Negative probabilities - what are two ordinary pgfs that correspond to the gf of a half-coin? In Half of a Coin: Negative Probabilities, author considers pgf of a fair coin represented by random variable, $X = 1_H$:
$$G_X(z) = E[z^X] = \sum_{x=0,1} z^xP(X=x) = (z^0)(1/2) + (z^1)(1/2) = \frac{z+1}{2}$$
Author defines gf of a half-coin (a coin with sides $n=0,1,2,...$ with some sides having negative probability):
$$G_{X_{0.5}}(z) = \sqrt{\frac{z+1}{2}} = \frac{1}{\sqrt{2}}\sum_{k=0}^{\infty} \binom{1/2}{k} z^k$$
That is, $$P(X = k) = \frac{1}{\sqrt{2}} \binom{1/2}{k}$$
If we flip two independent half-coins, their sum is 0 or 1 with probability 1/2. So it's like flipping one fair coin.
This is based on pgf of sums of independent random variables being the products of the pgfs of the random variables.
Later on, there's this fundamental theorem saying that for any gf $f$, there exists two pgfs $g, h$ such that
$$fg = h$$
How is that related to the half-coin?
What are possible $g$ and $h$ that correspond to $$f(z) = \sqrt{\frac{z+1}{2}}$$
?
I was expecting something like two gfs for one pgf as was done with the fair coin and the two half coins (product of two gfs for the half coins is the pgf for the fair coin).
 A: My previous answer, in much simplified and more explicit form: take $g(z):=\frac{1-\sqrt{1-z}}{z}$ and $$h(z):=f(z)g(z)=\frac{1-\sqrt{1-z}}{z}\,\sqrt{\frac{1+z}{2}}
=\frac{\sqrt{1+z}-\sqrt{1-z^2}}{z\sqrt2}.$$ 
Then $g$ and $h$ are pgf's. 
Indeed, as before, let 
\begin{equation}
 c_j:=\frac1{j2^{2j-1}} \binom{2 j-2}{j-1}>0. 
\end{equation} 
Then 
\begin{equation}
 \sqrt{1+z}=1+\sum_{j=1}^\infty(-1)^{j-1}c_j z^j,\quad 
 \sqrt{1-z^2}=1-\sum_{i=1}^\infty c_i z^{2i},
\end{equation}
\begin{equation}
 g(z)=\frac{1-\sqrt{1-z}}{z}=\sum_{j=1}^\infty c_j z^{j-1}, 
\end{equation}
\begin{equation}
 h(z)\sqrt2=\frac{\sqrt{1+z}-\sqrt{1-z^2}}{z}
 =\sum_{i=0}^\infty c_{2i+1} z^{2i}+\sum_{i=1}^\infty (c_i-c_{2i}) z^{2i-1}. 
\end{equation}
It remains to check that $c_i\ge c_{2i}$ for $i\ge1$. Let $r_i:=c_{2i}/c_i$. Then $r_{i+1}/r_i=\frac{16i^2-1}{16i^2-4}>1$ for $i\ge1$, so that $r_i$ is increasing in $i\ge1$ to $\lim_{i\to\infty}r_i=\frac1{2\sqrt2}<1$. So, $r_i<1$ for $i\ge1$, which confirms that $c_i\ge c_{2i}$. This completes the proof. 
I am retaining the previous answer, because it shows some of the process by which the second answer was obtained. 
A: Let $a_j:=\binom{1/2}j$. Then $a_0=1$ and $a_j=(-1)^{j-1}c_j$ for $j=1,2,\dots$, where 
\begin{equation}
 c_j:=\frac1{j2^{2j-1}} \binom{2 j-2}{j-1}>0. 
\end{equation}
Let 
\begin{equation}
 g(z):=\sum_{j=0}^\infty b_j z^j,\quad \text{where}\quad 
b_j:=c_{j+1}>0. 
\end{equation}
One may note that $g(1)=1$. 
Since $a_0=1>0$ and $b_j>0$ for all $j\ge0$, it is enough to show that for all natural $n$ 
\begin{equation}
 s_n:=\sum_{j=1}^n a_j b_{n-j}=\sum_{j=1}^n (-1)^{j-1}p_{n,j}\overset{\text{?}}\ge0,
\end{equation}
where $p_{n,j}:=c_j c_{n+1-j}$. Indeed, then for $h:=fg$ one has $h(z)=\frac1{\sqrt2}\sum_{n=0}^\infty (b_n+s_n) z^n$, where $s_0:=a_0b_0=1/2>0$. 
Obviously, $p_{n,j}=p_{n,n+1-j}$. So, $s_n=0$ if $n$ is even. 
Let now $n=2m+1$ be odd, so that $m\in\{0,1,\dots\}$. Then 
\begin{equation}
 s_n=s_{2m+1}\ge\sum_{0\le i\le(m-1)/2}(p_{2m+1,2i+1}-p_{2m+1,2i+2}). 
\end{equation}
So, it suffices to show that $p_{2m+1,j}\ge p_{2m+1,j+1}$ for $j=1,\dots,m$. But 
\begin{equation}
 \frac{p_{2m+1,j+1}}{p_{2m+1,j}}-1=-\frac{3 + 6 (m - j)}{(1 + j) (4 m + 1 - 2 j)}<0
\end{equation}
for $j=1,\dots,m$. This completes the proof. 
