Define a stochastic process ($X_n)_{n≥0}$ so that $X_n$ represents the number of individuals in the population at time $n$. Each generation gives birth to a random number of offspring that forms the next generation where the offspring distribution alternates between consecutive generations.Hence, with $X_0 = 1$, then $X_{n+1}$ = $\sum_{k=1}^{X_{n}}$ = $Z_k^{n}$ where $Z_k^{n}$ ∼ {$Z_1$ if $n$ is odd, and $Z_2$ if $n$ is even.} where $Z_1, Z_2$ are RVs on $N$ such that $E(Z_1)$ = $µ_1 < ∞$ and $E(Z_2) = µ_2 < ∞$.
a)Define $F_n(s) = E_1[s^{X_n}]$ and $G_i(s) = E_i[s^{Z_i}]$.
Show that for $0 ≤ s < 1$: $F_{2n}(s) = G_2(G_1(F_{2n-2}(s)))$
b)Suppose that $Z_1 ∼ Po(5)$ and $Z_2 ∼ Po(µ_2)$. Give a range of values for $µ_2$ that guarantee extinction for the population.
For a) I have a proof for when there is only one PGF for $Z_k$ but not for two and I am unable to adapt this.
For b) I yet again know how to do this with one PGF, but am unsure how to use that here.
1.)
Really this comes down to manipulating the tower property of conditional expectations. In substance this is the same thing you do with a 'regular' branching process, except this is an alternating process so you ultimately want a look the whole process 2 at a time. Since this is a recurrence that has $X_0$ we should stake a look at the beginning for a base case.
$X_{n+1}=\sum_{k=1}^{X_{n}}Z_k^{(n)}$
so
$X_{2}=\sum_{k=1}^{X_{1}}Z_k^{(1)}$ and
$X_{1}=\sum_{k=1}^{X_{0}}Z_k^{(0)} =Z_1^{(0)}$
note I assume all $Z_k^{(n)}$ are independent (and indeed iid for common n mod 2) as this is how branching processes (esp with this generating function) are structured. Technically this wasn't written in OP.
$F_{2n}(s) = E\Big[s^{X_{2n}}\Big]= E\Big[E\big[s^{X_{2n}}\big \vert X_{2n-1}\big]\Big] $
and we focus on the r.v.
$E\big[s^{X_{2n}}\big \vert X_{2n-1}\big] $
$= E\Big[s^{\sum_{k=1}^{X_{2n-1}}Z_k^{(2n-1)}}\big \vert X_{2n-1}\Big] $
$= E\Big[\prod_{k=1}^{X_{2n-1}} s^{Z_k^{(2n-1)}}\big \vert X_{2n-1}\Big] $
$= E\big[s^{Z_k^{(2n-1)}}\big]^{X_{2n-1}} $
i.e. considering $X_{2n-1}=r $ this reads
$ E\Big[\prod_{k=1}^{X_{2n-1}} s^{Z_k^{(2n-1)}}\big \vert X_{2n-1}=r\Big]= E\Big[\prod_{k=1}^{r} s^{Z_k^{(2n-1)}}\big \vert X_{2n-1}=r\Big] = E\Big[\prod_{k=1}^{r} s^{Z_k^{(2n-1)}}\Big] = E\Big[ s^{Z_k^{(2n-1)}}\Big]^r $
where for a fixed superscript, the $Z_k$ are iid. Now letting $t:=E\big[s^{Z_k^{(2n-1)}}\big] = G_1^{(2n-1)}(s)$
this reads
$F_{2n}(s) = E\Big[s^{X_{2n}}\Big]= E\Big[E\big[s^{X_{2n}}\big \vert X_{2n-1}\big]\Big]= E\Big[E\big[t^{X_{2n-1}}\big \vert X_{2n-1}\big]\Big]= E\Big[t^{X_{2n-1}}\Big]=E\Big[G_1^{(2n-1)}(s)^{X_{2n-1}}\Big] $
since $F_{0}(s)=s$, in terms of base case, consider $n=1$, we should get
$F_{2}(s) = G_2(G_1(F_{0}(s))) = G_2(G_1(s)) = G_2(E[s^{Z^{(1)}}]^{X_1})= E\big[E[s^{Z^{(1)}}]^{Z^{(0)}}\big]$
but there is an awful lot of symbol manipulation here with many different indices so there's a good chance of a bug or two.
To finish this off, re-run the argument for $F_{2n-1}(s)$ and induct backward
2.)
application of the chain rule tells you that
$\frac{d}{ds}F_{2n}(s) =G_2'(s)\cdot G_1'(s)\cdot \frac{d}{dx}F_{2n-2}(s)$
and to get the expected value evaluate at
$\frac{d}{ds}F_{2n}(s)_{\substack{s\uparrow 1}}$
(Abel's theorem in the background)
Considering the case of $n=1$ i.e. $2n=2$ gives you a base case value of $\mu_1\cdot \mu_2$ and you can guess the result and justify with induction from there.
Really the mean is enough to justify extinction probabilities for any $Z_i$ defined on the natural numbers so long as they have non-zero probability of being zero.
The Markov chain view (taken 2 generations at a time) is that you necessarily have a countable state transient chain since the chain is connected and state zero is absorbing. So all probabilities are either absorbed in state zero, or they run off to $\infty$. But if the mean of the process is $\leq 1$, you can apply Markov inequality (or better, but more work Kolmogorov upcrossing inequality) to show that the probabilities of $X_{2n}$ becoming arbitrarily large is arbitrarily small. A different view specializes to a mean $\lt 1$ and direct application of markov inequality pinches $Pr(X_{2n}\geq \delta) \leq \frac{(\mu_1\cdot \mu_2)^{2n}}{\delta}$ for any $\delta \gt 0$ which is arbitrarily small for large enough $n$. The case of mean = 1 being extinction WP1 then is implied by continuity of the generating function (and the fact that the extinction probability =1 is a closed set). There are several other ways to view this of course but the point is that the mean gives an awful lot of information in this case.