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I am trying to solve the following problem. In a branching process the number offspring per individual has a binomial distribution with parameters 2, p. Starting with a single individual, calculate the extinction probability.

I believe the solution to such a problem is evaluated using the equation $z=P(z)$ where of course $p(z)$ is pgf of the size of the nth generation.

$z = (p(z-1) + 1)^n$

Due to lack of enough appropriate examples i am unsure how to proceed from here. I believe due to the convolution of each generation depending on the last we have a recursive equation. So would it be correct to solve for $z_2$ using
$z_2 = (p(z_1-1) + 1)^2$ and $z_1 = (p(z_0-1) + 1)^1$.

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Here's a more direct solution: You have two attempts not to go extinct. Each succeeds if a) a descendant is produced with probability $p$ and b) that descendant's branch survives with probability $q$. So your survival probability $q$ must satisfy

$q=(pq)^2+2pq(1-pq)=pq(2-pq)\;.$

One solution is $q=0$, the other is $q=(2p-1)/p^2$. The crossover occurs when the two solutions coincide, i.e. at $p=1/2$. For $p\le1/2$, the survival probability is $0$ (which makes sense, since in that case the expected number of descendants is $\le1$), whereas for $p\gt1/2$ it is $(2p-1)p^2$, so the extinction probability is $1-(2p-1)/p^2=(p^2-2p+1)/p^2=((1-p)/p)^2$.

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    @Hardy: You can view it as the binomial expansion of $((pq) + (1 - pq))^2$, with the three terms $(pq)^2$, $2pq(1-pq)$ and $(1-pq)^2$ corresponding to $2$, $1$ and $0$ descendants surviving, respectively; we have to add the probabilities for $2$ and $1$ descendants.2012-08-05
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$ \begin{align} r &= \Pr(\text{survival}) \\[6pt] & = \Pr((0\text{ offspring & survival})\text{ or }(1\text{ offspring & survival})\text{ or }(2\text{ offspring & survival})) \\[6pt] & = \Pr(0\text{ offspring & survival}) + \Pr(1\text{ offspring & survival}) + \Pr(2\text{ offspring & survival}) \\[6pt] & = 0 + \Pr(1\text{ offspring})\Pr(\text{survival}\mid 1\text{ offspring}) + \Pr(2\text{ offspring})\Pr(\text{survival}\mid 2\text{ offspring}) \\[6pt] & = 0 + 2p(1-p)r + p^2 (1-\Pr(\text{extinction}\mid 2\text{ offspring})) \\[6pt] & = 2p(1-p)r + p^2 (1 - \Pr(\text{both lines die out})) \\[6pt] & = 2p(1-p)r + p^2 (1 - (1-r)^2). \end{align} $

So we have a quadratic equation in $r$: $ r = 2p(1-p)r + p^2 (1 - (1-r)^2) $ The two solutions are $r=0$ and $r=\dfrac{2p-1}{p^2}$.

The second one is negative if $p<1/2$, so the probability in that case must be the first solution. If $p=1/2$ then the two solutions are $0$. If $p>1/2$, then must the solution be the second one? It would be enough to show $r>0$ in those cases. Obviously $r=1$ when $p=1$.