For $p \in[0,1]$. Prove that the function $$\rho(s)=\begin{cases} 1-\sqrt{1-4p(1-p)s^{2}} & 0\leq s< 1 \\ 1 & s=1 \end{cases}$$ is a probability generating function of a distribution.
Calculate the mean and the variance.
Attempt of solution:
It is necessary to verify that the conditions of this Theorem are satisfied:
Suppose $F$ is the distribution function related to $\rho$.
I already proved that the function is a probability generating function of a distribution. $$ \begin{align*} F(\{\infty\})=1-\rho(1-)=&1-\lim_{s\rightarrow 1^{-}}1-\sqrt{1-4p(1-p)s^{2}}\\ &=1-(1-\sqrt{1-4p(1-p))}\\ &=\sqrt{1-4p(1-p))} \end{align*} $$ rewriting $$ \begin{align*} F(\{\infty\})&=\sqrt{1-4p(1-p))}\\ &=\frac{|2p-1|^{2}}{\sqrt{|(2p-1)|^{2}}}\\ &=|2p-1| \end{align*}$$
Now we have $F(\{k\})=\frac{\rho^{k}(0)}{k!}$. So for $0<k=2m<\infty$ with $m\in\mathbb{N }$
$$ F(\{k\})=\frac{\rho^{k}(0)}{k!}=4^{m}{\frac{m}{k}\choose{m}}[p(1-p)]^{m} $$ for $k$ odd y $k=0$, $F(\{k\})=0$.
I'm trying to calculate the mean as follows:
$$ \rho^{\prime}(1-)=\lim _{s \nearrow 1} \frac{1-\rho(s)}{1-s}=\lim_{s\nearrow 1}\frac{\sqrt{1-4p(1-p)s^{2}}}{1-s} $$
I want to know if I'm on the right track. The answer of the book is: The mean is infinite and thus the variance is undefined. How can I prove this?