How to recognize that a Poisson Process that is split has two separate Poisson Process through viewing it as a filtration? I read in a paper that if we have a Poisson Process with rate $\lambda$, and that each arrival is separated into two separate categories, $A$ and $B$ with probability $p$ and $1-p$, then each of the two categories are themselves an independent Poisson Process with rate $\lambda p$ and $\lambda (1-p)$. However, the paper I read said that this is easily seen without proof as each of the two processes are just a filtration of the overall process. I am not sure why this is the case. Does anyone know?
 A: Comment: It seems you are asking for intuitive arguments. Here are three relevant ones:
1) Suppose a radioactive source emits particles into a counter according
to a Poisson process with rate $\lambda.$ Now a piece of lead foil is placed
between the source and the counter that randomly absorbs half of the
particles. Does it make sense to you that that the counter now observes a Poisson
process with rate $\frac{1}{2}\lambda\, ?$
2) Suppose  you have two small clay-like pieces of radioactive ore emitting
particles into a counter at rates $p\lambda$ and $(1-p)\lambda$ respectively.
Now you smash the two pieces into one. Does it make sense to you that you
now have one larger piece of ore emitting particles into the counter at
rate $\lambda\, ?$
3) In a certain city stroke patients enter the local hospital ER at a
Poisson rate $\lambda.$ When they arrive it turns out that stokes are randomly of
two different kinds: A with probability $p$ and B with probability $(1-p)$.
Upon diagnosis, type A stroke patients are sent to one department and type B patients to another.
Does it make sense to you that the department receiving type A
strokes observes admissions at Poisson rate $p\lambda\,?$
A: The fact that this results in a Poisson process can be obtained without relying on filtering. Just consider that each of the two resulting processes have the Poisson properties (independence of occurences, stable in time, memoryless) and are therefore Poisson processes. The rates are easy to obtain, as a proportion $p$ of $\lambda$ goes to the first process, that therefore has a rate of $p.\lambda$. 
