region of integration $X+Y>2$ of a probability im not sure why this probability is giving me ∞.
$ e^{-(x+y)} $ for $x>0$ and $y>0$, and you have to find $P(X+Y>2)$ This is what i did:
$$ \int_2^∞ \int_{2-x}^∞ e^{-(x+y)}  \,dy\,dx$$
THANK YOU
 A: Draw a picture that includes the line $x+y=2$. You will see that the integral needs to be broken up as
$$\int_{x=0}^2 \int_{y=2-x}^\infty e^{-(x+y)}\,dy\,dx +\int_{x=2}^\infty \int_{y=0}^\infty e^{-(x+y)}\,dy\,dx.$$
If you want to avoid a split integral, first calculate $\Pr(X+Y\le 2)$. We have
$$\Pr(X+Y\le 2)=\int_{x=0}^2 \int_{y=0}^{2-x} e^{-(x+y)}\,dy\,dx.$$
We could alternately in each case integrate first with respect to $x$. By symmetry this just means interchanging the roles of $x$ and $y$ everywhere.
A: The integral should be over the region $x>0$, $y>0$, and $x+y>2$ (equivalently, $y>2-x$). That region is the first quadrant with a small triangle near the origin chopped off. The outer integral should have $0$ to $\infty$ as its limits. You need to split the inner integral into two integrals, one for $0<x<2$, and one for $x \geq 2$. This is because if $0<x<2$, then you have a constraint on what $y$ can be in terms of $x$ (in particular, it must be greater than $2-x$). However, if $x \geq 2$, then $2-x$ is negative (or non-positive), so it should not be the lower limit of the inner integral; instead, $0$ should be the lower limit.
I can expand if you want more explanation.
