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 $X_i$ with the following distribution:

$X_i Bin (1,p) = Be(p)$ 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:

$Z = \sum_{i=1}^N X_i$

From the info given we know that $N \sim Bin (M,q)$

We need to proof that $Z \sim Bin (M, pq)$ by the definition of binomial random variable then we need to show that:

$E(Z) = Mpq$

$Var (Z) = Mpq(1-pq)$

The deduction is based on the definition of independent random variables, we can do this:

$E(Z) = E(N) E(X) = Mq (p)= Mpq$

And for the variance of Z we can do this:

$Var(Z)_ = E(N) Var(X) + Var (N) [E(X)]^2$

$Var(Z) =Mpq [p(1-p)] + Mq(1-q) p^2$

And if we take common factor $Mpq$ we got:

$Var(Z) =Mpq [(1-p) + (1-q)p]= Mpq[1-p +p-pq]= Mpq[1-pq]$

And as we can see then we can conclude that $Z \sim Bin (M, pq)$

8 0

3 years ago

A box contains

Usimov [2.4K]

<span>the probability that a pen from the first box is selected
= total number of pens in 1st box/total number of objects in box 1
= 5/12

</span>the probability that a crayon from the second box is selected
= total number of crayons in 2nd box/total number of objects in box 2
= 6/8 = 3/4

7 0

3 years ago