TITLE: Recurrence relation in the total progeny of a branching process
QUESTION [6 upvotes]: Let $Y_{n}= Z_0+Z_1+\cdots+Z_n$ model a branching process as the total number of individuals up through generation $n$.  The total progeny can be described as $Y = \lim_{n \to \infty} Y_n$.
In order to analyze expected generation size, we can follow 
$$\lim_{n \to \infty} E(Z_n) = \lim_{n \to \infty}\mu^n= \begin{cases}
0, & \mu<1,\\
1, & \mu=1,\\
\infty, & \mu>1.
\end{cases}$$
It can be shown that for the critical case where $\mu =1$, 
$$E(Y) = \sum_{i=0}^n E(Z_i)=\infty.$$
Now let $\psi_n(s)= E(s^{Y_n})$ be the probability generating function of $Y_n$.
How would one show that $\psi_n$ satisfies the recurrence relation 
$\psi_n(s)=sG(\psi_{n-1}(s))$ for $n=1,2,\ldots,$
where $G(s)$ is the probability generating function of the offspring distribution?

REPLY [5 votes]: Let $Z_0 = 1$, then $Y_n$ conditioned on ${Z_1 = k}$ is the sum of  the total progeny of each child $(1,...,k)$:
$Y_n = 1 + \sum_{i=1}^k T_{i}.$
$T_{1}, T_{2},..., T_{k}$ are independent and identically distributed random variables with the same distribution of $Y_{n-1}$
$\psi_n(s)= \sum_{k=o}^\infty p_kE[s^{y_n} | Z_1 = k]  = \sum_{k=o}^\infty p_kE[s^{1 + \sum_{i=1}^k T_{i}}] = s\sum_{k=o}^\infty p_kE[\prod_{i=1}^k s^{T_{i}}]$
$= s\sum_{k=o}^\infty p_k[\psi_{n-1}(s)]^k =sG(\psi_{n-1}(s))$
The first equality comes from $E[X] = E[E[X|W]]$, the fourth from i.i.d.