= Probability generating function of an independent sum
{title2=$G_{X+Y}(s)=G_X(s)G_Y(s)$}
For nonnegative integer-valued <independent random variables> $X,Y$, factor the <expected value> of a product:
$$
G_{X+Y}(s)=\mathbb E[s^Xs^Y]=\mathbb E[s^X]\mathbb E[s^Y]=G_X(s)G_Y(s),\qquad |s|\leq1.
$$
The <Cauchy product> explains what this means for individual <probability masses>:
$$
\mathbb P(X+Y=n)=\sum_{j=0}^n\mathbb P(X=j)\mathbb P(Y=n-j).
$$
Every way of splitting the total contributes once. This is <convolution of independent random variables>; multiplying <probability generating functions> performs the bookkeeping automatically.
For an original example, take <independent> $X\sim\operatorname{Bernoulli}(0.3)$ and $Y\sim\operatorname{Binomial}(2,0.5)$, using the <Bernoulli distribution> and <binomial distribution>. Then
$$
G_X(s)=0.7+0.3s,\qquad G_Y(s)=\frac{(1+s)^2}{4},
$$
so
$$
G_{X+Y}(s)=\frac{(0.7+0.3s)(1+s)^2}{4}=0.175+0.425s+0.325s^2+0.075s^3.
$$
Direct enumeration of the six pairs $(x,y)$ gives the same <probability masses>. For example, total one collects $(0,1)$ and $(1,0)$, with contributions $0.7\cdot0.5+0.3\cdot0.25=0.425$. The <coefficients> sum to one. The <expected value> is $G_{X+Y}'(1)=1.3$; since $G_{X+Y}''(1)=1.1$, the <variance> is $1.1+1.3-1.3^2=0.71$, agreeing with $0.3\cdot0.7+2\cdot0.5\cdot0.5$.
\Image[/probability-generating-function-coefficient-convolution.png]
{title=Multiplying probability generating functions accounts for every pair with the same total}
{description=The six cells give the exact <joint probability mass function> for the chosen <independent> <Bernoulli distribution> and <binomial distribution>. Colors group equal totals; the stacked bars collect their contributions to the <coefficients> of the product <probability generating function>. These are calculated model probabilities.}
<Uncorrelated random variables> are insufficient for the product identity. For a concrete counterexample, let $U$ be uniform on $\{-1,0,1\}$ and set $A=U+1$, $B=U^2$. Then $\operatorname{Cov}(A,B)=\mathbb E[U^3]=0$, but $\mathbb P(A+B=0)=0$, whereas $G_A(0)G_B(0)=1/9$. Zero <covariance> controls one mixed <moment>, not the full <joint probability mass function>.
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