Mean and extinction probability of a Galton-Watson branching with pmf of offspring produced $P(Q=q) = (q+1)(1-r)^2r^q, 0Initial population is $X_0 = g$, ($g$ being a positive number or $0$) and the probability mass function of the number of offsprings $(q)$ produced by an individual is $P(Q=q) = (q+1)(1-r)^2r^q, 0<r<1$.
I'm trying to calculate the expected value of $X_n$ and the extinction probability. I'm stuck on both but here's how far I got.
Mean:
$E[X_n] = E[f(q)]^q(g)$ (I'm using a known formula for this. let me know if I've used it wrong). Assuming $X_0 =g $ isn't $0$, we will have to calculate:
$$E[f(q)] = \Sigma^\infty_{q=1} qP(Q=q) = \Sigma^\infty_{q=1} q(q+1)(1+r)^2r^q$$
Is the upper limit of the sum here correct? Should it be $\infty$, or $g$ as we are starting with $g$ people in the population
Extinction probability $(\pi_0)$: Assuming that my $E[f(q)]>1 \implies \pi_0 = \Sigma^\infty_{q=1} \pi^q_0P(Q=q)$.
$\pi^q_0$ being the probability that the population dies out given $X_0 = q$. This gives me:
$$\Sigma^\infty_{q=1} \pi^q_0(q+1)(1+r)^2r^q$$
In both these cases I have no idea how to proceed further. This isn't a distribution that I recognize. Is there something I'm missing? Did I do a step wrong? Or is there an easier way to approach this that I am not seeing.

It is straightforward to show that if the offspring distribution has finite mean, that is,
$$
\mathbb E[Q] = \sum_{k=0}^\infty k\cdot\mathbb P(Q=k) :=\mu <\infty
$$
then the expected population size at time $n$, conditioned on $\{X_0=1\}$ is given by
$$
\mathbb E[X_n\mid X_0=1] = \mu^n.
$$
If $g$ is a positive integer, then with some additional work we see that
$$
\mathbb E[X_n\mid X_0=g] = g\cdot\mu^n.
$$
(The intuition is that the process is equivalent to $g$ separate processes each starting with one individual.) We compute the mean of $Q$:
$$
\mu = \sum_{k=0}^\infty k\cdot(k+1)(1-r)^2 r^k = \frac{2r}{1-r},
$$
and hence
$$
\mathbb E[X_n\mid X_0=g] = g\cdot\left(\frac{2r}{1-r}\right)^n.
$$
For the extinction probability, I will only consider the case where $g=1$. Let
$$
\tau = \inf\{n>0:X_0=0\}.
$$
It is known that $\pi:=\mathbb P(\tau<\infty)=1$ if $\mu\leqslant1$ and is a positive number less than one if $\mu>1$. Since $0<r<1$, it is clear that
$$
0<\frac{2r}{1-r}\leqslant 1 \iff 0<r\leqslant\frac13,
$$
and so extinction occurs with probability one if $r\leqslant\frac 13$. If $\frac13<r<1$, then it is well known that $\pi$ satisfies the equation $P(\pi)=\pi$, where $P(\cdot)$ is the probability generating function of $Q$; indeed, $\pi$ is the unique solution to this equation on the interval $(0,1)$. Let $P(s):= \mathbb E[s^Q]$ for $s\in[0,1]$, then
$$
P(s) = \sum_{k=0}^\infty (k+1)(1-r)^2 r^ks^k = \left(\frac{1-r}{1-rs}\right)^2.
$$
The equation $P(\pi)=\pi$, i.e.
$$
\left(\frac{1-r}{1-r\pi}\right)^2 = \pi
$$
is a cubic, and so has three solutions:
\begin{align}
\pi &= \frac{2r-r^2-\sqrt{4 r^3-3 r^4}}{2 r^2}\tag1\\
\pi &= \frac{2r-r^2+\sqrt{4 r^3-3 r^4}}{2 r^2}\tag2\\
\pi &= 1\tag3.
\end{align}
By inspection, we see that $(1)$ is the correct choice, since it yields numbers between zero and one.
