Thread: Help with probability generating function

Let X1, X2, .... be a sequence of iid random variables with PMF:

P(X=x)= [a(1-p)^x]/x, x=1,2,3...., with 1>1-p>0 and a=-1/log(p)

If N is independent of the Xi and has a Poisson distribution with parameter u (au is an integer) then show that S, the sum from i=1 to N of the Xi has a Negative Binomial distribution with PGF:


Gs(S)=[p/1-s(1-p)]^au

Any hints how to do this?

I know that Gs(S)=GN(Gx(s))

I have tried this many times by trying first to calculate Gx(s), with the thought that I should get something like the PGF of a geometric but am getting nowhere.

Can anyone hint of how to get started etc

Hey darren86.

What did you get for the product of PGF's and the PGF itself?

Take a look here for more details (you need to get the PGF for one variable first using E[z^x])

Probability-generating function - Wikipedia, the free encyclopedia

Originally Posted by chiro

Hey darren86.

What did you get for the product of PGF's and the PGF itself?

Take a look here for more details (you need to get the PGF for one variable first using E[z^x])

Probability-generating function - Wikipedia, the free encyclopedia

This is where I'm having trouble, Gx(s)=E[s^x]=???

a*the sum of {[s(1-p)]^x}/x for x=1 to infinity.

I presume this is a common power series or something, but I cant find what it is????

Have you considered transforming the random variable to a geometric random variable and then using the property that the sum of independent geometric has a negative binomial?

(Hint: Transform it by transforming the parameter and not the distribution itself since mean of geometric is 1/p and variance is [1-p]/p^2)

If you transform the parameter and show that it has the distribution form you can take it from there.

Originally Posted by chiro

Have you considered transforming the random variable to a geometric random variable and then using the property that the sum of independent geometric has a negative binomial?

(Hint: Transform it by transforming the parameter and not the distribution itself since mean of geometric is 1/p and variance is [1-p]/p^2)

If you transform the parameter and show that it has the distribution form you can take it from there.

I have thought about showing that the Xi are geometric random variables, but how can this be done when there is a x in the denominator? I have thought about different parameters for a geometric random variable but cannot see how we can have that x in the denominator, any hints?

I just took a look at the Wiki site and the sum of geometric random variables is definitely a Negative Binomial.

Since distributions are unique it means that the only way a sum of IID random variables gives a Negative Binomial is that all random variables (the Xi's) are also geometric random variables or some other formulation which happens to be (from Wikipedia) the logarithmic distribution.

Negative binomial distribution - Wikipedia, the free encyclopedia

Logarithmic distribution - Wikipedia, the free encyclopedia

I don't think in all honesty that I would have seen the connection myself, but the wiki page outlines exactly what you need and why it is the case.