Find the distribution function of bivariate distribution Find the distribution function of 
$$f_{X,Y}(x,y)=\begin{cases}
e^{-y},  & \text{if $0< x<y < \infty$} \\
0, & \text{ otherwise}
\end{cases}$$
Trial : According to my calculation 
$$F_{X,Y}(x,y)=\int_{0}^{y}\int_{0}^{x} e^{-y} dx dy = x(1-e^{-y})$$
But this is wrong. Since $F(\infty, \infty) = \infty \neq 1$.
So, What will be correct DF. Thanks in advance.
 A: You have the limits of integration wrong. You should have:
$$
F_{X,Y}(x,y)=\int_{v=0}^{y}\int_{u=0}^{x} f(u,v)\; du\; dv
$$
But your density is zero when $u\ge v$, so your integral becomes:
$$
F_{X,Y}(x,y)=\int_{v=0}^{y}\int_{u=0}^{\mbox{min}(x,v)} e^{-v}\; du\; dv
$$
So:
$$
F_{X,Y}(\infty,\infty)=\int_{v=0}^{\infty}\int_{u=0}^{v} e^{-v}\; du\; dv=\int_{v=0}^{\infty}ve^{-v}\;dv=1
$$
A: Good attempt, but you have to consider your support (i.e., the region of integration) for this question.

The shaded region is the region in which $0 < x < y$. Now consider the following example:

The shaded region is
$$\mathbb{P}\left(X \leq 2, Y \leq 4\right) = \int_{0}^{2}\int_{x}^{4}f_{X,Y}(x,y)\text{ d}y\text{ d}x\text{.}$$
You can generalize this to find that
$$F_{X,Y}(x,y) = \int_{0}^{y}\int_{s}^{x}f_{X,Y}(s, t)\text{ d}t\text{ d}s = \int_{0}^{y}\int_{s}^{x}e^{-t}\text{ d}t\text{ d}s\text{.}$$
It's quite late over here, but I hope my integration is right:
$$\int_{0}^{y}\int_{s}^{x}e^{-t}\text{ d}t\text{ d}s = \int_{0}^{y}\left(e^{-s}-e^{-x}\right)\text{ d}s = 1 - e^{-y}-ye^{-x}\text{, } 0 < x < y < \infty\text{.}$$
Now defining $F_{X,Y}(x,y) = 1 - e^{-y}-ye^{-x}$, it can easily be verified that $F_{X,Y}(0,0) = 0$ and $F_{X,Y}(\infty, \infty) = 1$.
Hopefully my calculus is right.
