Proof regarding a probability generating function (Poisson) Let $f(s)$ be the probability generating function ($pgf$) of a non-negative, integer valued random variable. It is also given that $f(1-p+ps)f(p) = f(ps)$. Prove that $f(s) = e^{\lambda(s-1)}$ for some $\lambda > 0.$
Progress: Since the multiplication of the two instances function $f(s)$ results in another instance of $f(s)$, one can "guess" the answer to be a power function of some sort:$b^{ag(s)} \Rightarrow g(1-p+ps) + g(p) = g(ps)$ One can further guess that $g(s) = (1-s)$ instead of $(s-1)$. I have yet to prove that $a>0$ AND $b=e$ AND $g(s) =(s-1)$ AND not $(1-s)$. In fact, I have not proven anything, just guessed!
Background information: The $f(1-p+ps)f(p) = f(ps)$ identity is the result of the following information: It is given that the distribution of $Y$ and $X\mid (Y=X)$ are identical, where $Y$ is the number of successes in $X$ binomial trials (each having $p$ probability of success). $f(s)$ is the pgf of $X$.
 A: The only approach I can think of is to note that the functional equation is multiplicative, so this suggests taking the logarithm and defining $g(x) = \log f(x)$ to obtain the additive equation $$g(1-p+ps) + g(p) = g(ps).$$  Now consider when $s = 1$, we get $g(1) + g(p) = g(p)$, so $g(1) = 0$, hence $f(1) = 1$.  Next, setting $s = 1/p$ gives $g(2-p) + g(p) = g(1) = 0$, hence $g(1+p) + g(1-p) = 0$, indicating $g$ is an odd function about the line $x = 1$; we then write $h(x) = g(1+x)$, so $h$ is an odd function satisfying the relation $$h(p(s-1)) + h(p-1) = h(ps - 1).$$  Next, set $s = 2-1/p$, the unique solution to the condition $p(s-1) = p-1$, to get $$2h(p-1) = h(2(p-1)).$$ Thus an obvious family of solutions to this equation is if $h(x) = kx$ for some constant $k \in \mathbb R$, from which we find $g(x) = k(x-1)$, and $f(x) = e^{k(x-1)}$.  It is not difficult to do the additional checking to show that this choice of $h$ satisfies the original functional equation for all valid choices of $p, s$.  That said, we have not shown this is the only family of solutions; additional arguments are required, which I will allow others to furnish.  
