How is this probability generating function in regard to branching obtained Suppose we have a branching process $\{X_{n} : n=0,1,2,.. \}$ with $X_{0}=1$, pgf of these $\phi_{X_{1}}(z)=\phi(z)$
and 
$S_{n}=X_{0}+X_{1}+...+X_{n}$ , which we assume has pgf $\Phi_{n}(z)$
How to show that 
$\Phi_{n+1}(z)=z \phi[ \Phi_{n}(z)]$
My issue is that I dont understand where the z on the right hand side would be coming from. Also, is writing $\Phi_{n}$ equivalent to just writing $\Phi_{Sn}$  if so
$\Phi_{n} (z)=E[z^{Sn}]$
by total expectation
$= E_{X_{1}}[E[z^{Sn}|X_{1}]$
The main thing that I have been trying to work out is how knowing $X1$ allows us to rewrite $S_{n}$ and then how the generating functions represent that.
It is important to realise that this is not talking about how many are in the nth generation, but how many total, ie the sum of all the generations up to and before it
And yes it does seem that the $z$ factor will be as $X_{0}$ is fixed and hence indepdent of any generation size. But I still dont know how to use the generating functions to achieve the full result
so can anyone help me to see where I am going wrong and or where I am making a mistake in my understanding? Please let me know if there is some info that isnt clear
Thanks
 A: There are two ways to do the recursion.
I describe here the top-bottom approach.
For the bottom-up approach you may look at the comment above by Didier.
The important point is that given $X_j$ then $X_{j+1}$ comes
from a sum of $X_j$  independent copies of branchings
each following the law of $X_1$. In terms of  an auxiliary variable $s$ we may express this as follows:
$$ E\left(s^{X_{j+1}} \ | \ X_j\right) =
\phi(s)^{X_{j}}
$$
Let us write $\Phi_0(z)=z$ and for $k\geq 0$: 
$$\Phi_{k+1}(z) \ =z \ \phi \left[ \Phi_k(z) \right].$$
I then claim that for $n\geq k \geq 0$ we have the (inductive)  identity
(we set $S_{-1}=0$):
$$ E\left(z^{S_{n-k}} \Phi_{k}(z)^{X_{n-k+1}} \right) =
 E\left(z^{S_{n-k-1}} \Phi_{k+1}(z)^{X_{n-k}}\right) 
$$
To see this note that by conditioning on $X_{n-k}$ and using the first identity above with $s=\Phi_{k}(z)$ and $j=n-k$, the LHS equals:
$$ E\left(E\left(z^{S_{n-k-1}} \; 
 z^{X_{n-k}}\Phi_{k}(z)^{X_{n-k+1}}\  |\ X_{n-k}\right)\right) 
= E\left(E\left(z^{S_{n-k-1}} \; 
\ z^{X_{n-k}}\phi\left[ \Phi_{k}(z)\right]^{X_{n-k}}\  
|\ X_{n-k}\right)\right) 
$$
which reduces to the RHS of the previous equation given the recursive definition of $\Phi_{k+1}$. Thus,
carrying out the induction:
$$
E\left(z^{S_{n+1}}  \right) =
 E\left(z^{S_{n}} \Phi_{0}(z)^{X_{n+1}}\right)  = ... =
 E\left(z^{S_{-1}} \Phi_{n+1}(z)^{X_{0}}\right) =\Phi_{n+1}(z)
$$
A: For fixed $z$ define the map $g_z(w)=\phi(zw)$. For fixed but arbitrary $n$ we show that 
\begin{align*}
\tag{1} E(z^{S_{n+1}})=E(z^{S_{n+1-k}}(g_z^{(k)}(1))^{X_{n+1-k}})
\end{align*}
for all $k\leq n+1$, where $g_z^{(k)}$ is the $k$-fold composition of $g_z$. $g_z^{(0)}$ should be interpreted as the constant map identically equal to $1$. The $k=0$ case holds by definition. 
Assume $(1)$ holds for some $k\leq n$. Then, 
$
\begin{align*}
E(z^{S_{n+1}})&=E(z^{S_{n+1-k}}(g_z^{(k)}(1))^{X_{n+1-k}}) \\
&=E(E(z^{S_{n+1-k}}(g_z^{(k)}(1))^{X_{n+1-k}}\mid X_1,\dots, X_{n-k})) \\
&=E(z^{S_{n-k}}E(z^{X_{n+1-k}}(g_z^{(k)}(1))^{X_{n+1-k}}\mid X_1,\dots, X_{n-k})) \\
&=E(z^{S_{n-k}}\mid E((zg_z^{(k)}(1))^{X_{n+1-k}}\mid X_1,\dots, X_{n-k})) \\
&=E(z^{S_{n-k}}\phi(zg_z^{(k)}(1))^{X_{n-k}}) \\
&=E(z^{S_{n-k}}g_z^{(k+1)}(1)^{X_{n-k}})
\end{align*}
$
so formula $(1)$ holds also for $k+1$, showing that it holds true in general. We may therefore set $k=n+1$ in $(1)$ and use the fact that $S_0=X_0=1$ to get the equation $$E(z^{S_{n+1}})=zg_z^{(n+1)}(1)$$ which holds for all $n$. Note that this is actually the explicit solution.
Finally,
$
\begin{align*}
z\phi(E(z^{S_n}))&=z\phi(zg_z^{(n)}(1)) \\
&=zg_z^{(n+1)}(1) \\
&=E(z^{S_{n+1}}).
\end{align*}
$
