Suppose that $Y_{n;k}\sim\text{Bin}(2,p)$, where $1/2<p<1$ is a branching process. What is the probability of extinction?
I know that to find the probability of extinction, we use the generating function as $\mathcal{P}_Y(q)$ where $q$ is the probability of extinction and $\mathcal{P}(t) = [(1-p) + tp]^2$. My problem is how to correctly solve this. I thought of this $X_n = \sum_{k=1}^{X_1}Y_{n,k} = Y_{2,1} + Y_{2,X_{1}}$ but I don't how to proceed with this.
By definition $X_n = \sum_{k=1}^{X_{n-1}} Y_{n,k}$
Observe that $E[X_n] = m^n = (2p)^3 $ where $m = (2p) > 1$, thus, $\pi_0 < 1.$ So we find $\pi_0.$ \begin{align*} \pi_0 & = \sum_{j=0}^{\infty} \{\text{Population dies out $| X_1 = j$}\}\times p_j\\ & = \sum_{j=0}^{3} \pi_0^j \times p_j \\ 0 & = p^2\pi^2_0 + (2p-2p^2 - 1)\pi_0 + (1-p)^2 \end{align*} This is a quadratic equation. Thus, \begin{align*} \pi_0 & = \frac{(2p^2-2p + 1) \pm \sqrt{4p^2 - 4p +1}}{2p^2}\\ & \text{since } p \in (1/2, 1), \text{we let } p = 0.7.\\ & = \frac{0.58\pm 0.4}{0.98}\\ & = (0.18367, 1) \end{align*} We observed that $\forall$ $ p \in (1/2,1)$, $\pi_0$ has the smallest possible solution for which $\pi_0 < 1$. Thus the required probability is $$\pi_0 = \frac{(2p^2-2p + 1) - \sqrt{4p^2 - 4p +1}}{2p^2}.$$