Let $(X_j)$ be a sequence of r.v. with common distributions $(f_j)$, $N$ be a r.v. having a Poisson distribution with mean $\lambda t$. Let $S_N = X_1 + \cdots + X_N$. Then, $S_N$ has the compound Poisson distribution with a generating function $e^{-\lambda t + \lambda tf(s)}$, where $f(s) = \sum f_j s^j$ (a generating function of $(f_j)$). Let this generating function be $h_t(s) = e^{-\lambda t + \lambda tf(s)}$.
(Feller Vol.1, P.289-290) A probability generating function $h$ is called infinitely divisible if for each positive integer $n$ the $n$th root $\sqrt[n]{h}$ is again a probability generating function.
This follows that if $h_{t+r}(s) = h_t(s) h_r(s)$ for some positive integer $t$ and $r$, then $\sqrt[n]{h_t(s)} = h_{t/n}(s)$, and the right side is a probability generating function. Therefore, $h_t$ is infinitely divisible.
In addition, the author says that a probability generating function, which satisfy $h_{t+r}(s) = h_t(s) h_r(s)$ must follows the compound Poisson distribution (i.e. $h_t(s) = e^{-\lambda t + \lambda tf(s)}$).
Then, I have the following theorem, which basically show that the converse is also true. I am struggling with the proof of this theorem .
(2.2) is $h_t(s) = e^{-\lambda t + \lambda tf(s)}$.
I don't particularly understand the statement highlighted with the purple line. Maybe I can express $\sqrt[n]{h(s)}$ as $- \sum_{k=0}^\infty {1/n \choose k} (1-h(s))^k$. This implies that $\sqrt[n]{h(0)} =- \sum_{k=0}^\infty {1/n \choose k} (1-h_0)^k$. But, I don't understand why $\sqrt[n]{h}$ vanishes. I am also not sure of how the author derives this convergence $\sqrt[n]{h(s)} \to 1$.
I would greatly appreciate if you elaborate the proof. Let me know if you need more context.
Best Answer
I will write $p_{n,k}$ for the coefficient of $s^k$ in $\sqrt[n]{h(s)}$, that is $$\sqrt[n]{h(s)} = \sum_{k=0}^\infty p_{n,k} s^k.$$ Suppose that $h_0 = 0$. Then, since $h_0 = \left(\sqrt[n]{h(0)}\right)^n = p_{n,0}^n$, we would have $p_{n,0} = 0$. This is what Feller means by "the absolute term in the power series for $\sqrt[n]{h}$ would vanish". Then $$\sqrt[n]{h(s)} = p_{n,1}s + \sum_{k=2}^\infty p_{n,k}s^k,$$ and this would imply that $$h(s) = \left( p_{n,1}s + \sum_{k=2}^\infty p_{n,k}s^k\right)^n = p_{n,1}^n s^n + \cdots.$$ In other words, $h_0 = h_1 = \ldots = h_{n-1} = 0$. Since $n$ is arbitrary, we would then have $h_k = 0$ for every $k \geq 0$ which is obviously impossible as $h$ is a PGF. Thus $h_0 >0$.
From this, since $\sqrt[n]{h}$ is nondecreasing, we can deduce that for every $s \in [0,1]$ $$\sqrt[n]{h_0} \leq \sqrt[n]{h(s)} \leq \sqrt[n]{h(1)} = 1.$$ Letting $n \to \infty$, this yields $\sqrt[n]{h(s)} \to 1$.