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Math Help - Galton-Walton branching process

  1. #1
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    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
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  2. #2
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    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.
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    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.
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