A branching process can be viewed as a Markov Chain $\{Z_n: n\ge 0\}$ such that $Z_0=1$ and $$Z_{n+1}:=\sum_{i=1}^{Z_n}L_{i}^{(n+1)},$$ where $L_i^{(j)}$ are independent Poisson random variable $Po(\lambda)$, i.e., $\mathbb{P}(L_i^{(j)}=k)=e^{-\lambda}\lambda^k/k!$.
It is known that the probability that such a process gets extinct with probability 1, i.e., $\mathbb{P}(\exists n, Z_n=0)=1$, if and only if $\lambda\le 1$. And $\mathbb{E}(Z_n)=\lambda^n$.
Now if we consider the following Markov chain $\{Y_n:n\ge 0\}$: $Y_0=1$, $Y_1=L_{1}^{(1)}$, and
$$Y_{n+1}:=\sum_{i=1}^{2Y_n}L_i^{(n+1)} \text{ for $n\ge 1$},$$
where $L_i^{(j)}$ are independent Poisson $Po(\lambda)$.
I am wondering is it possible to find the sufficient and necessary condition when this chain gets extinct with probability 1? And can we compute $\mathbb{E}(Y_n)$?
Best Answer
We have $$ Y_{n+1} = \sum_{i=1}^{2Y_n}L_i^{(n+1)} = \sum_{j=1}^{Y_n} (L_{2j-1}^{(n+1)} + L_{2j}^{(n+1)}) $$ where $L_{2j-1}^{(n+1)} + L_{2j}^{(n+1)}$, being the sum of two independent $\text{Po}(\lambda)$ random variables, has a $\text{Po}(2\lambda)$ distribution.
So $Y_{n+1}$ is almost a branching process with rate $2\lambda$. The only defect is that $Y_1$ is defined to be $L_1^{(1)}$, breaking the pattern: it ought to be $Y_1 = L_1^{(1)} + L_2^{(1)}$ to match the definition of $Y_{n+1}$. This results in a branching process which has rate $\lambda$ in the first generation, and rate $2\lambda$ in all other generations.
This does not affect at which point the extinction probability is $1$, but it cuts the expected value in half compared to a branching process with rate $2\lambda$.
If you are interested in finding the extinction probability in cases when it's less than $1$, we can imagine a second process $(Y_n' : n \ge 1)$ which is an independent copy of $(Y_n : n \ge 1)$. If $Y_n$ goes extinct with probability $q$, then so does $Y_n'$, and so the sum $Y_n + Y_n'$ goes extinct with probability $q^2$. But the sum $Y_n + Y_n'$ is exactly a branching processes with rate $2\lambda$.