Probability of independent exponential random variables I'm working on a problem from Klenke's book on probability theory:
I want to prove that if $X$ and $Y$ are independent exponentially distributed random variables with parameters $\theta$ and $\rho$ respectively, then
$$
\mathrm{P}\left(X < Y\right) = {\theta \over \theta + \rho}
$$
I'm having trouble proving this; I want to avoid using conditional probability because Klenke introduces this problem way before conditional probability. I'm guessing the densities of $X$ or $Y$ may come into play but I'm not sure how. 
Any suggestions ?.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}\int_{0}^{\infty}
\pars{\theta\expo{-\theta x}}\pars{\rho\expo{-\rho y}}\bracks{x < y}
\,\dd x\,\dd y =
\theta\rho\int_{0}^{\infty}\expo{-\rho y}\
\overbrace{\int_{0}^{y}
\expo{-\theta x}\,\dd x}^{\ds{\expo{-\theta y} - 1 \over -\theta}}\
\,\dd y
\\[5mm] = &\
\rho\int_{0}^{\infty}\bracks{\expo{-\rho y} - \expo{-\pars{\rho + \theta}y}}
\dd y =
1 - {\rho \over \rho + \theta} = \bbx{\theta \over \rho + \theta}
\end{align}
A: Hint
Use the fact that for every borel measurable set $B\subset \mathbb{R}^2$ we have: 
\begin{align}
\mathbb{P}((X,Y)\in B)  = \iint_B f_{X, Y} (x, y) \, dx\,dy
\end{align}
Where $f_{X, Y} (x, y)$ is the joint density of $(X, Y) $. But we know $X$ and $Y$ are independent, so $f_{X, Y} (x, y)=f_X(x) f_Y(y) $.  
You must choose $B$ such that:
\begin{align} 
(X, Y) \in B \Longleftrightarrow X<Y
\end{align} 
I'll leave the choice of $B$ and the integration that comes after that for you. 
A: \begin{align}
\Pr(X<Y) & = \int_0^\infty \left( \int_x^\infty \rho\theta e^{-\theta x} e^{-\rho y} \, dy \right) \,dx = \int_0^\infty \left( \theta e^{-\theta x} e^{-\rho x}  \right) \, dx \\[10pt]
& = \theta \int_0^\infty e^{-(\theta+\rho)x} \,dx = \frac\theta{\theta+\rho}.
\end{align}
This is a bit simpler than evaluating the iteratted integral in the opposite order.
