Consider a trivariate probability distribution $P: \mathbb{R}^3\rightarrow [0,1]$. I have the following questions:
(1) Are there necessary conditions on the cumulative distribution function (CDF) associated with $P$ ensuring that $$ \exists \text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$} $$
(2) Are there necessary and sufficient conditions on the CDF associated with $P$ ensuring that $$ \exists \text{ a random vector $(X_1,X_2)$ such that $(X_1, X_2, X_1-X_2)$ has probability distribution $P$} $$
(3) The conditions that you propose can be "approximated" as a linear constraint on the CDF?
I'm providing more details on my question also thanks to/inspired by the answers below. The answers below help, but I'm still not satisfied. Please help if you can.
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 $c_1\leq a_1-b_2$ $$ \begin{cases} P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [a_1-b_2, c_2])\\ P([a_1,a_2], [b_1, b_2], [c_3, c_1])=0 & \forall c_3\leq c_1 \end{cases} $$
All the implications above can be re-written as linear function of the CDF associated with $P$.
However: are these implications also sufficient? If yes, I don't know how to prove it; If not, I don't know how to find a counterexample.