Problem is from here problem2 final examination.
Consider branching processes with the following offspring distributions:
- $P(0) = \frac14, P(2) = \frac12, P(4) = \frac14$
- $ P(n) = \frac14{(\frac34)}^n, n = 0, 1, 2,... $
- $P(n) = e^{-1}\frac1{n!}$
- $P(0) = \frac34, P(3) = \frac14$
(a) In each case, find the probability of eventual extinction assuming that the population
starts with one individual.
(b) In case 3, find the probability that the population survives at least 200 generations given
that it survives at least 100 generations.
I have partly solve it. Denote probability of extinction by $\pi_0$. Conditioning the offspring of the first individual, $\pi_0$ satisfies $$ \pi_0 = \sum_{n=0}^\infty P(n) \pi_0^n $$ Hence, part (a) can be solved by the above equation.
For part (b), I'm not sure whether I'm right and hope anyone can give hints.
Denote the event surviving at least $k$ generations by $S_k$. Hence,
$$
\begin{align}
P(S_{200} | S_{100}) &= \frac{P(S_{200}, S_{100})}{P(S_{100})} \\
&= \frac{P(S_{200})}{P(S_{100})}
\end{align}
$$
I tried to establish a relationship between $P(S_{k+1})$ and $P(S_{k})$,
$$
\begin{align}
P(S_{k+1}) &= \sum_{n=1}^\infty P(\text{n offspring of the first individual}, S_k) \\
&=\sum_{n=1}^\infty e^{-1}\frac1{n!}{P(S_k)}^n \\
&=e^{{P(S_k)} - 1} - e^{-1} \\ \tag{1}\label{1}
\end{align}
$$
The iteration above and $P(S_0) = 1$ make $P(S_{100})$ too complicated.
Can anyone check the answer?
Update:
Motivated by this answer, I find a flaw in the deduction of $P(S_{k+1})\eqref{1}$,
$$ \begin{align} P(S_{k+1}) &= \sum_{n=1}^\infty P(\text{n offspring of the first individual}, S_k) \\ &=\sum_{n=1}^\infty e^{-1}\frac1{n!} P(\text{one of the n offspring survive up to k generations}) \\ &=\sum_{n=1}^\infty e^{-1}\frac1{n!}\left( 1-\left(1 - P(S_k)\right)^n\right) \\ &=1 - e^{-P(S_k)} \end{align} $$ But it is still hard to calculate $P(S_{100})$.