Respuesta :
Answer:
See explanation below.
Step-by-step explanation:
Part a
For this case we can use the moment generating function for the bernoulli distribution for n trials, given by:
[tex] p(s)^n = (q+ps)^n =\sum_{r=0}^n (nCr) (ps)^r q^{n-r}[/tex]
Wehre p is the probability of success and [tex] P(s_ = q+ps[/tex]
Using this property we see that;
If we multiply the two generating functions we got:
[tex] p(s)^N = (q+ps)^N =\sum_{r=0}^N (NCr) (ps)^r q^{N-r}[/tex]
[tex] p(s)^M = (q+ps)^M =\sum_{r=0}^M (MCr) (ps)^r q^{M-r}[/tex]
[tex] p(s)^M p(s)^N = \sum_{r=0}^N (NCr) (ps)^r q^{N-r} \sum_{r=0}^M (MCr) (ps)^r q^{M-r}[/tex]
And the mass function would be given by:
[tex]= (N+M C r) p^{r} (1-p)^{N+M-r}[/tex]
So we see that follows a binomial random variable with parameters (N+M, p)
Part b
For this case we are assuming that [tex] X \sim Bin (N,p) , Y\sim Bin (M,p)[/tex] and for this case we can assume that [tex] 0 \leq k \leq N+M[/tex] for the proof.
We are interested on the random variable [tex] Z= X+Y[/tex] since the two random variables are independent we can write the probability mass function for Z like this:
[tex] P(Z = X+Y = k) =\sum_{i=0}^k P(X=i , Y=k-i)[/tex]
[tex] P(Z = X+Y = k) =\sum_{i=0}^k P(X=i) P(Y=k-i)[/tex]
And we can replace the mass function for X and Y
[tex] P(Z = X+Y = k) = \sum_{i=0}^k (NCi) p^i (1-p)^{N-i} \sum_{i=0}^k (M C i-1) p^{k-i} (1-p)^{M-K+i}[/tex]
And we can rewrite this like that:
[tex] P(Z = X+Y = k) = \sum_{i=0}^k (NCi) p^i (1-p)^{N-i} (M C i-1) p^{k-i} (1-p)^{M-K+i}[/tex]
[We can take out the constant p:
[tex] P(Z = X+Y = k) = p^k (1-p)^{N+M-k} \sum_{i=0}^k (NCi)(M C k-i)[/tex]
And using properties of the binomial formula we can write this like that:
[tex] P(Z = X+Y = k) = (N+M Ck) p^k (1-p)^{N+M-k} [/tex]
So then we see that [tex] Z= X+Y \sim Bin(N+M ,p)[/tex]
