Galton-Walton branching process

**FGT12** · July 30th 2011, 05:20 AM

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

**Moo** · July 30th 2011, 12:57 PM

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.

**FGT12** · August 11th 2011, 06:52 AM

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.

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