Cfrgtkky

Consider a trivariate probability distribution $P: mathbb{R}^3rightarrow [0,1]$. 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 $a_2geq b_2+c_2$
  $$
  begin{cases}
  P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1, b_2+c_2], [b_1, b_2], [c_1, c_2])\
  P([a_2, a_3], [b_1, b_2], [c_1, c_2])= 0 & forall a_3geq a_2\
  end{cases}
  $$




• If $b_1leq a_1-c_2$
  $$
  begin{cases}
  P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [a_1-c_2, b_2], [c_1, c_2])\
  P([a_1,a_2], [b_3, b_1], [c_1, c_2])=0 & forall b_3leq b_1\
  end{cases}
  $$




• 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 $b_2geq a_2-c_1$
  $$
  begin{cases}
  P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, a_2-c_1], [c_1, c_2])\
  P([a_1,a_2], [b_2, b_3], [c_1, c_2])=0 & forall b_3geq b_2
  end{cases}
  $$




• If $c_2 geq a_2-b_1$
  $$
  begin{cases}
  P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [c_1, a_2-b_1])\
  P([a_1,a_2], [b_1, b_2], [c_2, c_3])=0 & forall c_3geq c_2
  end{cases}
  $$




• If $c_1leq a_1-b_2$
  $$
  begin{cases}
  P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [a_1-b_2, c_2])\
  P([a_1,a_2], [b_1, b_2], [c_3, c_1])=0 & forall c_3leq c_1
  end{cases}
  $$

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.

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 $S={(a,b,c)inmathbb R^3 colon a=b+c}$ 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 $Yle b$ 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 $(X>b+c, Yle b, Z le c)$ 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 ${(a,b,c)inmathbb R^3 colon a > b+c}$, 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 $(Xle b+c, Y > b, Z > c)$, 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... :)

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 $Ssubset mathbb R^3$ given by
$$S={(a,b,c)inmathbb R^3 colon c=a-b},$$
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.