Computing probability distribution function for uniform random variables and Y=1-X X is uniform random variable in [0,1] and Y=1-X. How do I calculate the distribution function F(X,Y)? I can see that Y is also uniformly distributed and can draw the intervals. But I am unable to compute the distribution function when x+y>1 & x,y in [0,1]. How do I compute it for this specific case?
 A: $F(x,y) = P(X \le x, Y \le y)$
Using the fact that $Y= 1-X$ and after simplifying, the above can be re-written as:
$F(x,y) = P(1-y \le X \le x)$
Since, $X$ is a uniform between 0 and 1, it follows that:
$F(x,y) = x+y-1$
Update
The above is a bit sloppy as I did not specify the domain of the cdf where it is non-zero appropriately. Consider:
$F(x,y) = P(1-y \le X \le x)$
Thus,
$F(x,y) = x+y-1$
Two points about the above cdf:

*

*For the above cdf to be non-zero it must be the case that:
$x > 1-y$


*Also, note that the cdf attains the value of 1 when $x+y=2$.
Thus, the correct way to represent the cdf is:
$F(x,y) = 0$ if $x+y \le 1$
$F(x,y) = x+y-1$ if $1 \le x+y \le 2$
$F(x,y) = 1$ otherwise
A: I'm interpreting your question as concerning the cdf, which by definition is a function $F$ for which
$$F(x,y) - F(x,0) - F(0,y) + F(0,0) = \Pr(X \le x, Y \le y) \text{;}$$
$$F(0,0) = 0.$$
For $x + y \lt 1$, the right hand side is zero and the left hand side becomes a statement that $F$ is a bilinear function, implying (in conjunction with some of the initial values specified below) that the graph of $F$ is part of a plane.  For $x + y \gt 1$, the assumption of uniform distributions implies $F$ must be increasing at a unit rate in both $x$ and $y$, whence the graph of $F$ in this region is a part of a plane of the form $x + y = \text{constant}$.  From the evident restrictions
$$F(x, 0) = F(0, y) = 0,$$
$$0 \le x  \le 1, 0  \le y  \le 1,$$
it is geometrically obvious that the first piece of the graph must lie in the xy plane and the second piece must intersect the first along the line segment $x + y = 1, 0  \le x  \le 1$.  The full solution therefore is
$$F(x,y) = 0 \quad \text{if} \quad x + y  \le 1$$
$$F(x,y) = x + y - 1 \quad \text{if} \quad x + y \gt 1.$$

This result is the Fréchet–Hoeffding minimum copula $W(x,y)$.  Generally, a copula expresses a multivariate distribution after the marginal variables have been subjected to a probability integral transformation; that is, the marginals have been made uniform.  All 2D copulas must have values between this minimum $W$ and the maximum copula $M(x,y) = \min(x,y)$.  $W$ expresses maximum anticorrelation between the variables while $M$ expresses maximum correlation between them.  Follow the link (a Wikipedia article) for Mathematica plots of these copulas.
