Fix $(f_1,f_2,f_3)\in \mathbb{R}^3_{+}$ such that $f_1+f_2+f_3=1$. Consider a random vector $(X,Y,Z)$ such that $$ \begin{aligned} &(1) \quad f_1={\Pr}(X\geq 0, Z\geq 0)\\ &(2) \quad f_2={\Pr}(Y\geq 0,Z<0)\\ &(3) \quad f_3={\Pr}(X<0, Y<0)\\ \end{aligned} $$
Question: Given any any $(X,Y,Z)$ satisfying (1)-(3), can we always construct from such $(X,Y,Z)$ a vector $(W,H,Q)$ satisfying the following conditions: $$ \begin{aligned} &(4) \quad f_1={\Pr}(W\geq 0, Q\geq 0)\\ &(5) \quad f_2={\Pr}(H\geq 0,Q<0)\\ &(6) \quad f_3={\Pr}(W<0, H<0)\\ &(7) \quad {\Pr}(W\geq t, H<u, Q<t-u)=0 \quad \forall (t,u)\in \mathbb{R}^2\\ &(8) \quad {\Pr} (W<t,H\geq u, Q\geq t-u)=0 \quad \forall (t,u)\in \mathbb{R}^2 \end{aligned} $$
Further (perhaps) useful observations:
As noted in the comments below, constraints (7) and (8) are just requiring that the distribution of $(W,H,Q)$ has support on $$ \{(w,h,q)\in \mathbb{R}^3: q=w-h\}. $$
Note that constraints (1)-(3) imply $$ {\Pr}(X\geq 0,Y<0, Z<0)=0\\ {\Pr}(X<0, Y\geq 0 , Z\geq 0 )=0 $$
Some motivation behind the question: I have a problem in statistics/computer science where I need to verify the existence of a 3-d distribution function that satisfies constraints (4)-(6) and that is "degenerate" on the third dimension (constraints (7)-(8)). However, constraints (7)-(8) are computationally intractable to implement because they should be imposed for each 2-tuple $(t,u)\in \mathbb{R}^2$. Much simpler is to verify the existence of a 3-d distribution function that satisfies constraints (1)-(3) (which are equivalent to (4)-(6)) and, then, construct from such distribution a new distribution function that satisfies constraints (4)-(8)
Attempted answer: Take any random variable $\epsilon$ taking only positive values. Define $$ (W,H,Q)\equiv \begin{cases} (X,Y, X-Y) & \text{ if } X\geq 0, Z\geq 0, X-Y\geq 0\\ & \text{ or if } Y\geq 0, Z< 0, X-Y<0\\ & \text{ or if } X< 0, Y<0\\ (X,X-\epsilon,\epsilon) & \text{ if } X\geq 0, Z\geq 0, X-Y<0\\ (Y-\epsilon,Y,-\epsilon) & \text{ if } Y\geq 0, Z< 0, X-Y\geq 0\\ \end{cases} $$
Then, $$ \begin{aligned} &\Pr(W\geq 0, Q\geq 0)=\Pr(X\geq 0, Z\geq 0, X-Y\geq 0)+\Pr(X\geq 0, Z\geq 0, X-Y<0)=f_1\\ &\Pr(H\geq 0, Q< 0)=\Pr(Y\geq 0, Z<0, X-Y< 0)+\Pr(Y\geq 0, Z<0, X-Y\geq 0)=f_2\\ &\Pr(W<0 , H<0)=\Pr(X< 0, Y<0)=f_3 \end{aligned} $$ Hence, (4)-(6) are satisfied. Moreover, since $Q=W-H$, (7)-(8) are also satisfied.