# From trivariate cdf to the distribution of differences of random variables

Consider a trivariate cumulative distribution function (cdf) $$G$$.

• Is there a collection of necessary conditions on $$G$$ ensuring that $$\exists \text{ a random vector (X_1,X_2) such that (X_1, X_2, X_1-X_2) has cdf G}$$ ?

• Is there a collection of necessary and sufficient conditions on $$G$$ ensuring that $$\exists \text{ a random vector (X_1,X_2) such that (X_1, X_2, X_1-X_2) has cdf G}$$ ?

Update I: Let $$P$$ be the probability distribution associated with $$G$$. We can claim that: 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 $$\int_{(a,b,c)\in \mathbb{R}^3 \text{ s.t. } c=a-b} dP=1$$

• Is this condition also sufficient? I.e., can we claim that if $$\int_{(a,b,c)\in \mathbb{R}^3 \text{ s.t. } c=a-b} dP=1$$ then there exists a random vector $$(X_1,X_2)$$ such that $$(X_1, X_2, X_1-X_2)$$ has probability distribution $$P$$?

• Can we write $$\int_{(a,b,c)\in \mathbb{R}^3 \text{ s.t. } c=a-b} dP=1$$ by using the cdf $$G$$ ?

Update II:

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 $$c_1\leq a_1-b_2$$ $$\begin{cases} P([a_1,a_2], [b_1, b_2], [c_1, c_2])= P([a_1,a_2], [b_1, b_2], [a_1-b_2, c_2])\\ P([a_1,a_2], [b_1, b_2], [c_3, c_1])=0 & \forall c_3\leq c_1 \end{cases}$$ These implications can be written using $$G$$ (as I want!). However: are these implications also sufficient? I don't know how to prove or dis-prove it.

• You have supplied such conditions: to wit, when $(X_1,X_2,X_3)$ is a random variable with distribution $G,$ then almost surely $X_3=X_1-X_2.$ That's dead simple. What other form are you hoping to express these conditions in that would be any simpler or more useful? – whuber Nov 29 '18 at 15:17
• Thanks. I'm hoping for conditions directly imposed on $G$. – user3285148 Nov 29 '18 at 15:20

The condition in your first update is sufficient, because it implies that $$X_3=X_1-X_2$$ almost surely, so in particular $$(X_1,X_2,X_3)=^d(X_1,X_2,X_1-X_2)$$.
Edit: To be completely explicit about the dependence of $$G$$, this can be restated by requiring that $$P(B)=0$$ for any 3-dimensional box $$B$$ such that $$B\cap \{(x,y,z):z=x-y\}=\emptyset$$.
• Thanks. What about update II? Remember that I'm looking for conditions that can expressed using $G$. – user3285148 Dec 11 '18 at 18:36
• The measure $dP$ is defined in terms of $G$, so I'm not sure what else you want. – Mike Hawk Dec 11 '18 at 18:44
• I want necessary and sufficient conditions explicitly involving "boxes" in $\mathbb{R}^3$ – user3285148 Dec 11 '18 at 18:46
• Thanks. I think that "$P(B)=0$ for any 3-dimensional box $B$ such that $B\cap ...=\emptyset$" holds IF AND ONLY IF the bullet points in my update II hold. Is this correct? – user3285148 Dec 12 '18 at 11:19