Consider a Galton Watson process:
Let $Z_n$ be the number of the individuals in $n$'th generation . ( $Z_0 := 1 $) Let $X_{n,i}$ be the number of the offspring ( generation $n+1$ ) of the i'th individual in the n'th generation. The random variables $X_{n,i}$ are independent and identically distributed. For $ n \in \mathbb{N} $ we define:
$Z_n := \sum_{i=1}^{Z_{n-1}} X_{n-1,i} $
Let $g$ be the generating function of $X_{0,1}$ and $g_n$ the generating function of $Z_n$. We will write: $g = g^{(1)}$ and$ g^{(n)} = g^{(n-1)} \circ g $.
Show:
a) $g_n(t) = g^{(n)}(t)$
b)$ P( \exists n \in \mathbb{N} | Z_n = 0 ) = \lim_n g^{(n)}(0)$ , P is the extinction probability.
c) Calculate the extinction probability for:
$X_{0,1} := 0 $ with probability $1-p$ and $X_{0,1} := 2 $ with probability $p$.
So: I solved a) and b). but I'm stucked in c) in this exercise. Obviously we have to use b). That means that we have to compute $ \lim_n g^{(n)}(0)$. So what we have to do? First find $g$, then consider $g^{(n)}(t)$ then $g^{(n)}(0)$ and finally all we have to do is calculate $ \lim_n g^{(n)}(0)$. So first of all $X_{0,1}$ is bernoulli distributed. I think that $g = (1-p) + p \cdot t^2 $. Is this true? And if is: Im stucked to find $g^{(n)}$. Here is my problem.