Poisson process on nonintersecting sets

Question

I try to show that for a homogeneous Poisson process $N$ with intensity $\lambda$ and nonintersecting sets $A,B$ the amount of events happening in the two sets are independent. My idea is to prove the following $$P(N(A) = n, N(B) =z)=P(N(A\cup B)= n+z)=P(N(A) = n)P(N(B) =z)$$ the first equality comes from $A\cap B=\emptyset$ but I fail to show the next one because of the factorial $$P(N(A\cup B)= n+z)=\frac{(\lambda|A\cup B|)^{n+z}}{(n+z)!}e^{-\lambda(|A\cup B|)}$$

I would appreciate some help, thanks

Edit:

My definition of a HPP $N$:

1. For some $\lambda > 0$ and finite planar region $A$, we have a Poisson distribution, i.e. $$N(A)\sim \text{Poi}(\lambda \cdot |A|)$$
2. Given $N(A)=n$, the $n$ events in $A$ are an independent sample from the uniform distribution on $A$

Answer 1

For simplicity of notation let $X=N(A)$, $Y=N(B)$ and $Z=X+Y=N(A\cup B)=N(A)+N(B)$, let $a = EX = \lambda|A|$ and $b=EY$.

The probability that $(X,Y)=(k,l)$ can be found by conditioning on $Z$ as follows: $$\begin{align*} P((X,Y)=(k,l)) &= \sum_{n\ge0}P(Z=n)\times P((X,Y)=(k,l)|Z=n)\\ &=P(Z=k+l)\times P((X,Y)=(k,l)|Z=k+l)\tag{*}\\ &=\frac{(a+b)^{k+l}}{(k+l)!}e^{-(a+b)}\times P((X,Y)=(k,l)|Z=k+l)\\ &= \frac{(a+b)^{k+l}}{(k+l)!}e^{-(a+b)}\times \binom{k+l}k \left(\frac a{a+b}\right)^k \left(\frac b{a+b}\right)^l\\ &= \frac{(a+b)^{k+l}}{(k+l)!}e^{-(a+b)}\times \frac{(k+l)!}{k!l!} \left(\frac a{a+b}\right)^k \left(\frac b{a+b}\right)^l\\ &= \frac{a^k}{k!}e^{-a}\,\frac {b^l}{l!} e^{-b}\\ &= P(X=k)P(Y=l). \end{align*}$$ Here (*) holds because all terms $P((X,Y)=(k,l)|Z=n)$ vanish, except when $n=k+l$.

In words: the variables $X,Y,Z$ are definitely not independent. But when $Z$ is Poisson and $X$ conditional on $Z$ is binomial and conditional on $Z$ we have $Y=Z-X$, magically the right factorials cancel to make $X$ and $Y$ independent.