Probability that $\operatorname{Erlang}(2,\mu _2)$ is greater than $\exp(\mu _1)$

Question

I'm trying to work out that, given that $X\sim \exp(\mu_1)$ and $Y\sim \operatorname{Erlang}(2,\mu _2)$, what is $\mathbb{P}(X<Y)$? So far I have: $$\mathbb{P}(X<Y)=\int_0^{\infty}\mathbb{P}(X<t \mid Y\in dt)\mathbb{P}(Y \in dt)$$ $$=\int_0^{\infty}(1-e^{-\mu _1t}){\mu _2}^2te^{-\mu _2t}dt$$ I can do this integral, but it looks like it will be a great big mess. Am I on the right track?

Answer 1

The integral isn't so hard to compute:

\begin{align} \int_0^{\infty}(1-e^{-\mu _1t})\mu_2(\mu_2 t)e^{-\mu _2t}\, \mathsf dt &= \mu_2\int_0^\infty t\mu_2 e^{-\mu_2 t}\,\mathsf dt - \frac{\mu_2^2}{\mu_1+\mu_2}\int_0^\infty (\mu_1+\mu_2)te^{-(\mu_1+\mu_2)t}\,\mathsf dt\\ &=\mu_2\cdot\frac1{\mu_2} -\frac{\mu_2^2}{(\mu_1+\mu_2)^2}\\ &= \frac{(\mu_1+\mu_2)^2-\mu_2^2}{(\mu_1+\mu_2)^2}\\ &= \frac{\mu_1^2+2\mu_1\mu_2}{(\mu_1+\mu_2)^2}. \end{align}