The joint probability density function of the random variables $X$ and $Y$ is given by $$f_{X,Y}(x,y) = \begin{cases} e^{-x-y} & \text{for}\,\,(x,y)\in(0,+\infty)^{2}\\ 0 & \text{otherwise} \end{cases} $$
(a) Determine $\textbf{P}(X < Y)$
(b) $X$ and $Y$ are independent?
MY SOLUTION
(a) According to the support of the given joint probability density function, we have \begin{align*} \textbf{P}(X < Y) = \int_{0}^{\infty}\int_{0}^{y}e^{-x-y}\mathrm{d}x\mathrm{d}y = \int_{0}^{\infty}(e^{-y} - e^{-2y})\mathrm{d}y = \frac{1}{2} \end{align*}
(b) They are independent since $f_{X,Y}(x,y) = e^{-x}e^{-y} = f_{X}(x)f_{Y}(y)$. Indeed, we have \begin{align*} \begin{cases} \displaystyle f_{X}(x) = \int_{0}^{\infty}e^{-x-y}\mathrm{d}y = e^{-x}\int_{0}^{\infty}e^{-y}\mathrm{d}y = e^{-x}\\\\ \displaystyle f_{Y}(y) = \int_{0}^{\infty}e^{-x-y}\mathrm{d}x = e^{-y}\int_{0}^{\infty}e^{-x}\mathrm{d}x = e^{-y} \end{cases} \end{align*}