Say I have two distributions $P_A(t), P_B(t)$ representing the time intervals between events which occur randomly. For instance, if $P_A(t)$ corresponds to a Poisson process, then it is a negative exponential.
If I start the clock at $t=0$ and start recording events, what is the probability $a$ that an event of type A occurs before an event of type B?
Intuitively I would expect this to be $$\frac{\mu_A}{ \mu_A + \mu_B}$$ (where $\mu_i$ is $1/$mean of $P_i(t)$ for $i=A,B$) but I'm not sure how to prove this. Any help would be appreciated.
EDIT: I assume $A$ and $B$ are independent processes.
I'm not $100\%$ sure that this is what you want, so please let me know if you meant something different in your question. We have for two Poisson processes: \begin{align} P(t_A<t_B) &= \int_0^\infty P(t_A < t_B|t_B = t)P(t_B = t)\;dt \\ &= \int_0^\infty(1-e^{-\mu_A t})\mu_Be^{-\mu_B t}\;dt\\ &= \left[\frac{\mu_B}{\mu_A+\mu_B}e^{-(\mu_A+\mu_B)t}-e^{-\mu_bt}\right]_0^\infty \\ &=1 - \frac{\mu_B}{\mu_A + \mu_B} \\ &= \frac{\mu_A}{\mu_B + \mu_A} \end{align}