The joint probability function of $(X,Y)$ is given by:
$$f_{(X,Y)}(x,y) = e^{-x}$$
Which is defined for the values:
$$ 0 \le y\le x<\infty$$
$$0\text{ elsewhere}$$
How would I find the cumulative distribution function of $(X,Y)$?
I know that the area that I am integrating in is a infinite triangle(if drawn in a 2d plane) so I set up my integration as:
$$\int_0^\infty \int_y^\infty e^{-x}\,dx\,dy$$
After the inside integral is evaluated I get:
$$\int_0^\infty e^{-y}dy$$
Which then evaluates to 1.
But the answer is supposed to be:
$$ 0,\quad x<0 \quad \text{or} \quad \ y\ <0$$
$$1-e^{-y}-ye^{-x},\quad 0\le y\le x$$
$$1-e^{-x}-xe^{-x},\quad y>x\ge0$$
I have completely no idea how the answer came about and also why are these instances where y is greater than x even though the values specifically state that y is less than x?
Best Answer
A joint CDF $F_{X,Y}(x,y)$ gives the probability $$\Pr[(X \le x) \cap (Y \le y)].$$ Geometrically, what this means is that if you have a joint density $f_{X,Y}(x,y)$, then the CDF gives the total volume under the density over the region $(X \le x) \cap (Y \le y)$. That is to say, you are "cutting" the surface along $X = x$ and $Y = y$, and then discarding those pieces for which $X > x$ or $Y > y$. Here is a plot of the density:
Now you can see that if $x < 0$ or $y < 0$, then the point $(x,y)$ is in the L-shaped region to the left of the figure, and there is no volume in that region--the density is zero. That's the first part of the piecewise function in the answer. Now, if you're in the region $y > x > 0$, $(x,y)$ is in the flat triangular area just behind the curved wedge. But the rectangular region $(X \le x) \cap (Y \le y)$ for this point includes part of this wedge, but how much it includes does not depend on $y$ once $y$ is at least as large as $x$. That's the third part of the piecewise function: $$F_{X,Y}(x,y) = 1 - e^{-x} - xe^{-x}, \quad y > x > 0.$$ So $F(3,5) = F(3,10) = F(3,51147034)$. But if you choose a point inside the curved wedge; i.e., $0 < y < x$, then you can see that you're not only cutting away volume to the right, but also some volume to the back. So that's the second part of the piecewise CDF.
I won't go into more mathematical detail since I mainly wanted to give you a visual, intuitive explanation of what's going on. I find that this helps greatly when doing the actual computation.