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I hope someone would be able to shed some light as to why equation 1.33 is the way it is from first principles. It looks a lot like the case where the initial state is randomised but the final state to which the Markov chain is, is fixed. From the expression, there are two probability terms which isn't clear to me why this is so.

Any help is appreciated.

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2 Answers 2

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(1.33) is just the Law of Total Probabilities:

$$ P(A) = \sum_k \left[ P(B_k) P(A | B_k) \right], $$

where $\{B_k \;|\; k = 0, \ldots\}$ is a countable partition of the sample space.

In this case, clearly having $k$ children (in the first generation) is a countable partition of the sample space, and $P(B_k) = p_k$ is the probability for this.

Suppose there are $k$ children in the first generation. Then, under this model, they all peter out or not independently. Consequently, $P(A | B_k) = \rho^k$ is the probability of petering out assuming $k$ children in the first generation.

By the Law of Total Probabilities, the sum of this product is the overall probability of extinction.

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The reason this argument works is that, in a pure branching process, lineages don't interact in any way after they've branched off each other. It's as if every individual was alone in the world, never even seeing any others of its kind (except for its own parent, which it saw once when it was born, and its own offspring that also disappear somewhere far away immediately after being born).

What this means is that, if we initially have several (presumably widely separated) individuals living in the world, each of those individuals will be the founder of an independent lineage, and each of those lineages will either survive or die off independently of the others. Thus, if we know that a single lineage dies off with probability $\rho$, then the probability that $k$ separate lineages all die off independently of each other is simply $\underbrace{\rho \times \rho \times \dots \times \rho}_{k \text{ times}} = \rho^k$.

But if we know that the probability of a single individual producing $k$ offspring is $p_k$, then that means that the probability of the individual producing $k$ offspring and all the lineages of those offspring then dying off is $p_k \times \rho^k$. And we can add up these probabilities for all possible offspring counts $k$ to obtain the total probability $\sum_{k=0}^\infty p_k \times \rho^k$ of all the lineages of a single individual's offspring eventually dying off.

But the lineage of an individual dies off if and only if the lineages of all its offspring die off, so this total probability must be equal to just $\rho$! And thus we get the "equation 1.33" that you cite: $$\rho = \sum_{k=0}^\infty p_k \, \rho^k$$

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  • $\begingroup$ This explanation does solidifies my initially fuzzy intuition. I have a question though; if $P_{k}$ is the probability any individual (founders by their own right) produces k number of off springs, then for n number of generation, would this requires $P_{k}^{n}$, for an independent and identical distribution? $\endgroup$
    – Physkid
    Commented Mar 11, 2018 at 3:22
  • $\begingroup$ I'm not sure I can given any meaningful interpretation for $P_k^n$ (except "the probability that $n$ distinct individuals each produce exactly $k$ offspring in their lifetime", which is presumably not what you're looking for). I suppose something like "the probability that an individual produces exactly $k$ offspring, and the first of those offspring also produces exactly $k$ offspring, and so on for $n$ generations" would also yield the expression $P_k^n$; but that seems like a pretty pointless thing to calculate. $\endgroup$ Commented Mar 11, 2018 at 12:29
  • $\begingroup$ ... Anyway, the point is that in "equation 1.33" we don't really care how many third-generation offspring of their own each of the original individual's offspring have. All we care about is how many offspring the original individual has (the distribution of which is given by the $p_k$ values) and how likely the lineage descending from each of those offspring is to die off (which we call $\rho$ and assume, by the definition of the branching process model, to be the same for each individual, including the original one). $\endgroup$ Commented Mar 11, 2018 at 12:34

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