Answer:
See the proof below.
Step-by-step explanation:
Assuming this complete question: "For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed with parameters q and M. Formulate Z as a random sum and show that Z has a binomial distribution with parameters pq and M."
Solution to the problem
For this case we can assume that we have N independent variables [tex]X_i[/tex] with the following distribution:
[tex] X_i Bin (1,p) = Be(p)[/tex] bernoulli on this case with probability of success p, and all the N variables are independent distributed. We can define the random variable Z like this:
[tex] Z = \sum_{i=1}^N X_i[/tex]
From the info given we know that [tex] N \sim Bin (M,q) [/tex]
We need to proof that [tex] Z \sim Bin (M, pq)[/tex] by the definition of binomial random variable then we need to show that:
[tex] E(Z) = Mpq[/tex]
[tex] Var (Z) = Mpq(1-pq)[/tex]
The deduction is based on the definition of independent random variables, we can do this:
[tex] E(Z) = E(N) E(X) = Mq (p)= Mpq[/tex]
And for the variance of Z we can do this:
[tex] Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2 [/tex]
[tex] Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2[/tex]
And if we take common factor [tex]Mpq [/tex] we got:
[tex] Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq][/tex]
And as we can see then we can conclude that [tex] Z \sim Bin (M, pq)[/tex]