Negative probabilities - what are the two ordinary pgfs that correspond to the gf of a half-coin?

Question

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 the $g$ 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).

Answer 1

How is that related to the half-coin?

It provides intuitive interpretation for the half-coin and any other generalized (with negative probabilities) random variables, the way to think about them. This interpretation is given in the paragraph after the fundamental theorem. Namely, if $\mathcal{X}$ is a generalized random variable (with pgf $f$), there exist two ``classical'' independent random variables $Y$ and $Z$ (with pgfs $g$ anf $h$) such that $\mathcal{X}+Y=Z$ in distribution.

What are the $g$ and $h$ that correspond to $f(z)=\sqrt{(z+1)/2)}$?

I don't think the answer is simple and nice-looking. Otherwise author would give it. The fundamental theorem just says that there exists $g(x)=\sum_{n=0}^\infty p_nz^n$ and $h(x)=\sum_{n=0}^\infty q_nz^n$ such that $p_n\geq0$, $q_n\geq0$, and $f(z)g(x)=h(z)$.