Conditional Distribution - X|N Binomial where N is Binomial

Question

I would like to prove the following identity as a lemma to solve the following question on Conditional Distributions:

Qn: Let X have a binomial distribution with parameters p and N, where N has a binomial distribution with parameters q and M. What is the marginal distribution of X?

To prove:

$$ {\sum_{n=k}^M \frac{(M - k)!}{(n-k)!(M-n)!}(1 - p)^{n - k}(\frac{q}{1-q})^{n-k}= \left(1+\frac{q(1-p)}{1-q}\right)^{M-k}} $$

Would appreciate if anyone could advise on how to show this identity - have tried considering Binomial Expansion/GP but I can't land on the solution.

My Solution:

Since $$ f_{X|N}(k|n) = \frac{f_{X,N}(k,n)}{f_N(n)} $$ where we have the conditional probability $$ f_{X|N}(k|n) = P(X=k|N=n) = \left(\begin{array}{c} n \\ k \end{array}\right) p^{k}(1-p)^{n-k}, \quad k \leq n $$ and the marginal distribution $$ f_N(n) = P(N=n) =\left(\begin{array}{c} M \\ n \end{array}\right) q^{n}(1-q)^{M-n}, \quad n \leq M . $$

Then, the joint pmf is given by

$$ P(X=k,N=n) = {n \choose k} p^k(1-p)^{n-k} {M \choose n} q^n(1-q)^{M-n}. $$

Computing the marginal pmf, then

$$ \begin{aligned} &\operatorname{Pr}\{X=k\}=\sum_{n=0}^{M} \left\{ \frac{n !}{k !(n-k) !} p^{k}(1-p)^{n-k} \frac{M !}{n !(M-n) !} q^{n}(1-q)^{M-n} \right\} \\ &=\frac{M !}{k ! (M-k)!} p^{k}(1-q)^{M}\left(\frac{q}{1-q}\right)^{k} \sum_{n=k}^{M} \frac{(M-k)!}{(n-k) !(M-n) !}(1-p)^{n-k} \\ &\times\left(\frac{q}{1-q}\right)^{n-k} \\ &=\frac{M !}{k !(M-k) !}(p q)^{k}(1-q)^{M-k}\left[1+\frac{q(1-p)}{1-q}\right]^{M-k} \\ &=\frac{M !}{k !(M-k) !}(p q)^{k}(1-p q)^{M-k}, \quad k=0,1, \ldots, M \end{aligned} $$

i.e. X ~ Bin($M, pq$).

The sum I wanted to show is between the 2nd and 3rd step.

I would admit that I attempted the book-keeping on the sum with the end in mind that I wanted to get X as a Binomial with probability $pq$, so this may not be the best solution. Would greatly appreciate if you have any alternative solutions/ideas in mind! Thanks again!

Answer 1

$X$ represents a count for successes among $N$ independent Bernoulli trials with identical success rate $p$, where $N$ represents a count for successes among $M$ independent Bernoulli trials with identical success rate $q$.

Therefore $X$ represents a count for successes among $M$ independent Bernoulli trials with identical success rate $pq$, because, to be counted among the $X$ successes, a trial must succeed at both filters.

Thus you need to demonstrate that: $\mathsf P(X=x)=\dbinom{M}{x}(pq)^x(1-pq)^{M-x}$

To begin:

$$\begin{align}\mathsf P(X=x)&=\sum_{n=x}^M\mathsf P(N=n)\mathsf P(X=x\mid N=n)\\[1ex] & =\sum_{n=x}^M\dbinom{M}{n} q^n(1-q)^{M-n}\dbinom{n}{x}p^x(1-p)^{n-x}\\[1ex] &=\binom{M}{x}(pq)^x\sum_{n=x}^M\binom{M-x}{n-x}((1-p)q)^{n-x}(1-q)^{M-n}\\[1ex] &~~\vdots\\[1ex] &~~\vdots\\[1ex] &=\binom{M}{x}(pq)^x[1-pq]^{M-x}\end{align}$$

have tried considering Binomial Expansion

That's the way to go. Try it again.

Hint: $1-pq = (1-q)+(q-pq)$