Independence of two binomial variables I am trying to figure out whether these two variables are independent or not, and why.
In the situation of the problem I was given, the number of clients $N$ follows a Poisson distribution of mean $c$. The number of successful clients $X$ has a probability $p$. The number of events is unlimited and each client is independent.
I understand that both the number of successful clients $X$ and the number of unsuccessful clients $Y$ follow a binomial distribution, and that the total numbers of clients $N$ is the addition of those two variables:
$X\sim\operatorname{Bin}(N,p)$ ; $Y\sim\operatorname{Bin}(N,1-p)$ ; $N=X+Y$.
Are $X$ and $Y$ independent, and why?
 A: Rather than $X\sim\operatorname{Bin}(N,p)$ I would write $X\mid N\sim\operatorname{Bin}(N,p),$ and similarly $Y\mid N\sim\operatorname{Bin}(N,1-p).$
You have
\begin{align}
& \Pr(X=x\ \&\ Y=y) = \operatorname E(\Pr(X=x\ \&\ Y=y\mid N)) \\[8pt]
\text{and } & \Pr(X=x\ \&\ Y=y\mid N=n) = 0 \text{ unless } n = x+y. \\[8pt]
\text{So } & \operatorname E(\Pr(X=x\ \&\ Y=y\mid N)) \\[8pt]
= {} & \sum_{n=0}^\infty \Pr(X=x\ \&\ Y=y\mid N=n)\Pr(N=n) \\[8pt]
= {} & \Pr(X=x\ \&\ Y=y\mid N=x+y)\Pr(N=x+y) \\
& \text{(All other terms in the sum vanish.)} \\[8pt]
= & \Pr(X=x\mid N) \Pr(N=x+y) \\
& \text{(since conditional on $N=x+y$, the} \\
& \phantom{(}\text{events $X=x$ and $Y=y$ are the same)} \\[8pt]
= {} & \binom{x+y} x p^x(1-p)^y \cdot \frac{\lambda^{x+y} e^{-\lambda}}{(x+y)!} \\[8pt]
= {} & \frac{(p\lambda)^2 e^{-p\lambda}}{x!} \cdot \frac{((1-p)\lambda)^y e^{(1-p) \lambda}}{y!}
\end{align}
So $X,Y$ are independent Poisson-distributed random variables with expected values $p\lambda$ and $(1-p)\lambda.$
