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I have two poisson processes: $N_t$ with rate $\lambda$ and $M_t$ with rate $\mu$. I have to calculate the average number of arrivals for $N_t$ before the first arrival of $M_t$.

This is my reasoning: The average time for the first arrival of $M_t$ is $\mu$, and the average time between arrivals of $N_t$ is $\lambda$, independently of the moment I start to measure the time and counting the arrivals. So in a interval $[0, \mu]$ the average number of arrivals for $N_t$ is $\lambda\mu$ because for an arbitrary $t$ the average number of arrival for $N_t$ is $\lambda t$.

¿Am I right?

Edit: I applied the same reasoning to calculate the average arrivals for $M_t$ before the first arival form $N_t$ and it gives $\lambda\mu$ too, there's something wrong here...

Another edit: facepalm the average time between arrivals for $M_t$ is $\frac{1}{\mu}$, so the number of arrivals for $N_t$ in the interval $[0, \frac{1}{\mu}]$ would be $\frac{\lambda}{\mu}$.

On the other hand, the average number of arrivals for $M_t$ before the first arrival of $N_t$ would be $\frac{\mu}{\lambda}$

I think this looks better now.

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Just get

$$E[N(\tau_1)]=\sum_{n=0}^\infty nP(N(\tau_1)=n)=\sum_{n=0}^\infty n\int_0^\infty P(N(t)=n)\mu e^{-t\mu}dt=\frac{\lambda}{\mu}$$

Where $\tau_1$ is the first arrival time of $M(t)$.

Your answer is correct.

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  • $\begingroup$ Love the use of the density function of the interarrival time for $M_t$ :D $\endgroup$ – Broken_Window Dec 7 '16 at 14:36

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