Double integral for exponential function

Question

I am trying to find the cdf of this function $f(x,y)=e^{-y}$ for $0<x<y<\infty$.

My attempt: $$F(x,y)=\int_0^x\int_u^x e^{-v}dvdu$$

Am I on the right path? Can someone help me with how to go about it?

Answer 1

There is a typo in one of the upper limits (there should be a dependency on $y$ somewhere). If $y < x$, we can project the point $(x, y)$ onto the boundary of the sector $0 < x < y$: $$\operatorname P(X < x, Y < y) = \int_0^{\min(x, y)} \int_u^y e^{-v} dv du \,[x > 0 \land y > 0] = \\ (1 - e^{-\min(x, y)} - \min(x, y) e^{-y}) \,[x > 0 \land y > 0].$$