@automaticallyGenerated has helped me to understand a difficult problem I was struggling with. I want to do another question to make sure that I have understood the concept correctly. Please check my work, especially (c). I am pretty sure (a), (b) are correct. Thanks in advance!
Let 0 ≤ p ≤ $\frac{1}{3}$. Consider a branching process where the number X of offspring of an individual is zero with probability $p$, one with probability $2p$, and two with probability $1 − 3p$. Initially there is one individual.
(a) For what values of p will the branching process become extinct after a finite number of generations with probability 1? Explain.
(b) For the case $p = \frac{1}{5}$, calculate the probability that the branching process survives forever.
(c) Assume again that $p = \frac{1}{5}$. Given that there are three individuals in the second generation, find the probability that the branching process survives forever.
My answer:
(a) Calculating the mean of the first generation, $\mu$,the process goes extinct when $\mu \leq 1$. $\mu = 2p + 2 \cdot (1-3p)$. Therefore, when $\frac{1}{4} \leq p \leq \frac{1}{3}$, the process goes extinct.
(b) The generating function of the branching process is $G(s) = p + 2p\cdot s + (1-3p)s^2$. Using $s = G(s)$, I've got P(extinction) = 0.5. Therefore, the probability that the process survives forever = 0.5.
(c) $P(\text{three individuals' branching processes not dying off}) = 1 - (1-\frac{1}{2})^3 = \frac{7}{8}$.
Let the probability of the branching process survive forever be $x$. It can be seen that $$x = 0*p + x*(2p) + (1-3p)*(1-(1-x)^2)$$ as there is $0$ probability of it surviving if there are $0$ offspring, $x$ probability if there is $1$ offspring, and $1-(1-x)^2$ if there are $2$ offspring (at least one has to stay alive).
Simplifying: $$x = 2px+(1-3p)(2x-x^2) = 2px+2x-x^2-6px+3px^2$$
Solving for $x$: $$(3p-1)x^2+(-4p+1)x=0 \rightarrow x = 0, \frac{4p-1}{3p-1}$$
For the branching process to end with probability $1$, $\frac{4p-1}{3p-1}$ must be $ \le 0$. This means that $$\frac{1}{4} \le p \le \frac{1}{3}$$ which is what you got.
For (b), simply plug in $p = \frac{1}{5}$ into the above formula, giving $\frac{1}{2}$ probability of surviving forever.
For (c), simply find the complement of the probability of all three processes dying off, giving $$1-(1-\frac{1}{2})^3 = \frac{7}{8}$$