I have been studying hitting times on random walks and I am trying to understand if the following is true.
Let $S_n$ be a random walk such that $\Bbb{P}(S_{n-1}=S_{n}+1)=p$ and $\Bbb{P}(S_{n-1}=S_{n}-1)=1-p$.
Define $\tau_{0,i}=\inf\{ n>0,\text{ such that } S_n=i|S_0=0\}$
Then we have that $\tau_{0,2}=\tau_{0,1}+\tau_{1,2}$ in distribution and due to the properties of the random walk $\tau_{0,1}$ and $\tau_{1,2}$ are i.i.d. random variables.
Then is the following true?
i) $\Bbb{P}( \tau_{0,2}<\infty)= \Bbb{P}( \tau_{0,1} + \tau_{1,2}< \infty)$
I got the next statement from the first due to $\tau_{0,2}=\tau_{0,1}+\tau_{1,2}$ in distribution.
ii) $\Bbb{P}( \tau_{0,2} < \infty)= \Bbb{P}( 2\tau_{0,1}<\infty)=\Bbb{P}( \tau_{0,1} < \infty)$
iii) $\Bbb{P}( \tau_{0,2} < \infty)=\Bbb{P}( \tau_{0,1} < \infty \text{ and } \tau_{1,2} < \infty) = \Bbb{P}( \tau_{0,1}<\infty)\Bbb{P}( \tau_{1,2}<\infty)=(\Bbb{P}( \tau_{1,2}<\infty))^2$
Where I separated the 'and' statement by independence and then got the power due to the random variables being identical in distribution.
I believe at least one of these is false as they are leading to contradictions with known results.
EDIT: After thinking more about it I can see that ii) is incorrect.
To clarify your definitions: it is better just to define $\tau_i = \inf \{n \geq 0: S_n = i\}$ to be the hitting time of $i$ for the random walk $S_n$ (it is infinite if the walk never reaches $i$.) Then, your $\tau_{i,j}$, for $i < j$, can be thought of as
$\tau_{i,j} = \inf \{k > 0: S_{\tau_i+k} = j\}$.
Then it is easy to check $\tau_2 = \tau_{0,1}+\tau_{1,2}$, so (i) is true. As you say, (ii) is incorrect: this is because the two stopping times $\tau_{0,1}$ and $\tau_{1,2}$ are independent and identically distributed, but not equal!
The observation (iii) holds by independence. These probabilities can be computed explicitly, using the optional stopping theorem: one can show that, for $p \leq 1/2$,
$P(\sup_n S_n \geq m) = (p/q)^m,$
where $q = 1-p$. Thus, since $\tau_m < \infty$ iff $\sup_n S_n \geq m$,
$P(\tau_2 < \infty) = (p/q)^2 = P(\tau_1 < \infty)^2.$
If you want to learn more, this problem is usually called 'gambler's ruin.'