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My question is related to this question here.

This says that if $S_{n}$ is a simple random walk (with steps $+1$ or $-1$ with probability $p$ and $q$ respectively) started at $S_{0}=1$, and if $T=\min\{k\geq 0:S_{k}=0\}$ then $$Z_k = \sum _{n=0}^{T-1} \textbf{1}_ {\{ S_n= k, S_{n+1}=k+1 \}}\,, Z_{0}=1 $$ is a Galton Watson Branching Process with Geometric Offspring Distribution.

My question is that how is $Z_{1}$ distributed as a Geometric $q$ variate (with the convention that the variate can take value $0$, i.e. the second description as given in Wikipedia)?

Now I can see as given in the comments that $Z_{2}=\sum_{k=1}^{Z_{1}}\xi_{k}$ where $\xi_{k}$ are iid Geometric $q$ variates. This precisely counts the random "upcrossings" from $2$ to $3$ strictly in between the $Z_{1}$ many upcrossings from $1$ to $2$. Now inductively, I can show that $Z_{n+1}=\sum_{k=1}^{Z_{n}}\xi_{k}^{(n)}$ where $\xi_{k}^{(n)}$ are iid Geometric $q$ variates which would prove that $Z_{n}$ is a branching process.

But how do I show that $Z_{1}$ has Geometric distribution?

I can see that $P(Z_{1}=k)=\sum_{m=0}^{\infty}P(Z_{1}=k|T=2m+1)P(T=2m+1)$ but this just complicates matters more as I would end up with many different ways of crossing from $1$ to $2$ for large $m$. Also I would have to multiply by the probability that $T=2m+1$ which would be $(pq)^{m}\binom{2m}{m}$.

Maybe I am missing something very trivial. Can anyone help me out with this?

Blitzkrieg
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    Well here's a non rigorous way to think about this. Forget $T$ for now and consider the whole random walk. Suppose you are visiting $1$ for the kth time, i.e. you have visited $1$, $k-1$ times before. Then, the probability that you reach $2$ is $\frac{1}{2}$ independent of the past by the markov property(i.e. Z_{1} will get incremented by $1$). This is precisely the memoryless property. Thus $Z_{1}$ has to have a geometric distribution. Now, it is clear that $T<\infty$ and hence one need look at the stopped process to get your tree. Else, you'll end up exploring the forest one by one. – Mr. Gandalf Sauron Nov 21 '23 at 17:52
  • Thanks @Mr.GandalfSauron I think I understand what you mean. But I don't think I can do the entire rigorous proof from here by my own. Any help is appreciated. – Blitzkrieg Nov 21 '23 at 18:17

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