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Say we have the following PMF for a simple branching process and want to find the eventual extinction probability,

$$ P(Z_{1,1} = 0) = 0.25$$ $$ P(Z_{1,1} = 1) = 0.25$$ $$ P(Z_{1,1} = 2) = 0.50$$

Then, we can easily find $$G_1(s)=0.50s^2+0.25s+0.25$$ From there, we can also find $$G_2(s) = G_1(G_1(s)) = 0.50(0.50s^2+0.25s+0.25)^2+0.25(0.50s^2+0.25s+0.25)+0.25$$

So I get to this step, and I think I can generalize to find $G_n(s) = as^2 + bs + c$ where $a$, $b$, and $c$ are simply the results of the geometric series simplification $\sum\limits_{k=0}^{n-1} ar^k=a\frac{1-r^n}{1-r}$ (or do we do the infinite sum?). Then setting $G_n(s)=s$, we can find the eventual extinction probability. Is this correct? I am unsure about the generalization of $G_n(s)$

user130912
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1 Answers1

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Actually, each $G_n(s)$ is a polynomial of degree $2^n$ and quite impossible to write down. Fortunately, one never (well, very rarely) rely on an explicit formula for $G_n(s)$ to compute the probability $q$ of extinction. Instead, as your textbook most probably mentions, one uses the fact that $q$ is the smallest root in $[0,1]$ of the identity $q=G_1(q)$. In the present case, $q=\frac12$.

Did
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  • Wow, I can not believe I overlooked that, I guess I should have done the third expansion. After re-reading my textbook, I realized I completely misunderstood setting $q=G_1(q)$, I thought that was the extinction probability at that generation instead of the eventual extinction probability. Thanks! – user130912 Feb 25 '14 at 20:01