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Find the extinction probability for a branching process with offspring distribution $a =(1∕6, 1∕2, 1∕3)$.

Solution

The mean of the offspring distribution is
$\mu = 0(1∕6)+ 1(1∕2)+ 2(1∕3)= 7∕6 > 1$,
so this is the supercritical case.

The offspring generating function is $G(s)= 1/6 + s/2 + s^2/3$.

Solving $s = G(s)= 1/6 + s/2 + s^2/3$ gives the quadratic equation $s^2∕3 − s∕2 + 1∕6 = 0$, with roots $s = 1$ and $s = 1∕2$.

The smallest positive root is the probability of eventual extinction $e = 1∕2$.

I have several questions.

  1. What would be the offspring generating function in critical and sub-critical cases?
  2. Why is $s = G(s)$?
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  • $\begingroup$ What do you know about Branching processes? The second property is a basic theorem on extinction probability. The generating function of a critical or sub-critical BP depends on the offsping distribution , so what type of answer can be given to 1)? $\endgroup$ – Kavi Rama Murthy May 11 at 0:10
  • $\begingroup$ @KaviRamaMurthy, The generating function of a critical or sub-critical BP depends on the offspring distribution - I want to see what would PGF look like in each of those cases. $\endgroup$ – user366312 May 11 at 0:31
  • $\begingroup$ @KaviRamaMurthy, The second property is a basic theorem on extinction probability. - kindly, supply me some material to study. $\endgroup$ – user366312 May 11 at 0:32
  • $\begingroup$ @KaviRamaMurthy, What do you know about Branching processes? - I just started it today. $\endgroup$ – user366312 May 11 at 0:32

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