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Suppose $f$ is the probability generating function for the Galton-Watson branching process.

What intuition makes the fact that $f(s) = s$ is the probability of extinction obvious? Moreover, can one find a meaningful interpretation for the "dummy variable", $s$? By meaningful, I mean an interpretation of $s$ which would allow us to write down that the root of $f(s) = s$ yields the extinction probability without needing to expand $f(s)$ to see the result.

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  • $\begingroup$ As far as I know generating functions are defined as $f(s) = \mathbb{E}s^X$ for random variables $X$ (often with values in $\mathbb{N}_0$). What is the random variable $X$ in your case? Then $f(s) = s$ implies that $\mathbb{P}(X = 1) = 1$. Is this what you want? $\endgroup$
    – Dieter Kadelka
    Commented Feb 6, 2019 at 11:29
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    $\begingroup$ Please omit the last two sentences in my first comment. You asked for the implications of $f(s) = s$ for some (not all) $0 < s < 1$. $\endgroup$
    – Dieter Kadelka
    Commented Feb 6, 2019 at 11:59
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    $\begingroup$ If $p$ is the extinction probability and the first individual has $n$ children, then the conditional extinction probability is $p^n$. Therefore, $p=\mathbb{E}[p^n]$. With a proof that simple, what more intuition are you asking for? $\endgroup$
    – Kostya_I
    Commented Jul 6, 2019 at 18:31
  • $\begingroup$ I'm a bit late to the party, but this resource is very helpful - stats.libretexts.org/Bookshelves/Probability_Theory/…. The key insight is that the probability of extinction $q$ for each generation is greater than the last, but less than 1, so $q_n \to q $ as $n \to \infty $. $\endgroup$
    – JasTonAChair
    Commented Oct 4 at 1:25

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I'm not quite shure, look in J.R. Norris, Markov Chains, Cambridge University Press, (1997), Example 5.1.1 (Branching Processes), in particular p. 172. Here you find that $s$ is the extinction probability, if $s = \inf \{t \in [0,1] \colon f(t) = t\}$. Here $f$ is the generating function of the random number $N$ of offsprings for each individual.

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