Let $N, X_1, \dots , X_n$ be independent random variables.
$N \sim P(\lambda) \quad (\text{Poisson distribution})$, while $X_k \sim B(p)$ (Bernoulli)
Let us consider the "random" sum $S = X_1 + \dots + X_N$, which is random also because of the index $N$
(e.g. if $N=0$ then $ S = 0$, if $N = 3$ then $ S = X_1 + X_2 + X_3$, and so forth)
Question 1 (theoretical question)
How do I prove that $S$ and $N - S$ are random variables?
Question 2 (more pratical)
I've determined that $S \sim P(\lambda p)$.
How do I calculate the distribution of $N - S$?
They are both Poisson, but they are not independent (it suffice to notice that $N=0 \implies S = 0$)
What I've tried is: $$P(N-S = k) = \sum_{j=0}^\infty P(S=j)\cdot P(N=j+k)$$
But I got stuck at $$P(N-S = k) = e^{-\lambda (p+1)} \lambda^k \sum_{j=0}^\infty \frac{\left(\lambda^2p\right)^j}{j!(\ j+k)!}$$
I don't know how to simplify that. And I'm not even that sure that the way I'm doing is right.
Any help is greatly appreciated! :)