Galton Watson and generating function

Question

In the treatment of the Galton-Watson process one usually introduces the "generating function" of the random variable $Z_n$ as $\psi_{Z_n}(s)=E[s^{Z_n}]$. Why would $\psi_{Z_n}(0)=P(Z_n=0)$? It seems to me that this identity being true would hinge on $0^0=1$ and $0^k=0$ for $k\geq 1$. Is it usually assumed that $0^0=1$ in this context? Is there another way to see this?

Answer 1

$Es^{X}$ is always interpreted as $P(X=0)+sP(X=1)+s^{2}P(X=2)+\cdots$. This is the correct definition of generating function and it avoids the indeterminate form $0^{0}$. When $s=0$ the sum becomes $P(X=0)$.

Answer 2

In this context, $s^0$ uses $0$ as a monomial index and hence a non-negative integer, and the $s=0$ case is just the empty product $1$. (You can think of this as the hypervolume of a $0$-dimensional cube of side length $s$; in $0$ dimensions, hypervolume just counts the single point present.) The "$0^0$ is an indeterminate form" concern is a separate issue that results when the exponent can vary continuously, e.g. $\lim_{x\to0^+}x^x\ne\lim_{x\to0^+}(e^{-1/x^2})^{x^2}$.