Construct a random vector as a function of another random vector ASSUMPTION 1: there exists a continuous random vector $(X,Y,Z)$ such that
$$
\begin{cases}
p_1=\Pr(X\geq 0, Z\geq 0)\\
p_2=\Pr(Y\geq 0, Z< 0)\\
p_3=\Pr(X< 0, Y<0)\\
\end{cases}
$$
where $(p_1,p_2,p_3)\in [0,1]^3$ and $p_1+p_2+p_3=1$. Further, the marginal distribution of each of $X,Y,Z$ are symmetric around 0.
Note that such a random vector $(X,Y,Z)$ may not exist for some values of $(p_1,p_2,p_3)$. This is why here I assume that it exists.
QUESTION: Does Assumption 1 imply that there exists (i.e., we can construct from $(X,Y,Z)$) a continuous random vector $(W,H,Q)$ such that:

*

*it holds that
$$
\begin{cases}
\Pr(W\geq 0, Q\geq 0)=p_1\\
\Pr(H\geq 0, Q< 0)= p_2\\
\Pr(W< 0, H<0)= p_3\\
\end{cases}
$$


*the marginal distribution of each of $W,H,Q$ are symmetric around 0.


*$Q=W-H$.

Note that the map from $(X,Y,Z)$ to $(W,H,Q)$ does not need to be deterministic. For instance, it could be that $W=X+\epsilon$ where $\epsilon$ is another well defined random variable.

SOME DISCUSSION ON THE 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 1-3. However, constraint 3 is computationally intractable to implement because   infinite-dimensional. Much simpler is to verify   the existence of a 3-d distribution function that satisfies constraints 1-2 and, then, use the construction I'm investigating about (if it exists!) to conclude about the existence of the originally desired distribution.

ATTEMPTED ANSWER (with questions):
Let $(X,Y,Z)$ and $(W,H,Q)$ be defined on the same probability space $(\Omega,\mathcal{F}, \Pr)$.
Define $(W,H,Q)$ as follows:

*

*For each $\omega \in \Omega$ such that $X(\omega)\geq 0$ and $Z(\omega)\geq 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(2X(\omega), X(\omega), X(\omega)\big)$$

*

*For each $\omega \in \Omega$ such that $Y(\omega)\geq 0$ and $Z(\omega)< 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(Y(\omega), 2Y(\omega), -Y(\omega)\big)$$

*

*Complete the definition of $(W,H,Q)$ with negative values for $W$ and $H$ in such a way that the marginals are symmetric around zero and that $\Pr(W<0, H<0)=p_3$. Hence, we will need to have: $-2X, -Y$ for $W$; $-X, -2X$ for $H$; $-X, Y$ for $Q$.  I'm not sure we can always do this, though. Can we?

 A: I will begin with a reformulation of your question which makes it not only more symmetric, by also (at least for me) more natural and interesting. I will pass from your variables $(W,H,Q)$ to new variables $(X,Y,Z)$ (which are not your original $X,Y,Z$) by putting $X=W, Y=-Q, Z=-H$. Then your condition (3) becomes just $X+Y+Z=0$. Therefore you are asking about a probability distribution $\pi$ on the plane
$$\mathcal P=\{x+y+z=0\}\subset\mathbb R^3$$ such that all its one-dimensional marginals are symmetric, and you are interested in the possible values of
$$
\begin{aligned}
p_1=\pi\{x>0, y<0\} \;, \\
p_2=\pi\{y>0, z<0\} \;, \\
p_3=\pi\{z>0,x<0\} \;.
\end{aligned}
$$
Although it's not completely clear to me what is the continuity of random vectors you are referring to, I will presume that it implies that the probability $\pi$ of the line $\ell_x=\mathcal P \cap\{x=0\}$ and of the analogous lines $\ell_y,\ell_z\subset\mathcal P$ are all 0, so that replacing non-strict inequalities with strict ones doesn't change anything (the general case can also be treated in pretty much the same way). For a better visualization one can draw a picture of the plane $\mathcal P$ and the lines $\ell_x,\ell_y,\ell_z$ in the coordinates $x,y$ (so that $z=-x-y$).
Since all the domains from the definitions of the numbers $p_i$ are pairwise disjoint, and the complement of their union consists of the zero measure lines $\ell_x,\ell_y,\ell_z$, indeed $\sum p_1=1$. Moreover, the symmetry condition on the marginals means that all numbers $p_i$ satisfy the conditions $p_i\le \frac12$ (and, of course $p_i\ge 0$).
I claim that, conversely, any collection $(p_i)$ like this can be realized by a measure $\pi$ satisfying the above conditions. Indeed, take for $\pi$ the distribution with two atoms at the points $(1,-2,1)$ and $(-1,2,-1)$ and the weight $\frac12$. Then the corresponding $p$-vector $(p_1,p_2,p_3)$ is
$$
v_3 = \left(\tfrac12, \tfrac12, 0\right) \;.
$$
If you want to have an absolutely continuous measure $\pi$ with the same property, you can just smoothen the above distribution by taking the normalized sum of the uniform distributions on, say, side $\varepsilon$ squares centered at $(1,-2)$ and $(-1,2)$ in the coordinates $(x,y)$.
By symmetry, there are also measures $\pi$ with the $p$-vectors $v_2=\left(\tfrac12, 0, \tfrac12\right)$ and $v_1=\left(0, \tfrac12, \tfrac12\right)$. By taking convex combinations of the associated measures $\pi$ one can realize any $p$-vector from the convex hull of $v_1,v_2,v_3$, which is precisely the set determined by the above conditions on $p_i$ (for visualization one can draw a picture in the coordinates $p_1,p_2,1-p_1-p_2$).
