Necessary and sufficient condition for $G(z)$ to be a probabilty generating function. Given a function $G(z)$, what are the sufficient and necessary conditions for being a $G(z)$ to be a probability generating function.
Few necessary condition which I know of are


*

*$G(1) = 1$

*All the coefficients of $z^n$ must be positive.


Thanks in advance! 
 A: Sufficient and necessary condition for $G$ to be a generating function is that $G$ is smooth on $(-1,1)$ with every derivative $G^{(n)}(0)$ at $0$ nonnegative and $\sum\limits_{n=0}^\infty\frac1{n!}G^{(n)}(0)=1$.
A: Regarding "Answer", although this is a necessary condition it does not seem to be sufficient. For example take $G$ satisfying these conditions and add to it the function $g(x)=e^{-1/x}1_{\{x>0\}}$. This function is known to be smooth (in the sense of infinitely differentiable) with all derivatives at zero being zero and thus it's Taylor series expansion around zero is zero so it cannot be equal to the function for any $x>0$. This means that if $G$ satisfies the requirement in "Answer" then so does $G(x)+g(x)$. However $G(x)+g(x)>G(x)$ for $x\in(0,1]$ cannot be a probability generating function. In particular $G(1)+g(1)>1+e^{-1}>1$.
Although I can't point to a reference which I would be very surprise if it doesn't exist (I proved it for myself, hopefully without errors), I believe that a necessary and sufficient condition for a function $G:(0,1]\to\mathbb{R}$ to be a probability generating function (of an almost surely finite nonnegative integer valued random variable) is:
$G$ is smooth on $(0,1)$ with $G^{(n)}(z)\ge 0$ for all $n\ge 0$ and all $z\in(0,1)$ ($G^{(0)}\equiv G$) and $\exists\lim_{z\uparrow 1}G(z)=G(1)=1$.
Note that the limit in this statement exists since $G'(z)\ge 0$ so that $G$ is nondecreasing on $(0,1)$.
Also note that any p.g.f. must satisfy these conditions so that the challenge is to prove that this condition is sufficient. One needs to show two things. The first is that there exists some a.s. nonnegative and finite random variable $X$ that satisfies $G(z)=Ez^X$ for $z\in(0,1]$ and the other is to show that necessarily $P(X\in\mathbb{Z}_+)=1$.
Finally, if you are restricting yourself to functions of the form $\sum_{n=0}^\infty a_nz^n$ that are absolutely convergent on $[-1,1]$ (thus clearly smooth on $(-1,1)$, then of course a necessary and sufficient condition is that $a_n\ge 0$ and $\sum_{n=0}^\infty a_n=1$. This is a far easier question.
