# Waiting time for two independent poisson processes

Order of Events in Poisson Processes

Assume that you have two independent Poisson process, $N_1(t)$ with rate $\lambda_1$ and $N_2(t)$ with rate $\lambda_2$. The probability that $n$ events occur in the first process before $m$ events occur in the second process is $$\sum_{k = n}^{n+m-1}\binom{n+m-1}{k}\left(\frac{\lambda_1}{\lambda_1+\lambda_2}\right)^k\left(\frac{\lambda_2}{\lambda_1+\lambda_2}\right)^{n+m-1-k}.$$

Source.

I'm trying to figure out what the expected amount of time would be for at least one event to happen for both processes. I know that the waiting time for Poisson Process in general has an exponential distribution, but I'm not sure how to find the waiting time for two distributions.

## 1 Answer

Let the inter arrival times be represented by $S\sim\mathcal{Exp}(\lambda)$ and $T\sim\mathcal{Exp}(\mu)$

The probability that at least one of each Poisson process occurs during some interval of length $t$ is:

\begin{align}\mathsf P(\max(S,T)\leq t) & = \mathsf P(S\leq t)~\mathsf P(T\leq t) \\[1ex] & = (1-\mathsf e^{-\lambda t})(1-\mathsf e^{-\mu t}) \end{align}