What is the basic lemma on composition of probability generating functions and how is it most clearly proved?
(I'm posting this mainly to see if I can write an answer as clearly and simply as possible, but maybe other people know things about this that I don't and can post answers from their points of view.)
On
Suppose a random variable $X$ takes values in the set $\{0,1,2,3,\ldots\},$ each finite cardinal number having a probability assigned to it, their sum being $1.$
The ordinary generating function of this sequence of probabilities is $$ g_X(s) = \sum_{x=0}^\infty \Pr(X=x) s^x = \operatorname E(s^X). $$ It is routine to see that if $X,Y$ are independent random variables then \begin{align*} & g_{X+Y}(s) = \operatorname E(s^{X+Y}) \\[8pt] = {} & \operatorname E(s^X) \operatorname E(s^Y) & & \text{by independence} \\[8pt] = {} & g_X(s) g_Y(s). \end{align*} (This simplicity of this argument is something that I actually missed in a recent stackexchange posting because my attention at that moment was on other aspects of my topic.)
Lemma: Suppose $X$ is as above and $Y_1,Y_2,Y_3,\ldots$ are identically distributed, also taking values in $\{0,1,2,3,\ldots\}$ and $X, Y_1,Y_2, Y_3,\ldots$ are independent. Let $g_Y$ denote the probability generating function of any of $Y_1,Y_2,Y_3,\ldots\,.$ Let $$ Z = \sum_{n=1}^X Y_n. $$ Then $g_Z(s) = g_X(g_Y(s)).$
Proof: \begin{align} g_Z(s) & = \operatorname E(s^Z) = \operatorname E(\operatorname E(s^Z \mid X)) \\[8pt] & = \operatorname E( \operatorname E(s^{Y_1+\cdots+Y_X} \mid X)) \\[8pt] & = \operatorname E( g_Y(s)^X ) & &\text{(This follows from} \\ & & & \phantom{\text{(}}\text{the “routine'' thing} \\ & & & \phantom{\text{(}}\text{mentioned above.)} \\[8pt] & = g_X(g_Y(s)). \end{align}
Do you mean this? Let $S = \sum_{i\le N} Y_i$ where $N, Y_1, Y_2, \ldots$ are independent integer-valued random variables where $N\ge 0$ a.s., and $\mathbb E[z^N] = f(z)$ while $\mathbb E[z^{Y_i}] = g(z)$, then $$ \mathbb E[z^S] = f(g(z))$$