1. ## Galton-Walton branching process

In a Galton-Walton branching process, the offspring distribution is given by

$\mathbb{P}(X = r) = p_0 \ if \ r=0, (1-p_0)pq^{r-1}\if\ r=1,2,...$

where $0 < p < 1, q=1-p, and \ 0 < p_0 < 1$

Find the probability of extinction (assuming $N_0=1$)

i did:

find the generating function first G(s).

$G(s)= \sum_{i=0}^{\infty} s^k.p_k$
$= p_0 + (1-p_0)ps\frac{1}{1-qs}$

now to find e so i need to find smallest non negative solution of s=G(s)

$s=p_0 + (1-p_0)ps\frac{1}{1-qs}$

and i get:
$s=\frac{1}{4}\frac{q^2+1-p_0q-p_0^2}{(1-p)^2}$

however the next part of the question asks you to find, supposing $N_0$ has a poisson distribution with mean $\lambda$. Supposing also that $p_0 < q$ the probability of ultimate extinction is

$exp({-\lambda(1-p_0/q)})$

I know that if $N_0=m$ then it will be $s^m$

however i cannot see how to get another distribution into the answer

2. ## Re: Galton-Walton branching process

Hello,

Awww your notations are really awful !!! Pay more attention to what you are writing, you're not making any effort...

First simplify your solution to the first equation, because q=1-p, you simply get that s is 0 or 1.

If $N_0$ follows a $P(\lambda)$, then compute the new generating function, by first conditioning it by $N_0$, then by taking the expectation of the result. That's one basic property of conditional expectations.

3. ## Re: Galton-Walton branching process

How would you condition on $N_0$ to get another generating function?

(To make things clearer I found the following answers to the first bit of the question)

Generating function when $N_0=1$: $G(s)=p_0+\frac{sp(1-p_0)}{1-sq}$

Probability of ultimate extinction in this case: $\frac{p_0}{q}$ if $p_0 < q$ otherwise 1

Probability of ultimate extinction when $N_0=m$: ${\frac{p_0}{q}}^m$ or 1.