I see that a similar question was asked last year, but I am still confused.
I have $f(x,y) = 2e^{-x-y}$, $ 0 < x < y < \infty $ and need to find the joint CDF.
I have a solution that separates the CDF into 3 cases:
-
For $x<0$ or $y<0,$ $F(x,y) = 0.$ This one is obvious.
-
For $0<y\le x$, $F(x,y) = \ldots$
I managed to solve it with bounds $0<x<y$ and $0<y<x$. So isn't this the case $y=x$? From what we are given, $y$ can never be equal to $x$. I don't understand how we got this case.
- For $0<x<y$, $F(x,y) = \ldots$
I can't figure out what the bounds should be. I tried $0<x<y$ and $x<y<\infty$ but it's not working out. Also, isn't "$0<x<y$" true for the entire range? I don't understand how we got this case either.
Can someone please explain why we have the second and third case(why do they need to be separate?), and how to determine what the bounds should be?
Best Answer
We always have $X<Y$. So if $x > y$ then
$$F(x,y)-F(y,y)=\int_y^x \int_{-\infty}^y f(u,v) du dv = \int_y^x \int_{-\infty}^y 0 du dv = 0.$$
In other words, if $x>y$ then $F(x,y)=F(y,y)$. So that reduces the problem to case 3. In case 3 you just have to calculate the integral:
$$\int_{-\infty}^y \int_{-\infty}^{\min \{ v,x \}} 2 e^{-u-v} du dv.$$
I got the limits here by rephrasing the region of integration defined as $\{ (u,v) : u \leq x, \, v \leq y, \, u \leq v \}$