If $(N_t)_{t\geq 0}$ is a Poisson Process w/ parameter $\lambda$, then $(N_{t+s} - N_s)_{t\geq 0}$ is also PP$(\lambda)$

Question

All of the notes I've seen on Poisson Processes just hand wave this proof and say that it just immediately follows from the memory-less property of Exponential RVs. Could someone walk though how one could show this? Thanks.

Answer 1

Let $N(t)$ be a Poisson process. Then

1. Then $N(0)=0$.
2. The increments are independent, e.g. $N(t_1,t_1+h_1),N(t_2,t_2+h_2)$ are independent as long as they don't overlap.
3. $N(h,h+t)$ follows $Poi(t\lambda)$.

Now consider $N^*(t)=N(t+s)-N(s)=N(s,s+t)$.

First $$N^*(0)=N(t-0)-N(t)=0$$.

Second for any $i$ when the intervals don't overlap $$N^*(t_i,t_i+h_i)=N(t_i+s,t_i+h_i+s),$$

so that the increments are independent by 2.

Finally $N^*(h,h+t)=N(s+t)-N(s+h)=N(s+h,s+h+t)$, which follows $Poi(\lambda t)$ by 3.