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Given absolutely continuous dependent random variables $X$ and $Y$:
$X = B - A$
$Y = C - A$

$A, B,$ and $C$ are i.i.d. absolutely continuous random variables. $X$ and $Y$ are both include $A$, so they are dependent.

The following CDFs are known:
$F_X(x)$, $F_Y(y)$, $F_A(a)$, $F_B(b)$, $F_C(c)$

What is the joint cumulative distribution $F(x,y) = P(\{X \leqslant x\} \bigcap \{Y \leqslant y\})$ in terms of the known CDFs?

Ideally, I'd like to find a solution that avoids integration and utilizes $F_X(x)$, $F_Y(y)$ rather than $F_A(a)$, $F_B(b)$, $F_C(c)$.

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Your notation $F(x,y) = P(F_X(x) \bigcap F_Y(y))$ makes no sense since for given numbers $x$ and $y$, $F_X(x)$ and $F_Y(y)$ are numbers, not events or sets. What is the probability of the intersection of two numbers? What you mean is $$F(x,y) = P(X \leq x, Y \leq y) = P(\{X \leq x\} \cap \{Y \leq y\}).$$ In general, different joint distributions can result in the same marginal distributions, and so your ideal solution is not possible. However, it is possible to determine $F(x,y)$ using the relationship that you state and the common CDF of the iid random variables $A, B$, and $C$. –  Dilip Sarwate May 16 '12 at 10:43
    
@Dilip I've edited the question to correct the notation error that you pointed out. Thanks. Given that the relationship between X, Y and A, B, and C are known how can F(x,y) be determined? A solution that uses the CDFs for A, B, and C is OK, since these CDFs are known. –  Greg May 16 '12 at 15:55
    
An mechanical method is to compute the joint density function of $[Z,X,Y] = [A, B-A, C-A]$ which is an invertible linear transformation from $[A, B, C]$, and whose joint density $f_{Z,X,Y}(z,x,y)$ is thus given by a canned formula using Jacobians. Then, $$F_{X,Y}(a,b) = \int_{-\infty}^a \int_{-\infty}^b \int_{-\infty}^{\infty}f_{Z,X,Y}(z,x,y)\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$ –  Dilip Sarwate May 16 '12 at 16:01
    
@Dilip The equation makes sense, but I don't know how to calculate $f_{Z,X,Y}(z,x,y)$ given that Z, X, and Y are dependent. I'm guessing that the joint PDF can somehow be expressed as the product of the PDFs of independent variables A, B, and C, but I can't figure out a substitution that can do this. Also, I'm wondering if its possible to utilize the CDFs directly rather than differentiating the CDFs to determine the PDFs, and then re-integrating to get back to CDFs again. Thanks. –  Greg May 17 '12 at 5:29

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