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I have arrived to this equation in several contexts within branching processes. It arises from textbook exercises, so it must be solvable somehow. Here $f$ is a probability generating function which is known, and $g$ is one which I want to know. Here is a specific case:

$1-g\left(b+\frac{(1-b)(1-a)s}{1-as}\right)=\left(\frac{1-b}{1-a}\right)(1-g(s))$

Where $a$ and $b$ are fixed real numbers.

In this case $g$ should be the pgf of a geometric distribution.

I would write my ideas but I have none. I wouldn't know how to start solving this, and I'm pretty sure there's a small set of solutions for $g$. Thanks in advance for your hints or solutions!

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Let $h(s) = 1 - g(s)$, so your equation says $h \circ f = M \circ h$, where $M$ is multiplication by $m$. Then $h \circ f^k = M^k h$ (where the exponent refers to composition, not multiplication). In your example, $f$ is a fractional linear transformation, and these can be embedded in a continuous group: $ f^k(s) = {\frac { \left( as-b \right) + \left( -sb+b \right) r ^{k}}{ \left( as-b \right) + \left( -as+a \right) r ^{k}}} $ where $r = (1-b)/(1-a)$. Thus $f^{k(x)}(0) = x$ where $k(x) = \log \left( \dfrac{xb-b}{xb-a}\right)/\log(r)$, and so $h(x) = h (f^{k(x)}(0)) = m^{k(x)} h(0) $