Let $\{X_n\}$ be a branching process, that is,
$X_0=k\quad k\in \mathbb{N}$
$X_{n+1}=\sum_{r=1}^{X_n}\xi_r^{(n)}$ where $\xi_1^{(n)}, \xi_2^{(n)}, \cdots $ are non-negative i.i.d. random variables having distribution $P(\xi=i)=p_i$
Let $\phi_n(s)$ and $\phi(s)$ be p.g.f.s of $X_n$ and $\xi$, respectively.
Then we have to show that $\phi_n(s)=\phi_{n-1}[\phi(s)]$ and $\phi_{n}(s)=\phi[\phi_{n-1}(s)]$
The proof in the text I am following given as
For $n\ge 1,$
$\begin{array}{ll}
\text{we have } \phi_n(s) & =E[s^{X_{n}}]\\
& = \sum_{j=0}^\infty s^j P(X_n= j) \\
& = \sum_{j=0}^\infty \sum_{i=0}^\infty s^j P(X_n= j \>|\> X_{n-1}=i) P(X_{n-1}=i)\\
& = \sum_{j=0}^\infty \sum_{i=0}^\infty s^j P(\xi_1^{(n-1)}+\xi_2^{(n-1)}+\cdots +\xi_i^{(n-1)}= j) P(X_{n-1}=i)\\
& = \sum_{i=0}^\infty P(X_{n-1}=i) \cdot (\sum_{j=0}^\infty s^j P(\xi_1^{(n-1)}+\xi_2^{(n-1)}+\cdots +\xi_i^{(n-1)}= j)\\
& = \sum_{i=0}^\infty P(X_{n-1}=i) \cdot (\sum_{j=0}^\infty s^j P(\xi_1^{(n-1)}+\xi_2^{(n-1)}+\cdots +\xi_i^{(n-1)}= j))\\\\
& [\text{Since }\xi_1^{(n-1)}, \cdots , \xi_i^{(n-1)}\text{ are i.i.d. r.v.s p.g.f. of their sum}\\
&\text{would be product of their p.g.f.s]}\\
& = \sum_{i=0}^\infty P(X_{n-1}=i) \cdot (\phi(s))^i\\
& = E[(\phi(s))^{X_{n-1}}]\\
& = \phi_{n-1}(\phi(s))
\end{array}$
Now I am confused when $X_{n-1}=0$, then the expression $P(\xi_1^{(n-1)}+\cdots +\xi_0^{(n-1)})$ does not make any sense as the subscripts would only start from $1$.
We can argue that $P(X_n=j|X_{n-1}=0)=0$ since there is no one in the $(n-1)$-th generation to produce offspring and we can take the summation from $i=1$
But in that case, without starting from $i=0$, how can we say that the expression gives us the pgf in the end?
Also, it seems like the text only showed the proof for $X_0=1$ and not proved it for $X_0>1$
How can we prove that? Or will there be different relation for $X_0>1$?
