Poisson processes

**Moo** · May 10th 2010, 03:15 PM

Hi,

Next time I promise I won't learn a new chapter just the day before the exam

We have two Poisson processes $N_1(t)$ and $N_2(t)$ which are independent (with parameters $\lambda,\mu$ )

Then $N(t)=N_1(t)+N_2(t)$ is a Poisson process with parameter $\lambda+\mu$ . Okay with that.

And it follows (how ??) that $N_2(t)-N_1(t)$ have the same distribution as $\sum_{i=1}^{N(t)} \epsilon_i$ , where $P(\epsilon_i=1)=\frac{\lambda}{\lambda+\mu}$ and $P(\epsilon_i=-1)=\frac{\mu}{\lambda+\mu}$

I don't understand at all where the last part comes from

Thanks for any help

**bcvaidyanath** · May 16th 2010, 11:34 AM

If you got how the step came, please post it. I have my Random process exam tom, it will be really helpful.

Thanks in advance

Thread: Poisson processes

LinkBack

Thread Tools

Display