Prove independence of Y,Z

Question

Consider $Y \sim Binom(n, p)$ and $N \sim Poisson(\lambda)$ Let $Z = N-Y$. Prove that $Y,Z$ are independent.

Attempt:

\begin{align*} P(Z=z|Y=y) &= \frac{P(Y=y, Z=z)}{P(Y=y)}\\ &= \frac{P(N=y+z)}{P(Y=y)}\\ &= \frac{\frac{\lambda^{y+z}e^{-y-z}}{(y+z)!}}{{n \choose y}p^y(1-p)^{n-y}} \end{align*}

Answer 1

Do you mean: $Y\mid N\sim \mathcal {Binom}(N, p)$ , $N\sim \mathcal {Pois}(\lambda)$ , and $Z=N-Y$ ?

Then $\mathsf P(Y= y, Z=z) ~{= \mathsf P(Y=y, N=y+z) \\ = \mathsf P(Y=y\mid N=z+y)~\mathsf P(N=y+z) \\ = \binom{z+y}{y}p^y(1-p)^{z}\cdot \dfrac{\lambda^{z+y}e^{-\lambda}}{(z+y)!} \\ =\dfrac{(y+z)!\;p^y\;(1-p)^z\,\lambda^y\,\lambda^z\,e^{-\lambda p}\, e^{-\lambda(1-p)}}{y!~ z!~(y+z)!} \\ = \dfrac{(\lambda p)^y }{y!}\cdot\dfrac{(\lambda(1-p))^z }{z!}\cdot e^{-\lambda} }$

Hmm, that almost looks like the product of two independent Poisson distributions, but the first two terms would need to be multiplied by $e^{-\lambda p}$ and $e^{-\lambda(1-p)}$ respectively.   There is also that extra $e^{-\lambda}$ term.   Oh, what to do?

Then, of course, you would need to show that $Y$ and $Z$ do actually have such Poisson distributions.

$$\mathsf P(Y=y)~{=\sum_{z=0}^\infty \mathsf P(Y=y\mid N=y+z)\mathsf P(N=y+z) \\\ddots}$$

And such.