In a tutorial exercise I am asked to determine:
$\mathbb P(X_1 < X_2)$, given that $X_1$ and $X_2$ are both independently exponentially distributed with the same parameter $\lambda$.
I would be inclined to just write $\mathbb P(X_1-X_2 <0) = \mathbb P(Y<0)$, where $Y=X_1-X_2$, but would this not simply be $0$?
What would change if they would each come with their own parameters $\lambda_1, \lambda_2$?
What I know: $$\mathbb P(X_1 \geq x)= \lambda e^{-\lambda x}.$$ The hint in the books says to condition on the value of $X_2$, I'm not completely following what is meant.
When they have the same distribution, $P(X_1\lt X_2)=P(X_2\lt X_1)=\frac{1}{2}$. Because the distributions are continuous, $P(X_1=X_2)=0$