Moment generating Function Problem

Question

My task is to find a fitting probability distribution function.

For $X$, the moment generating function is given as $$mx(s) = \left(\frac{1-p}{1-pe^s}\right)^2$$

$p\in(0,1)$ is a parameter.

I don't know where to start.

Answer 1

It is enough to find the distribution that belongs to the generating function $$M_U(s)=\frac{1-p}{1-pe^s} $$ When $s$ is chosen such that $|pe^s|<1$, we can write: $$M_U(s)=(1-p) \sum_{k=0}^\infty (pe^s) ^k=\sum_{k=0}^\infty e^{sk}p^k(1-p)$$ But that is just the one for geometric distribution. Since MGF is unique we can conclude $U\sim \operatorname{Geo}((1-p))$ with probability mass function: $$f_U(k)=p^k(1-p)\ \ \ \ \ k=0,1,2,...$$ Notice that some define the geometric distribution in another way, but that is not a problem.

The original $M_X(s)$ is just the moment generating function of $U+U'$ with $U$ and $U'$ i.i.d.

Such distribution has a name, namely negative binomial distribution. I think from here you can pick the robe to finish it, right?