Hello guys I have the following problem and cannot figure out the computation of $\mathbb E(I(X-Y)_{(0, +inf)} )$
Given $X∼Exp(λ)$ and $ Y∼Exp(θ)$.
Let $X$ represent the length of the red petals.
Let $Y$ represent the length of the yellow petals.
Finally let $p=Pr(Y>X)$ be the probability that on average the yellow petals are greater in length than red petals.
The task is to show that $p$ can obtained in function of $λ$ and $θ$.
I'm stuck here:
$\mathbb E(I(X-Y)_{(0, +inf)} )$
I know the indicator function is distributed as a Bernoulli distribution, but I cant figure out how to get to the expression of p in function of lambda and theta.
Any thoughts?
Assuming that $X$ and $Y$ are independent the value is $P(X>Y)=\int_0^{\infty}\int_y^{\infty} \lambda e^{-\lambda x}dx \theta e^{-\theta y} dy=\frac {\theta} {\lambda+\theta}$.
I have used the fact that $EI_A=P(A)$ where $A=(X-Y \in (0,\infty)$. Note that $A$ is same as the event $(X>Y)$