Basic probabilities: one poisson then 2 binomial (Wasserman 2.14 - 11) I am self-refreshing some stats concepts by reading "All of Statistics" (Wasserman 2004) and am puzzled by the following problem (section 2.14, exercise 11):

Let N ~ Poisson($\lambda$) and suppose we toss a coin N times and let $p$ be the probability of heads. Let $X$ and $Y$ be the number of heads and tails. Show that $X$ and $Y$ are independent. 

It seems obvious that $X$ and $Y$ are dependent when conditioned by $N$ since they must sum up to N, so $f(X,Y|N) \ne f_X(X|N) f_Y(Y|N)$ but the question is about showing that $f_{XY}(X,Y) = f_X(X) f_Y(Y)$. 
I tried obtaining $f(X,Y)$ by first expressing $f(X,Y,N) = f(N)f(X|N)f(Y|X,N)$ and then marginalising over $N$, the first and second factors are clearly a poisson and binomial respectively but the last one appears to be $1$ for $Y=N-X$ and $0$ otherwise (and I'm stuck there ^__^).
Any help or comment welcome! Thanks!

Let us first partition the event $(Y=y,X=x)$ into disjoint events indexed by $N$:
$$\begin{align}
\mathbb{P}(X=x,Y=y)&=\sum_n \mathbb{P}(X=x,Y=y,N=n),\\
&=\sum_n \mathbb{P}(X=x,Y=y|N=n)\mathbb{P}(N=n),\\
&=\sum_n \mathbb{P}(Y=y|X=x,N=n)\mathbb{P}(X=x|N=n)\mathbb{P}(N=n).
\end{align}$$
Up to this point this is exactly what you did. Now recall that you wish to calculate the probability that $X=x$ and $Y=y$, that is, $x$ and $y$ are fixed. Now since $y$ is fixed, the factor $\mathbb{P}(Y=y|N=n,X=x)$ will take out the values of the summation for which $n\neq x+y$ (given $x$ and $y$, there will be a value of $n$ for which this term doesn't vanish, namely $n=x+y$), this yields:
$$\begin{align}
\mathbb{P}(X=x,Y=y)&=\mathbb{P}(X=x|N=x+y)\mathbb{P}(N=x+y),\\
&=\frac{\lambda^{x+y}}{x! y!}e^{-\lambda}p^x (1-p)^{y}.
\end{align}$$
Calculating the marginals through $f_X(x)=\sum_{n=x}^\infty f(x|n)f(n)$ (and the analogous formula for $Y$) and multiplying them together produces the density stated above (I checked it), thus $X$ and $Y$ are independent.
