What is Probability generating function?

Example 23.2 (First reflexes)

A constant variable $X = c$ has $G_X(t) = t^c$; a shift obeys $G_{X+c}(t) = t^c\,G_X(t)$; and evaluating at special points reads off information without any expansion: $G_X(0) = \P(X = 0)$, $G_X(1) = 1$, and $G_X(-1) = \P(X\text{ even}) - \P(X\text{ odd})$, the parity balance exploited in Exercise 23.10. These one-liners are used silently everywhere below — and the evaluation $G_X(0)$ is exactly how extinction probabilities will be extracted from iterated generating functions at the end of the chapter.

Example 23.5 (Integrating the generating function)

Derivatives of $G_X$ at $1$ give positive moments; the integral gives a negative one. From $\int_0^1t^k\dd t = \frac1{k+1}$ and term-by-term integration (normal convergence on $\intcc01$):

For $X \sim \mathcal P(\lambda)$:

recovering in one line the series computation of Example 22.10. The generating function is a two-way instrument: differentiate at $1$ for the moments $\E(X)$, $\E(X(X-1))$, integrate over $\intcc01$ for $\E\bigl(\frac1{1+X}\bigr)$ — one analytic object, queried in whichever direction the problem needs.

Example 23.6 (A law with radius exactly one)

Let $\P(X = k) = \dfrac{6}{\pi^2k^2}$ for $k \geq 1$ — a probability law by Basel’s identity (Example 14.12). Its generating function $G(t) = \frac6{\pi^2}\sum_{k\geq1}\frac{t^k}{k^2}$ has radius of convergence exactly $1$: the general bound “radius $\geq 1$” of Proposition 23.3 cannot be improved. And the mean is

$G$ is continuous on $\intcc{-1}1$, smooth inside, but its derivative blows up at $1^-$ — the graph arrives at the point $(1, 1)$ with a vertical tangent. Heavy tails are visible geometrically on the generating function, at the single point $t = 1$; the moments theorem below makes this correspondence exact.