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Math Help - PGF of a branching process (very hard)

  1. #1
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    PGF of a branching process (very hard)

    Let X_0, X_1, ... be a branching process with \mathbb{P} (X_0 = 1) = 1, and offspring distribution \mathbb{P} (Y= k ) = qp^k \ \mbox{for k} = 0,1,2,... where q = 1- p \ \mbox{and} \ p \in (0,1).

    (1) Prove that the pgf  G_n \ \mbox{of} \ X_n is

    G_n (s) = \begin{cases} \frac{n-(n-1)s}{n+1-ns} & p = q \\ \frac{q(p^n-q^n-ps(p^{n-1} - q^{n-1}))}{p^{n+1} - q^{n+1} - ps(p^n - q^n)} & p \neq p \end{cases}

    and

    (2) Determine (explicitly) the pdf  H_n of  X_n conditioned on eventual extinction.

    For part two, I have seen that the probability of eventual extension, \beta, is given by

    \beta = I_{\{p \leq \frac{1}{2} \}} + \alpha I_{\{p > \frac{1}{2}\}} \ \mbox{where} \ \alpha = \frac{q}{p}
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  2. #2
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    Quote Originally Posted by RoyalFlush View Post
    Let X_0, X_1, ... be a branching process with \mathbb{P} (X_0 = 1) = 1, and offspring distribution \mathbb{P} (Y= k ) = qp^k \ \mbox{for k} = 0,1,2,... where q = 1- p \ \mbox{and} \ p \in (0,1).

    (1) Prove that the pgf  G_n \ \mbox{of} \ X_n is

    G_n (s) = \begin{cases} \frac{n-(n-1)s}{n+1-ns} & p = q \\ \frac{q(p^n-q^n-ps(p^{n-1} - q^{n-1}))}{p^{n+1} - q^{n+1} - ps(p^n - q^n)} & p \neq p \end{cases}

    and

    (2) Determine (explicitly) the pdf  H_n of  X_n conditioned on eventual extinction.

    For part two, I have seen that the probability of eventual extension, \beta, is given by

    \beta = I_{\{p \leq \frac{1}{2} \}} + \alpha I_{\{p > \frac{1}{2}\}} \ \mbox{where} \ \alpha = \frac{q}{p}
    The pgf of X_1 (or Y, equivalently) is easily found to be f(s)=G_1(s)=\frac{q}{1-ps}.
    Then you probably know that the pgf of X_2 is f(f(s)), etc., in general G_{n+1}(s)=f(G_n(s)). The formulas from the text can then be verified by induction; it's just tedious computation.

    As for (2), your computation of the extinction probability is correct. When p\leq 1/2, the probability is 1, hence conditioning to extinction is like doing nothing: H_n=G_n ( H_n being the pgf of X_n conditioned to extinction). When p>1/2, this probability is positive ( \alpha=\frac{q}{p}). In this case, you can deduce the pgf of the offspring distribution of the process conditioned to extinction: from your lecture (very likely), it is \widetilde{f}(s)=\widetilde{G}_1(s)=\frac{1}{\alph  a}G_1(\alpha s) = \frac{p}{1-qs}. As you notice, this is the same as G_1 except that p and q are swapped. The pgf H_n is therefore obtained by swapping p and q in G_n. Notice that this swapping resumes to the situation when p\leq 1/2, where the extinction is almost sure: that's consistent with the fact that the process conditioned to extinction reaches extinction eventually.
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  3. #3
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    Thank-you so very much Laurent.
    Last edited by RoyalFlush; September 23rd 2009 at 07:55 AM.
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  4. #4
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    Quote Originally Posted by Laurent View Post
    The pgf of X_1 (or Y, equivalently) is easily found to be f(s)=G_1(s)=\frac{q}{1-ps}.
    Then you probably know that the pgf of X_2 is f(f(s)), etc., in general G_{n+1}(s)=f(G_n(s)). The formulas from the text can then be verified by induction; it's just tedious computation.
    I dont understand how you did this. Whenever I try I get something completely different.
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