Poisson processes

Moo

MHF Hall of Honor
Mar 2008
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2,802
P(I'm here)=1/3, P(I'm there)=t+1/3
Hi,

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

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

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

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

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

Thanks for any help
 
May 2010
6
1
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