Q:

How to compute $P(X>Y)$ if $X$ and $Y$ are independent and exponentially distributed

$X$ and $Y$ are independent and exponentially ditributed with parameters $a,b$, respectively. Calculate $P(X>Y)$
1)$P(X>Y)=1-F_{X-Y}(0)$. Then I try to calculate density  of $X-Y$ from characteristic functions but I cannot deduce distribution of $X-Y$ from this.
2)ok I can also calculate density of $Z=X-Y$ from transformation of RV formula and then marginalize
3) Is there any simpler way to calculate $P(X>Y)$ than 2)?

A:

Given that the random variables  $X$ and $Y$ are independent. Further, $X\sim Exp(a)$ and $Y\sim Exp(b)$. We want the probability of the event $\{X>Y\}$. The random variable $X$ takes values larger than those of the  random variable $Y$. Now,
\begin{eqnarray*}
P\{X>Y\} &=& \int_{0}^{\infty}\int_{y}^{\infty} (a\cdot e^{-ax})(b\cdot e^{-by}) dx\cdot dy\\
&=& \int_{0}^{\infty} b\cdot e^{-by}\left( \int_{y}^{\infty} a\cdot e^{-ax} dx\right)\cdot dy\\
&=&  \int_{0}^{\infty} b\cdot e^{-by}\left( 1-\int_{0}^{y} a\cdot e^{-ax} dx\right)\cdot dy\\
&=&  \int_{0}^{\infty} b\cdot e^{-by}\left( 1-(1-e^{-ay}) \right)\cdot dy\\
&=& \int_{0}^{\infty} b\cdot e^{-(a+b)y} dy\\
&=&\dfrac{b}{a+b}\int_{0}^{\infty} (a+b)\cdot e^{-(a+b)y} dy\\
&=&\dfrac{b}{a+b}
\end{eqnarray*}

