Conditional binomials

Question

I'm trying to proof this, If X ~ B(n, p) and, conditional on X, Y ~ B(X, q), then Y is a simple binomial variable with distribution Y ~ B(n, pq) . Can someone show me how or link some reference.

Answer 1

For me no particular reference comes to mind immediately. Let's try it: $$ \begin{align} \Pr(Y=y) & = \mathbb E(\Pr(Y=y\mid X)) \\[10pt] & = \sum_{x=0}^n \Pr(Y=y\mid X=x)\Pr(X=x) \\[10pt] & = \sum_{x=0}^n \binom x y q^y(1-q)^{x-y} \binom n x p^x(1-p)^{n-x} \\[10pt] & = \sum_{x=y}^n\cdots\cdots\text{ditto}\cdots\cdots \qquad(\text{discarding the zero terms}) \\[10pt] & = \sum_{w=0}^{n-y} \binom{w+y}{y} q^y (1-q)^{w} \binom{n}{w+y} p^{w+y}(1-p)^{n-y-w} \\[10pt] & = (pq)^y \sum_{w=0}^{n-y} \binom{n}{w+y} \binom{w+y}{w} ((1-q)p)^w (1-p)^{(n-y)-w} \\[10pt] & = (pq)^y \binom{n}{y} \sum_{w=0}^{n-y} \binom{n-y}{w} ((1-q)p)^w (1-p)^{(n-y)-w} \text{ (see comment below)} \\[10pt] & \dots\text{and now apply the binomial theorem in a routine way:} \\[10pt] & = \binom{n}{y} (pq)^y \Big( (1-q)p + 1-p \Big)^{n-y} \\[10pt] & \dots\text{then some simple algebraic simplifications:} \\[10pt] & = \binom{n}{y} (pq)^y (1-pq)^{n-y}, \end{align} $$ as predicted.

Comment: The problem now is why is it true that $$ \binom{n}{w+y} \binom{w+y}{w} = \binom{n}{y} \binom{n-y}{w}. $$ A bijective argument about choosing two subcommittees will do that.