$ ?

  • Is there a collection of necessary and sufficient conditions on $G$ ensuring that $$ \exists \text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has cdf $G$} $$ ?


Update I: Let $P$ be the probability distribution associated with $G$. We can claim that: if there exists a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$, then $$ \int_{(a,b,c)\in \mathbb{R}^3 \text{ s.t. } c=a-b} dP=1 $$

  • Is this condition also sufficient? I.e., can we claim that if $$ \int_{(a,b,c)\in \mathbb{R}^3 \text{ s.t. } c=a-b} dP=1 $$ then there exists a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$?

  • Can we write $$ \int_{(a,b,c)\in \mathbb{R}^3 \text{ s.t. } c=a-b} dP=1 $$ by using the cdf $G$ ?


Update II:

If there exists a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$, then $P$ should satisfy: for every $\begin{pmatrix} a_1\\ b_1\\ c_1 \end{pmatrix}\leq \begin{pmatrix} a_2\\ b_2\\ c_2 \end{pmatrix}$

  • If $a_2\geq b_2+c_2$ $$ \begin{cases} P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1, b_2+c_2], [b_1, b_2], [c_1, c_2])\\ P([a_2, a_3], [b_1, b_2], [c_1, c_2])= 0 & \forall a_3\geq a_2\\ \end{cases} $$

  • If $b_1\leq a_1-c_2$ $$ \begin{cases} P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [a_1-c_2, b_2], [c_1, c_2])\\ P([a_1,a_2], [b_3, b_1], [c_1, c_2])=0 & \forall b_3\leq b_1\\ \end{cases} $$

  • If $a_1 \leq b_1+c_1$ $$ \begin{cases} P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([b_1+c_1,a_2],[b_1,b_2],[c_1,c_2])\\ P([a_3,a_1], [b_1, b_2], [c_1, c_2])=0 & \forall a_3 \leq a_1 \end{cases} $$

  • If $b_2\geq a_2-c_1$ $$ \begin{cases} P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, a_2-c_1], [c_1, c_2])\\ P([a_1,a_2], [b_2, b_3], [c_1, c_2])=0 & \forall b_3\geq b_2 \end{cases} $$

  • If $c_2 \geq a_2-b_1$ $$ \begin{cases} P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [c_1, a_2-b_1])\\ P([a_1,a_2], [b_1, b_2], [c_2, c_3])=0 & \forall c_3\geq c_2 \end{cases} $$

  • If $c_1\leq a_1-b_2$ $$ \begin{cases} P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [a_1-b_2, c_2])\\ P([a_1,a_2], [b_1, b_2], [c_3, c_1])=0 & \forall c_3\leq c_1 \end{cases} $$ These implications can be written using $G$ (as I want!). However: are these implications also sufficient? I don't know how to prove or dis-prove it.

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    $\begingroup$ You have supplied such conditions: to wit, when $(X_1,X_2,X_3)$ is a random variable with distribution $G,$ then almost surely $X_3=X_1-X_2.$ That's dead simple. What other form are you hoping to express these conditions in that would be any simpler or more useful? $\endgroup$ – whuber Nov 29 '18 at 15:17
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    $\begingroup$ Thanks. I'm hoping for conditions directly imposed on $G$. $\endgroup$ – user3285148 Nov 29 '18 at 15:20
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The condition in your first update is sufficient, because it implies that $X_3=X_1-X_2$ almost surely, so in particular $(X_1,X_2,X_3)=^d(X_1,X_2,X_1-X_2)$.

Edit: To be completely explicit about the dependence of $G$, this can be restated by requiring that $P(B)=0$ for any 3-dimensional box $B$ such that $B\cap \{(x,y,z):z=x-y\}=\emptyset$.

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  • $\begingroup$ Thanks. What about update II? Remember that I'm looking for conditions that can expressed using $G$. $\endgroup$ – user3285148 Dec 11 '18 at 18:36
  • $\begingroup$ The measure $dP$ is defined in terms of $G$, so I'm not sure what else you want. $\endgroup$ – Mike Hawk Dec 11 '18 at 18:44
  • $\begingroup$ I want necessary and sufficient conditions explicitly involving "boxes" in $\mathbb{R}^3$ $\endgroup$ – user3285148 Dec 11 '18 at 18:46
  • $\begingroup$ Thanks. I think that "$P(B)=0$ for any 3-dimensional box $B$ such that $B\cap ...=\emptyset$" holds IF AND ONLY IF the bullet points in my update II hold. Is this correct? $\endgroup$ – user3285148 Dec 12 '18 at 11:19

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