$P(X_1<X_2)$ where $X_i$ is exponential and independent - not sure of my solution.

Question

We have two independent random variables $X_1$, $X_2$, with law $Exp(\rho_i)$ respectively. I want to find the probability of the following event $\{X_1<X_2\}$.

Is the following correct?

$P(X_1<X_2)=\int_0^\infty\int_0^{x_2=x_1}\rho_1\rho_2 e^{-(\rho_1x_1+\rho_2x_2)}dx_2dx_1=\int_0^\infty-\rho_1[e^{-(\rho_1x_1+\rho_2x_2)}]_0^{x_2=x_1}dx_1=-\rho_1[-\frac{1}{\rho_1+\rho_2}e^{-(\rho_1+\rho_2)x_1}+\frac{1}{\rho_1}e^{-\rho_1x_1}]^\infty_0=1-\frac{\rho_1}{\rho_1+\rho_2}$

Is this solution correct?

Answer 1

The answer is $\frac 1 2$ if $\rho_1=\rho_2$, not in general. ALlso you computed $P(X_2 <X_1)$ instead of $P(X_1<X_2)$. So the correct answer is $\frac {\rho_1} {\rho_1+\rho_2}$.