Generating function of a sum of independent random variables

Statement

Let $X_1, \dots, X_n$ be independent random variables with values in $\{0, 1, 2, \dots\}$, and let $S = X_1 + \dots + X_n$. Then for each complex $z$ with $|z| \le 1$, the probability-generating functions satisfy $$G_S(z) = G_{X_1}(z) \, G_{X_2}(z) \cdots G_{X_n}(z).$$

Remarks

Proof: $z^S = z^{X_1} \cdots z^{X_n}$. The factors are independent, so the product rule for expectations applies. Example: a sum of $n$ independent Bernoulli($p$) variables has the generating function $(1 - p + pz)^n$. This is the generating function of $\operatorname{Bin}(n, p)$. For a random number $N$ of i.i.d. terms $X_i$ with $N$ independent of the $X_i$, the sum $S_N = X_1 + \dots + X_N$ has $G_{S_N}(z) = G_N(G_{X_1}(z))$.

Explanations

L1 human

Adding independent counts corresponds to multiplying their generating functions. This turns the hard computation of the law of a sum into a product of power series.