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?

  • $\begingroup$ Given that you let $Y= \lim_{n\to\infty} Y_n,$ the p.g.f. for $Y_n$ should be denoted $\operatorname E(s^{Y_n})$ rather than $\operatorname E(s^Y). \qquad$ $\endgroup$ Mar 25, 2018 at 20:56

1 Answer 1


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.

  • $\begingroup$ Why is it that : $T_1,T_2,...,T_k $ have " the same distribution of $ Y_{n−1}$" $\endgroup$
    – geoffrey
    Mar 28, 2022 at 20:25

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