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?

Thanks for your time.

K. Kamal

  • 3
    $\begingroup$ To be hones I don't understand the sentence "with P1 is the probability generating function Qn of the family-size distribution". Could you write a more formal definition? $\endgroup$ – Kore-N Mar 13 at 11:29
  • $\begingroup$ 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. $\endgroup$ – Augusto Santos Mar 16 at 20:59
  • $\begingroup$ 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. $\endgroup$ – Augusto Santos Mar 16 at 21:05

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