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.
Best Answer
(1) -(3) each have one unrestricted dimension. As suggested in comments under the question, place constructed points on the plane $W - H - Q = 0$:
$ \\ (W,H,Q) = \begin{cases}X\ge 0,~Z\ge 0:&W = X,&Q=Z,&H=W-Q\\ \\ Y\ge 0,~Z<0:&H=Y,&Q=Z,&W=H+Q\\ \\ X<0,~Y<0:&W=X,&H=Y,&Q=W-H\\ \end{cases} $
This satisfies (4)-(6).
Treating (7) and (8) as logical constraints:
$\begin{align}(7) \quad\neg&\left( (W \ge t) ~\cap~(H<u)~\cap~(Q<t-u) \right)~=\\ ~\neg&(W\ge t) ~\cup~ \neg(H<u) ~\cup~ \neg(Q<t-u) ~=\\ &(W<t) ~\cup~ (H \ge u) ~\cup~ (Q \ge t-u) \end{align}$
$\begin{align}(8) \quad\neg&\left( (W < t) ~\cap~(H\ge u)~\cap~(Q\ge t-u) \right)~=\\ ~\neg&(W< t) ~\cup~ \neg(H \ge u) ~\cup~ \neg(Q \ge t-u) ~=\\ &(W \ge t) ~\cup~ (H < u) ~\cup~ (Q < t-u) \end{align}$
The truth table for the assertions $(W \ge t)$ and $(H \ge u)$
and implications for the satisfaction of (7) and (8):
$ \begin{matrix} W \ge t & H \ge u & Q \ge t-u & (7)~\text{and}~(8)?\\ \texttt{false}(7) & \texttt{false}(8) & & \checkmark \\ \texttt{false}(7) & \texttt{true}(7) & \texttt{false}(8) & \checkmark \\ \texttt{true}(8) & \texttt{false}(8) & \texttt{true}(7) & \checkmark \\ \texttt{true}(8) & \texttt{true}(7) & & \checkmark \\ \end{matrix} $
Answering the OP's question, given any $\begin{bmatrix}X\\Y\\Z\end{bmatrix}$ satisfying (1)-(3), you can always construct $\begin{bmatrix}W\\H\\Q\\\end{bmatrix}$ satisfying conditions (4) - (8).