Poisson Variable is Independent of sum of Bernoulli Variables

Question

Let $(X_n)_{n\geq 1}$ be i.i.d. Bernoulli random variables with parameter $p\in(0,1)$. Let $N$ be a Poisson random variable with parameter $\lambda>0$. Assume $N$ is independent from $(X_n)_{n\geq 1}$.

Let $S = \sum_{i=1}^N X_i$, $D = N - S$.

1. What is the joint distribution of $(S, N)$?

2. Prove that $S$ and $D$ are independent.

If I had that all of these objects were independent, I know what to do here. But how do I prove that $N, S$, and the second question - $S, D$ are independent?

Answer 1

You can write the joint pmf

\begin{eqnarray*} P\left(S=k,N=n\right) &=& P\left(S=k\mid N=n\right)P\left(N=n\right)\\ &=& {n\choose k}p^k\left(1-p\right)^{n-k}\cdot e^{-\lambda}\frac{\lambda^n}{n!}\\ &=& \frac{1}{k!}e^{-\lambda p}\left(\lambda p\right)^k\cdot \frac{1}{\left(n-k\right)!}e^{-\lambda\left(1-p\right)}\left(\lambda\left(1-p\right)\right)^{n-k} \end{eqnarray*}

which is exactly the product of two independent Poisson pmfs, which are exactly the pmfs of $S$ and $D$ because $P\left(S=k,N=n\right) = P\left(S=k,D=n-k\right)$. Thus, $S$ and $N$ are not independent but $S$ and $D$ are.