$$
f(x,y) =
\begin{cases}
e^{-x-y}, & \text{for x>0 and y>0} \\
0, & \text{elsewhere}
\end{cases}
$$
I found this joint probability density by solving a previous problem that gave me the joint distribution function of
$
F(x,y) =
\begin{cases}
1-e^{-x}-e^{-y}+e^{-x-y}, & \text{for x>0 and y>0} \\
0, & \text{elsewhere}
\end{cases}
$
Okay so I took the double integral of the joint probability density to find P(X+Y>3)
$$\int_3^{\infty} \int_3^{\infty} e^{-x-y} dxdy$$
I get $e^{-6}$ as my answer however the book gets $(e^{-2}-e^{-4})^{2}$
Can someone explain what I'm doing wrong?
Best Answer
$$ \Bbb P (X+Y>3)=\Bbb P (Y>3-X)=1-\Bbb P(Y<3-X) \\=1-\int_0^{3}\int_0^{3-y}e^{-x-y}dx~dy=4e^{-3} $$
We are looking for the region to the right of the triangle enclosed between the x axis, y axis and the line, which should be the same as the converse of the region to the left, im not sure why your answer is like that, unless there is some other piece of information that is missing from the question, or maybe I've made a mistake.