I'm having trouble with the following problem:
Let $\xi_i$ be i.i.d random variables with distribution,
$$P(\xi_i = k) = \begin{cases}\frac{1}{2} & k = 0 \\ \frac{1}{4} & k = 1 \\ \frac{1}{4} & k = 2\end{cases}$$
Let $X_n = \sum\limits_{k=1}^{X_{n-1}}\xi_k$ and assume $X_0 = 0$. In a typical branching process, the state $0$ is absorbing, i.e $P(X_{n+1} = 0 | X_n = 0) = 1$. However, suppose we modify this so that there is a positive probability of "regenerating" after extinction. Specifically, suppose that,$$P(X_{n+1} = 1 | X_n = 0) = \frac{1}{3} = 1 - P(X_{n+1} = 0 | X_n = 0)$$
Let $T = \inf\{n \geq 1: X_n = 0\}$ be the first hitting time of state $0$. Find the expectation, $E[T]$.
Finding this expectation is tricky because there doesn't seem to be a nice way of computing $P(T = n)$ for any $n$. The $n = 1$ case is nice because,
$$ P(T = 1) = P(X_1 = 0 | X_0 = 0) = \dfrac{1}{3} $$
Similarly, the $n = 2$ case isn't too bad,
$$ P(T = 2) = P(X_2 = 0 | X_1 = 1)P(X_1 = 1 | X_0 = 0) = \dfrac{1}{5}\dfrac{1}{3} = \dfrac{1}{15} $$
However, things start to become quite tedious even for $n = 3$. Is there a more efficient way of computing this?
If $X_1=0$ then $T=1$. Else the typical branching process starts with initial unit number of particles, and for the time $T'$ of its extinction, $$\mathbb E[T]=1+\frac13\mathbb E[T'|X_1=1]$$ But the extinction probability is $1/2$, so $\mathbb E[T'|X_1=1]=\infty$ since $T'=\infty$ w.p. $1/2$.