Explicit distribution of B which has conditioned binomial distribution

Question

I have a question concerning the following task:

Let $A \sim$ Bin($n, p$) and let $B$ conditioned on {A = m} have binomial distribution with parameters $m$ and $q$, that is

$P(B = k|A = m) =$ ${m} \choose {k}$ $q^k(1-q)^{m-k}$, $k = 0, ..., m$

What is the distribution of B?

What I already got is the following:

$P(B = k) = \sum_{m=0}^n P(B = k|A = m) P(A = m) =$ ...

$= \sum_{m=0}^n$ ${m} \choose {k}$ $q^k(1-q)^{m-k}$ ${n} \choose {m}$ $p^m(1-p)^{n-m} = \sum_{m=k}^n$ ${m} \choose {k}$ $q^k(1-q)^{m-k}$ ${n} \choose {m}$ $p^m(1-p)^{n-m}$

Has anyone a hint how i could simplify this sum?

Answer 1

We have \begin{align} \Pr(B = k) &= \sum_{m= 0}^n \binom{m}{k}q^k(1-q)^{m-k}\binom{n}{m}p^m(1-p)^{n-m} \\ &=\left(\frac{q}{1-q}\right)^k(1-p)^n\sum_{m=0}^n \binom{m}{k}\binom{n}{m}\left(\frac{p(1-q)}{1-p}\right)^m \\ &=\left(\frac{q}{1-q}\right)^k(1-p)^n\sum_{m=0}^n\binom{n}{k}\binom{n-k}{n-m}\left(\frac{p(1-q)}{1-p}\right)^m \\ &=\left(\frac{q}{1-q}\right)^k(1-p)^n\binom{n}{k}\left(\frac{p(1-q)}{1-p}\right)^n\sum_{m=0}^n \binom{n-k}{n-m}\left(\frac{1-p}{p(1-q)}\right)^{n-m} \\ &=\left(\frac{q}{1-q}\right)^k(1-p)^n\binom{n}{k}\left(\frac{p(1-q)}{1-p}\right)^n\left(1 + \frac{1-p}{p(1-q)}\right)^{n-k} \\ &=\binom{n}{k}(1-pq)^{n-k}(pq)^k \end{align}

Answer 2

Different look on it:

Take independent $X_1,\dots, X_n, Y_1,\dots,Y_n$ with $X_i\sim\text{Ber}(p)$ and $Y_i\sim\text{Ber}(q)$ for $i=1,\dots,n$.

Setting: $$A:=X_1+\cdots+X_n$$ and: $$B:=X_1Y_1+\cdots+X_nY_n$$ we are exactly in the situation described in your question.

Evidently the $X_iY_i$ are independent with $X_iY_i\sim\text{Ber}(pq)$ for $i=1,\dots n$ so that $B\sim\text{Bin}(n,pq)$.

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Actually this does not answer your question ("how to simplify this sum"), but shows that the distribution can be found without using the sum.