# Distribution of time of Poisson process

Say we have two Poisson process $N_1$ and $N_2$ with parameter $\tau_1$ and $\tau_2$.

How can I derive the distribution of time of the 1st arrival? I know that for a single Poisson process $X(t)$ the the time of the first arrival, is exponential with parameter $\tau$ but what will be the scenario when there will be two process? Any idea?

• The sum of two independent Poisson processes $N_1(t)$ and $N_2(t)$ (with rates $\lambda_1$ and $\lambda_2$) is again Poisson with the sum of the rates. That is, $N_1(t) + N_2(t)$ is a Poisson process with rate $\lambda_1+\lambda_2$. That is not necessarily true without independence. – Michael Jul 31 '15 at 5:56

I will assume $N_1$ and $N_2$ are independent. We want to find the distribution of the minimum of two independent exponentially distributed random variables.
The min $W$ is $\gt w$ if both exponentials are greater than $w$. Thus if $w\gt 0$, then $\Pr(W\gt w)=e^{-\tau_1w}e^{-\tau_2 w}$. So the cdf of $W$, for $w\gt 0$, is $1-e^{-(\tau_1+\tau_2)w}$. We recognize this as the cdf of the exponential with parameter $\tau_1+\tau_2$.
Another way: The sum of the two independent Poisson random variables is Poisson with parameter $\tau_1+\tau_2$. So the time to first arrival is exponential, same parameter.