A problem from All of Statistics pg. 45:
Let $N \sim \text{Poisson}(\lambda)$ and suppose we toss a coin $N$ times. Let $X$ and $Y$ be the number of heads and tails. Show that $X$ and $Y$ are independent.
If we let $f$ denote the pmf, I know that independence holds iff
$$ f(X = k) = f(X = k | Y = j) = \frac{f(X = k, Y = j)}{Y = j} $$
I know also that $N \sim \text{Poisson}(\lambda)$ implies that
$$ f(X = i) = \frac{\lambda^i e^{-\lambda}}{i!} $$
for all $i \in \mathbb{N}$. The problem is that I don't know how in this instance to compute $f(X = k, Y = j)$.
We show how to compute the probability that $X=k$ and $Y=j$ that you asked about. This is the probability that $X+Y=k+j$ times the probability that $X=k$ given that $X+Y=k+j$. We assume the coin is fair. A similar calculation can be made for unfair coins.
We have $$\Pr(X+Y=k+j)=e^{-\lambda}\frac{\lambda^{k+j}}{(k+j)!}.$$ Also, $$\Pr(X=k\mid X+Y=k+j)=\binom{k+j}{k}(1/2)^{k+j},$$ since the conditional distribution of $X$ given that $N=n$ is binomial. It follows that $$\Pr(X=k\cap Y=j)=e^{-\lambda}\frac{\lambda^{k+j}}{(k+j)!}\binom{k+j}{k}(1/2)^{k+j}.$$ This simplifies to $$e^{-\lambda} \frac{1}{k!j!}(\lambda/2)^k(\lambda/2)^j.$$
For the independence calculation, the simplest approach now is to note that the above expression factors as a function of $k$ times a function of $j$.