If $X\sim \exp(\lambda)$ and $Y\sim \exp(\mu)$ then $P(X\leq Y)=\frac{\lambda}{\lambda+\mu}$. Is there an intuitive interpretation for this fact? I can verify this via double integrals, but I'm wondering if this can be put in the context of a Poisson process or something to give it an obvious meaning. I can't think of exactly how it would work.
 A: Here is a possible intuition:
Suppose you have two poisson processes, one of parameter $\lambda$ (meaning that time within arrivals in $exp(\lambda)$) and another one of parameter $\mu$. The parameter of a Poisson process is also called the "intensity" of the process since exponentials with higher parameters have lower means. Now suppose you combine the processes, that is, you don't distinguish between the $X$ type and the $Y$ type. Then the waiting time is $\min(X,Y)$. It is well known that the minimum of two exponentials is distributed as an exponential of parameter the sum of their parameters, that is, $\lambda + \mu$. So the combined process is a Poisson process of intensity $\lambda+\mu$.
So what the probability $P(X\leq Y)=\lambda/(\lambda+\mu)$ is telling you is how much does the intensity of the $X$ type process contributes to the intensity of the whole process. For instance, if that probability is .6 you can say that  $60\%$ of the intensity is due to the arrival of $X$-types.
A: $\mathbb P(X\leq Y)$ is the probability that the first arrival in the Poisson process corresponding to $X$ (call it “$X$-process”) happens before the first arrival in the Poisson process corresponding to $Y$ (call it “$Y$-process”). For a given value of $X$, this probability is equivalent to no arrivals in the $Y$-process in the time interval $[0,X]$, because the first arrival in the $Y$-process must occur later than in the $X$-process if it is to be the case that $X\leq Y$. In turn, the number of arrivals in the $Y$-process in the time interval $[0,X]$ follows a Poisson distribution with mean $\mu X$. Therefore, the probability that the $Y$-process has no arrivals in the time interval $[0,X]$ $$\frac{(\mu X)^0}{0!}\exp(-\mu X)=\exp(-\mu X),$$
conditional on $X$. Now, we know that $X$, the time of the first arrival in the $X$-process, is exponentially distributed, so that the unconditional probability is $$\int_0^{\infty}\exp(-\mu x)\,f_X(x)\,\mathrm d x=\int_0^{\infty}\exp(-\mu x)\,\lambda \exp(-\lambda x)\,\mathrm d x=\frac{\lambda}{\lambda+\mu}.$$
A: Recall that the combination of two independent Poisson processes of intensities λ and μ is a Poisson process of intensity λ+μ. The construction going in the other direction might be even more illuminating: start from a single Poisson process with intensity λ+μ, independently color each event in red or blue with probability λ/(λ+μ) and μ/(λ+μ). Then the red events are Poisson λ, the blue events are Poisson μ, and the red and blue processes are independent. This decomposition procedure extends to more than two colors and is called thinning.
