# Branching process probability generating function

I'm trying to solve the following exercise but I can't seem to solve it.

A branching process $$(X_n :n \geq 0)$$ has $$P(X_0 = 1) = 1$$. Let the total number of individuals in the first $$n$$ generations of the process $$Z_n$$, with probability generating function $$Q_n$$. Prove that for $$n \geq 2$$

$$Q_n(s) = sP_1(Q_{n-1}(s))$$ with $$P_1$$ is the probability generating function $$Q_n$$ of the family-size distribution.

What I immediately understood, was that we have to relate $$Z_n$$, the number of individuals in the first $$n$$ generations, to $$Z_{n-1}$$, which is the number of individuals in the first $$n-1$$ generations. Logically: $$Z_n = Z_{n-1} + C$$ , where we call $$C$$ the number of individuals in the $$n$$'th generation. Note that $$Z_{n-1}$$ and $$C$$ are independent.

$$Q_n(s) = E(s^{Z_n}) = E(s^{Z_{n-1}+C}) = E(s^{Z_{n-1}})E(C) = Q_{n-1}(s)E(C)$$.

I believe this is not what I should be doing, does anyone have any hints on how to attack this problem?

• If $X_n$ is the Galton–Watson process with $X_{n+1}=\sum_{j=1}^{X_n}\xi_j^{(n)}$ and $\xi_j^{(n)}$ i.i.d.; $Z_n:=\sum_{i=0}^n X_i$; $Q_n(s):=E\left[s^{Z_n}\right]$; and $P_1(s):=E\left[s^{\xi}\right]$, then the relationship $Q_n(s)=s P_1(Q_{n-1}(s))$ makes no sense. This relationship is telling us, in words, that the individuals at generation $n$ are not only descendant from individuals at generation $n-1$, but also from all individuals $Z_{n-1}$ comprising all previous generations. – Augusto Santos Mar 16 at 20:59
• More formally, under the assumption that your process $(X_n)_{n\geq0}$ is rather defined as $X_{n+1}=\sum_{j=1}^{Z_{n}}\xi_j^{(n)}$, then the relation holds. Please, define the processes in your question and/or point to a reference. – Augusto Santos Mar 16 at 21:05