TITLE: How do I sum two Poisson processes?
QUESTION [8 upvotes]: If we have a Poisson Process $Y$ with intensity $\lambda$ and a Poisson Process $X$ with intensity $\mu$, where $X$ and $Y$ are two independent Poisson processes. How can I find the process $Z=Y+X$?

I know it should involve convolutions in some way but I dont know how to construct it!

REPLY [5 votes]: Hint: $$P(Z=z)=P(X+Y=z)=\sum_{x=0}^z f_X(x)f_Y(z-x)$$
So, we have 
$$
f_Z(z)=\sum_{x=0}^z e^{-\mu} \dfrac{\mu^x}{x!}e^{-\lambda} \cdot\dfrac{\lambda^{z-x}}{(z-x)!} \\ =e^{-(\lambda+\mu)}\sum_{x=0}^z \dfrac{\mu^x}{x!} \cdot\dfrac{\lambda^{z-x}}{(z-x)!}\\=e^{-(\lambda+\mu)}\cdot \dfrac{(\lambda+\mu)^z}{z!}[\text{Use of binomial formula}]
$$