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.

  • $\begingroup$ Begin by simplifying the problem. For instance, by replacing $(X,Y,Z)$ by $(X+a,Y+b,Z+a-b)$ you may assume, with no loss of generality, that $a=b=a-b=0,$ thereby eliminating these superfluous variables. It would also help to motivate this question by explaining how it arose and what statistical application or interest it might have. Otherwise, its complexity (and the unusual, abstract notation) will likely discourage most potential solvers from even reading it. $\endgroup$
    – whuber
    Nov 3, 2021 at 17:41
  • $\begingroup$ Here's one idea: ask the question you really need answered. That would appear to have nothing to do with $F$ and everything with $G.$ In other words, rather than requiring that $G$ be constructed from such an $F,$ wouldn't it suffice just to exhibit a single example of a $G$ satisfying your criteria? Another idea: aren't (7) and (8) just a way of requiring that $G$ be supported on the set where $h+q=w$? A third: use conventional mathematical notation for $G$ (which is simpler and clearer). That is, define $G(w,h,q)=\Pr(W\le w,H\le h,Q\le q)$ and state your question in those terms. $\endgroup$
    – whuber
    Nov 4, 2021 at 11:44
  • $\begingroup$ Still regarding your comment 3) on notation: I have slightly modified my notation along your lines (although without completely following your suggestions for the reasons mentioned above). $\endgroup$
    – user339651
    Nov 4, 2021 at 12:00
  • $\begingroup$ I view the introduction of $F$ as a complication, not a simplification. For instance, your criteria $(4)-(8)$ for $G$ are directly satisfied--if I read your criteria correctly--by a distribution assigning probability $f_1$ to $(w,h,q)=(2,1,1),$ probability $f_2$ to $(0,1,-1),$ and probability $f_3$ to $(-2,-1,-1).$ If this is the sort of thing you are looking for, it is easy to characterize all such distributions. $\endgroup$
    – whuber
    Nov 4, 2021 at 13:37
  • $\begingroup$ It's a probability mass function and it does satisfy criteria $(4)-(6),$ as is easy to check. $\endgroup$
    – whuber
    Nov 4, 2021 at 13:49

1 Answer 1


(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).

  • $\begingroup$ Thanks for your answer. I'm having troubles in understanding your notation and terminology, sorry for that. Could you expand a bit? For example, what do you mean by a truth table? What is the link between the truth table and your conclusion? $\endgroup$
    – user339651
    Nov 6, 2021 at 16:29
  • $\begingroup$ I interpreted the two zero probability statements as conditions that must always be false / their negation always true. The assertions $W\ge t$ is either true or false. Similarly, $H \ge u$ is either true or false. So listing the four possible combinations FF, FT, TF, and TT, I evaluated the implications for $Q \ge t-u$. Since I reduced the logical statements for (7) and (8) to $\cup$ ("or") statements, I could prove (7) and (8) row by row in one of three ways. See en.wikipedia.org/wiki/Truth_table $\endgroup$
    – krkeane
    Nov 6, 2021 at 16:43
  • $\begingroup$ The conditions have to hold for each u,t. Where can I see this in your answer? $\endgroup$
    – user339651
    Nov 6, 2021 at 16:55
  • $\begingroup$ I attempted to show (7) and (8) were satisfied for each possible outcome (the four rows). The condition met is noted after each Boolean value. In each row, I demonstrated (7) and (8) were satisfied in some way. The earlier matrices state W=X; H=Y; Q=X-Y. Let me know if anything else in my approach is unclear. $\endgroup$
    – krkeane
    Nov 6, 2021 at 16:56
  • $\begingroup$ "hold for each u,t" - I don't know what u or t will be thrown at me, but I know for instance $W \ge t$ will be either true or false. The implications of $W \ge t$ being false appear on the first two rows in the truth table, the implications of $H \ge u$ being false appear on rows one and three. ... All possible outcomes are covered. $\endgroup$
    – krkeane
    Nov 6, 2021 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.