Consider a branching process $\{ X_n , n \geq 0 \}$ in which the offspring distribution is binomial $(k,p)$. Find probability of ultimate extinction when $k = 3$.
So I've tried this:
$P_k = P(\lambda = t) = {3 \choose t} p^t (1-p)^{3-t} $
$ \phi(s) = \sum_{k=0}^\infty p_k s^k = p_0 s^0 + p_1s^1 +p_2s^2 + p_3s^3 $ (as $k =3$)
$ \phi(s) = (1-p)^3 + 3p(1-p)^2 s + 3p^2(1-p)s^2 + p^3s^3$
I've gotten till here but I'm unsure if I'm on the right track and if right, how do I proceed?
Thanks!
Let $\mu=3p$ be the mean of the offspring distribution. If $\mu\leqslant 1$, then the probability $\pi$ of extinction is one. If $\mu>1$, then $\pi$ is the unique solution to the equation $\varphi(s)=s$ on the interval $(0,1)$. Solving $$ (1-p)^3 + 3p(1-p)^2 s + 3p^2(1-p)s^2 + p^3s^3 = s $$ yields the solutions \begin{align} s &= 1\tag1\\ s &= \frac{2 p^3-3 p^2-\sqrt{4 p^3-3 p^4}}{2 p^3}\tag2\\ s &= \frac{2 p^3-3 p^2+\sqrt{4 p^3-3 p^4}}{2 p^3}.\tag3 \end{align} By inspection, $(3)$ is the correct solution (since it yields numbers between zero and one when $\frac13<p<1$).