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Thread: PGF of a branching process (very hard)

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

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

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

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

    and

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

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

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

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

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

    and

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

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

    $\displaystyle \beta = I_{\{p \leq \frac{1}{2} \}} + \alpha I_{\{p > \frac{1}{2}\}} \ \mbox{where} \ \alpha = \frac{q}{p} $
    The pgf of $\displaystyle X_1$ (or $\displaystyle Y$, equivalently) is easily found to be $\displaystyle f(s)=G_1(s)=\frac{q}{1-ps}$.
    Then you probably know that the pgf of $\displaystyle X_2$ is $\displaystyle f(f(s))$, etc., in general $\displaystyle G_{n+1}(s)=f(G_n(s))$. 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 $\displaystyle p\leq 1/2$, the probability is 1, hence conditioning to extinction is like doing nothing: $\displaystyle H_n=G_n$ ($\displaystyle H_n$ being the pgf of $\displaystyle X_n$ conditioned to extinction). When $\displaystyle p>1/2$, this probability is positive ($\displaystyle \alpha=\frac{q}{p}$). 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 $\displaystyle \widetilde{f}(s)=\widetilde{G}_1(s)=\frac{1}{\alph a}G_1(\alpha s) = \frac{p}{1-qs}$. As you notice, this is the same as $\displaystyle G_1$ except that $\displaystyle p$ and $\displaystyle q$ are swapped. The pgf $\displaystyle H_n$ is therefore obtained by swapping $\displaystyle p$ and $\displaystyle q$ in $\displaystyle G_n$. Notice that this swapping resumes to the situation when $\displaystyle p\leq 1/2$, 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; Sep 23rd 2009 at 06:55 AM.
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  4. #4
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    Quote Originally Posted by Laurent View Post
    The pgf of $\displaystyle X_1$ (or $\displaystyle Y$, equivalently) is easily found to be $\displaystyle f(s)=G_1(s)=\frac{q}{1-ps}$.
    Then you probably know that the pgf of $\displaystyle X_2$ is $\displaystyle f(f(s))$, etc., in general $\displaystyle G_{n+1}(s)=f(G_n(s))$. 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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