Let be $\{Z_n\} $ a Galton Watson process.
If the distribution of the quantity of the descendants $B(2,p),p>0.5$ and the distribution of $Z_0$ is Pois($\lambda$), how can I calculate the probability that the population will be died?
I only know that
$$var(Z_n|Z_0=1)=\begin{cases} \sigma^2 \mu^{n-1} \frac{\mu^n-1}{\mu-1} &, \mu\neq 1\\ n\sigma^2 &, \mu=1 \end{cases}. $$
The probability of extinction is the smallest positive root of $$G_O(z)=z$$ Where $O$ denotes the offspring distribution, and $G_O(z)$ its generating function at $z$.
It is easily seen that $G_O(0)$ is the probability of extinction in the first generation.
Second, if you know about generating functions, then you know that the sum: , where $X$ is independent from the iid $Y_j$'s: $$ Z:=\sum_{j=1}^X Y_j \text{ has generating function } G_Z(z)=G_X(G_Y(z)) $$ Thus $G_{Z_n}(z)=\underbrace{G_O\circ\ldots\circ G_O}_{n \text{ times}}z)$. Also, $G_O^{(n)}(z)$ is increasing - and strictly increasing if $\exists m>n$ such that $\mathbb{P}(O=m)>0$. So we iterate $G_0(z)$, starting from $0$. - It can be verified that it will converge to the smallest fixed point of $G_O(z)$.
So all you have to do is calculate $$G_O(z)=\mathbb{P}(Bin(2,p)=0)+\mathbb{P}(Bin(2,p)=1)z+\mathbb{P}(Bin(2,p)=2)z^2=$$ $$ (1-p)^2+2p(1-p)z+p^2z^2=z\Leftrightarrow (z-1)\left(z-\frac{(p-1)^2}{p^2}\right) $$ Now since $p>\frac{1}{2}$, $\left|\frac{p-1}{p}\right|<1$, thus the result is $$ \lim_{n\to\infty}\mathbb{P}(Z_n>0)=\frac{(p-1)^2}{p^2} $$