I'm a bit stuck trying to make sense of an exam review problem in my probability course. The problem is this:
Consider the following joint density function: $$ f_{X,Y}(x,y) = \cases{ e^{-y} & $ 0<x<y<\infty$ \\ 0 & otherwise } $$ Find $P(X+Y\ge 1)$
Here is my attempt:
\begin{align} P(X+Y\ge 1) & = P(Y\ge 1-X) \\ & = \int_0^\infty \int_0^{1-x} e^{-y} \ \mathrm dy \ \mathrm dx \\ \end{align}
But, this integral does not converge. So, I decided to look at the solution to figure out why my initial approach was incorrect. The solution states:
\begin{align} P(X+Y\ge 1) & = P(Y\ge 1-X) \\ & = \int_0^{1/2} \int_{1-x}^\infty e^{-y} \ \mathrm dy \ \mathrm dx + \int_{1/2}^\infty \int_{x}^\infty e^{-y} \ \mathrm dy \ \mathrm dx \\ & = \frac{\sqrt e - 1}{e} + \frac{1}{\sqrt e} \end{align}
Apparently, we must introduce the line $y=x$ here, and I'm not sure why this was introduced into the problem. My confusion with these problems is setting up the bounds on the integrals, so any advice for this and future problems would be appreciated.
Any help would be appreciated, thank you.
If we examine the support carefully, we can see that we need $x< y$, hence the result.
$$ f_{X,Y}(x,y) = \cases{ e^{-y} & $ 0<\color{red}{x<y}<\infty$ \\ 0 & otherwise } $$
\begin{align} P(Y \ge 1-X) &= \int_0^\infty \int_{\max(x,1-x)}^\infty \exp(-y) \,\, dy dx \\ &= \int_0^\frac12 \int_{1-x}^\infty \exp(-y) \, \, dydx + \int_\frac12^\infty \int_{x}^\infty \exp(-y) \, \, dydx \end{align}