# Mean and extinction probability of a Galton-Watson branching with pmf of offspring produced $P(Q=q) = (q+1)(1-r)^2r^q, 0<r<1$

Initial 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.

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 $$0, it is clear that $$0<\frac{2r}{1-r}\leqslant 1 \iff 0 and so extinction occurs with probability one if $$r\leqslant\frac 13$$. If $$\frac13, then it is well known that $$\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.