Survival probability up to time $n$ in a branching process.

Question

Let $\{Z_n : n=0,1,2,\ldots\}$ be a Galton-Watson branching process with time-homogeneous offspring distribution $$\mathbb P(Z_{n,j} = 0) = 1-p = 1 - \mathbb P(Z_{n,j}=2), $$ where $0<p<1$. That is, $Z_0 = 1$ and $$Z_n = \sum_{j=1}^{Z_{n-1}}Z_{n,j}$$ for $n\geqslant 1$. Let $$T=\inf\{n : Z_n=0\} $$ be the extinction time of the process. I want to find $$t_n := \mathbb P(T>n), $$ i.e. the probability that the process survives up to time $n$. I found the following recurrence:

$$t_{n+1} = p(2t_n - t_n^2)$$ (with $t_0=1$) from this mathoverflow question: https://mathoverflow.net/questions/87199/branching-process-survival-probability

I checked the recurrence for small values of $n$, and confirmed that $t:=\lim_{n\to\infty} t_n$ satisfies $$ t = p(2t - t^2),$$ both in the case where $t=0$ $\left(p\leqslant\frac12\right)$ and where $t=2-\frac1p$ $\left(p>\frac12\right)$. So I'm fairly confident this recurrence is valid. However, I have no idea how to solve it.

For context, this problem comes from Adventures in Stochastic Processes by Sidney Resnick:

From @Did's comment, it appears to be intractable to find a closed form for $\mathbb P(T>n)$. I find it curious that the question would be asked were it not, though.

Answer 1

Consider the first generation in the branching process: either (1) we have $0$ offspring or (2) $2$ offsprings. Suppose $t_n$ is known. We would like to compute $t_{n+1}$, the probability that $T > n+1$. In order to have $T > n+1$, the first generation has to have $2$ offsprings, say $A$ and $B$. Now one of those two offsprings has to have $>n$ generations (think of two branching processes starting with $A$ and $B$ respectively). Therefore $$t_{n+1}=p\operatorname{Pr}\{A \text{ or }B \text{ (or both) has more than }n\text{ generations of offsprings}\} = p(2t_n-t_n^2).$$

In order to solve the recurrence, one may take advantage of $P(s) = q + ps^2$. Note that $$1-t_{n+1} = p(1-t_n)^2 + (1-p) = P(1-t_n).$$ Therefore $1-t_n = P^{(n)}(0)$ and so $t_n = 1-P^{(n)}(0)$.

Answer 2

For p approaching 1, there is a solution, and I'll post a proof later, at least this will get the ball rolling. For p=1 the recurrence relation solves to $$t_n =1-e^{c*2^{n}}$$ For p=2, certainly not applicable to this case, the recurrence relation solves to $$t_n=1-\cos(c2^n)$$

Proof

The original recurrence relation is given by $$t_{n+1}=p(2t_n-{t_n}^2)$$ I'll prove the first solution. For $p=1$ assume that the first equation is a solution for a number n. Substitute the solution into the relation. $$1-e^{2c2^n}=2-2e^{c2^n}-(1-e^{c2^n})^2$$ Expand and simplify

$$1-e^{2c2^n}=2-2e^{c2^n}-1+2e^{c2^n}-e^{2c2^n}=1-e^{2c2^n}$$ The second solution is proved in much the same way, you'd just need to know some trig identities.

Suggestions Use a computer for most of the calculations and do interpolation between the points. If you feel adventurous, find a derivation for the $p=1$ solution, and perform perturbation analysis, you may or may not get a convergent solution. I'll repost if I get anything interesting.