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.


2 Answers 2


One necessary condition is that $P$ is a degenerate distribution, since the vector $(X_1,X_2,X_1-X_2)$ takes values only on the surface $S\subset \mathbb R^3$ given by $$S=\{(a,b,c)\in\mathbb R^3 \colon c=a-b\},$$ which is a two dimensional subspace of $\mathbb R^3$ (a plane through the $\vec 0$).

If the probability is not concentrated on this plane, that is if $$\int_S dP<1,$$ then $P$ cannot be the probability distribution of such a random vector.

  • $\begingroup$ Thanks. Is there a way to the requirement "$P$ is a degenerate distributions" as a mathematical constraint on $P$ (equality or inequality)? Let me explain: I have an algorithm looking for a $P$ lying in a certain function space. I would like to add some conditions necessary (or in an even better scenario NECESSARY AND SUFFICIENT) for $P$ being the prob distribution of a random vector of the type $(X,Y, X-Y)$ $\endgroup$
    – Star
    Commented Nov 30, 2018 at 17:26
  • $\begingroup$ Is it a discrete random vector? $\endgroup$ Commented Nov 30, 2018 at 17:28
  • $\begingroup$ No, it is a continuous random vector $\endgroup$
    – Star
    Commented Nov 30, 2018 at 17:28
  • $\begingroup$ And the study of the distribution is based on the probability density function? $\endgroup$ Commented Nov 30, 2018 at 17:30
  • 1
    $\begingroup$ Ok, that's fine, since if it had a density on $\mathbb R^3$ it wouldn't be degenerate. I'll think a little about it. $\endgroup$ Commented Nov 30, 2018 at 17:31

Edited to fix a bug, and expand. Still only a rough outline...

If you can work with PDF then I think the answer by @AlejandroNasifSalum is necessary and sufficient -- with possible exceptions having zero probability, not sure if you care.

For the rest, I assume you do not care about zero-probability events. If you do care about zero-probability violations, my guess is it will be very hard (impossible?) to exclude such violations using CDFs.

Anyway, let $F(a,b,c) = Prob(X \le a, Y \le b, Z \le c)$ be the CDF. Using the CDF and varying the $3$ inputs you can draw out "boxes" aligned along the $3$ axes, and any such box not intersecting $S=\{(a,b,c)\in\mathbb R^3 \colon a=b+c\}$ must have probability $0$. So you can easily come up one Necessary Condition:

[NC1] $\forall b, c: F(\infty,b,c) = F(b+c,b,c)$, because given $Y\le b$ and $Z \le c$ this restricts $X \le b+c$ and so increasing the range of $X$ beyond $b+c$ does not increase the probability. [I will use $\infty$ to denote positive infinity.]

Geometrically, [NC1] basically uses a box $B(b,c)$ defined by $(X>b+c, Y\le b, Z \le c)$ and is saying $P(B(b,c)) = F(\infty,b,c) - F(b+c,b,c)= 0$.

The question is whether by using enough boxes of this and similar forms, one can rigorously prove what you want, i.e., $X = Y + Z$ (with prob $1$).

The rest of this answer is speculative / a rough outline. I don't actually have enough rigorous probability/measure theory background for a rigorous proof. Where I have doubt I will write (?) to denote my doubt. Anyway, here are some geometric-inspired arguments.

First of all, observe that [NC1] stipulates $P(B(b,c))=0$ for all $b, c$. Therefore (?) the union of all such boxes $\bigcup_{b,c} B(b,c)$ also have $P(\bigcup_{b,c} B(b,c)) = 0$. The union $\bigcup_{b,c} B(b,c)$ is (?) in fact $\{(a,b,c)\in\mathbb R^3 \colon a > b+c\}$, i.e. the open region above the plane $S$.

So my idea is to similarly exclude the open region below the plane $S$. To do this, we will use a box $Q(b,c)$ of the form $(X\le b+c, Y > b, Z > c)$, i.e. the exact "opposite" of the format of $B$. By inclusion-exclusion, we can use CDF to write

  • $P(Q(b,c)) = F(b+c, \infty, \infty) - F(b+c, b, \infty) - F(b+c, \infty, c) + F(b+c, b, c)$.

This directly gives us another Necessary Condition:

[NC2] $\forall b, c: P(Q(b,c)) = 0$.

I think it is non-controversial that both [NC1] and [NC2] are necessary. Further, I think (?) together they are sufficient, but I am less sure about that, partly because I am less sure that $\bigcup_{b,c} Q(b,c)$ is (?) the open region below plane $S$. There may be some subtleties I am missing here with the equal signs...?

Anyway, this is the best I can do. :) If all the (?) above turn out to be OK, then the two conditions together would imply $P(S) = 1$, i.e. $X = Y+Z$ with probability $1$. But it will take a better-trained theoretician than me to verify all of the above... :)

  • $\begingroup$ By PDF you mean probability distribution function or probability density function? $\endgroup$
    – Star
    Commented Dec 10, 2018 at 10:20
  • $\begingroup$ Thanks. (1) How do you get the second bullet point? If $X\leq a, \text{ }Z\equiv X-Y\leq c$, then we have that $X\leq a, \text{ }Y \geq X-c$ which does not imply $Y\leq a-c$. $\endgroup$
    – Star
    Commented Dec 10, 2018 at 10:30
  • $\begingroup$ (2) I'm really looking for a formalisation of the arguments. Hence, can you explain better this part "etc. I'd guess that if you through in enough of these, the conditions together would be sufficient, but I'm not sure about that"? Which other conditions do you have in mind when you write "etc."? $\endgroup$
    – Star
    Commented Dec 10, 2018 at 10:32
  • $\begingroup$ Please see my update on the question. I still need more details, sorry. $\endgroup$
    – Star
    Commented Dec 11, 2018 at 17:00
  • $\begingroup$ @STF - you are absolutely right that my original 2nd bullet is wrong. Hopefully my new answer is more correct (?) or at least more intuitively useful. Good luck! $\endgroup$
    – antkam
    Commented Dec 11, 2018 at 22:53

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