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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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2 Answers 2

up vote 4 down vote accepted

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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I am unsure if i follow how you compiled your first equation. Is there a missing step before that equation ? –  Hardy Aug 5 '12 at 8:52
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@Hardy: One way to derive that equation is via the inclusion-exclusion principle: the probability of at least one branch surviving is the sum of the probabilities of each branch surviving (i.e. $pq+pq=2pq$) minus the probability of both surviving (i.e. $(pq)^2$), giving $q=2pq-(pq)^2=pq(2-pq)$. Or, as joriki seems to have done it, $p$ also equals $(pq)^2$ plus twice the probability $pq(1-pq)$ of one (specified) branch surviving and the other dying. Note that, in both cases, we're implicitly making use of the assumption that the survival probabilities are equal and independent of each other. –  Ilmari Karonen Aug 5 '12 at 9:51
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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. –  joriki Aug 5 '12 at 15:34

$$ \begin{align} r &= \Pr(\text{survival}) \\[6pt] & = \Pr((0\text{ offspring & survival})\text{ or }(1\text{ offspring & survival})\text{ or }(2\text{ offspring & survival})) \\[6pt] & = \Pr(0\text{ offspring & survival}) + \Pr(1\text{ offspring & survival}) + \Pr(2\text{ offspring & survival}) \\[6pt] & = 0 + \Pr(1\text{ offspring})\Pr(\text{survival}\mid 1\text{ offspring}) + \Pr(2\text{ offspring})\Pr(\text{survival}\mid 2\text{ offspring}) \\[6pt] & = 0 + 2p(1-p)r + p^2 (1-\Pr(\text{extinction}\mid 2\text{ offspring})) \\[6pt] & = 2p(1-p)r + p^2 (1 - \Pr(\text{both lines die out})) \\[6pt] & = 2p(1-p)r + p^2 (1 - (1-r)^2). \end{align} $$

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 $p=1$.

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