Galton-Walton branching process

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• Jul 30th 2011, 05:20 AM
FGT12
Galton-Walton branching process
In a Galton-Walton branching process, the offspring distribution is given by

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

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

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

i did:

find the generating function first G(s).

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

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

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

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

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

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

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

however i cannot see how to get another distribution into the answer
• Jul 30th 2011, 12:57 PM
Moo
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 $\displaystyle N_0$ follows a $\displaystyle P(\lambda)$, then compute the new generating function, by first conditioning it by $\displaystyle N_0$, then by taking the expectation of the result. That's one basic property of conditional expectations.
• Aug 11th 2011, 06:52 AM
FGT12
Re: Galton-Walton branching process
How would you condition on $\displaystyle N_0$ to get another generating function?

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

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

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

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