Branching Process and Random Walk

Question

Consider this version of branching process: each individual has 2 offspring with probability p, and 0 offspring with probability 1-p. Let the 0th generation has 1 individual, then I managed to find the extinction probability q is \begin{equation} q= \left\{ \begin{array}{**lr**} 1 &{p\leq0.5}\\ \frac{1-p}{p} &{p\geq0.5} \end{array} \right. \end{equation} Consider the random walk that starts from site 1, and in each stage it increase by 1 with probability p and decrease by 1 with probability p-1. By calculation and taking limits, the probability that the probability that the walk ever hits site 0, is exactly the same with the extinction probability above.
How does we derive a mapping between these two processes to understand this equality?

Answer 1

You list the individuals (initially just one).

You then process the list: if an individual at the start of the list produces children, you add these to the end of list; in any case you then delete the individual from the start of the list and move on to the next in the list. If the list is ever empty then you have extinction and you stop.

• Looking at the length of the list after processing each individual gives you your random walk. (Extinction is the equivalent of hitting $0$.)

• Looking at the length of the list after processing each generation gives you your original question.

In both cases the probability of extinction or hitting $0$ is the probability the list is ever empty.

Answer 2

Consider a continuous time version of your branching process, when the times between your birth and the time when you either split in two or vanish are iid exponential. Then this is a continuous time Markov process $X(t)$ with rates of jumps, once you are at $n$, given by $q_{n,n+1}=pn$ and $q_{n,n-1}=(1-p)n$. The embedded Markov chain is exactly your random walk and thus the survival is equivalent to the recurrence.