Hi,

I've shown that the probability generating function of the RV X, which has a negative binomial dist with parameters k € N and p € (0,1) is:
GX(s)= (p/1-(1-p)s)^k

b)I am told that Y is another RV independent of X with a negative binomial dist with parameters k' € N and p € (0,1).
To show that X+Y has a negative binomial dist with parameters k+k' and p do I write:

GX+Y(s)=GX(s).GY(s) (independence)
=(p/1-(1-p)s)^k.(p/1-(1-p)s)^k'=(p/1-(1-p)s)^(k+k')?

c) Then I have to show that the distribution of X can always be equal X1+....+Xk where the Xi's are RVs independent and have the same distribution.

From part b can I say that the Xis must each be NegBin(1,p) which is P(X=n) = (1-p)^n * p because X1+...+Xk of a NegBin(1,p) dist would be NegBin(1+1+...,p) = NegBin(k,p) Then what do I need to do?

Thanks for your help