Calculate the expectation of $\tau$=inf{$n\in\mathbb{N}_0|\exists{k}\lt{n}:S_{k}\le-1, S_n=1$}

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Let $(Z_k)_{k\in\mathbb{N}} \text{ be a sequence of i.i.d. random variables on } (\Omega,\mathscr{F},\mathbb{P}) \text{ with }$ $$\mathbb{P}(Z_1=-1)=\frac{1}{2}=\mathbb{P}(Z_i=1)$$ and symmetric random walk $$S=(S_n)_{n\in\mathbb{N}_0},S_0=0,S_n=\sum_{k=0}^{n}Z_k.$$ It's a martingale with respect to filtration $\mathscr{F}_n:=\sigma(Z_1,...,Z_n) \text{ with } \mathscr{F}_0:=\{\emptyset,\Omega\}$.

Also $$\operatorname{\tau}(\omega):=\inf\{n\in\mathbb{N}_0|S_{n-2}(\omega)=-1,S_{n-1}(\omega)=0,S_n(\omega)=1\},\omega\in\Omega.$$ Questions:

a).Prove:$\tau$ is a stopping time.

b).Show that $\tau$=inf{$n\in\mathbb{N}_0|\exists{k}\lt{n}:S_{k}\le-1, S_n=1$}.

c).Assume that $\lim\limits_{n\to\infty} \inf S_n=-\infty\text{ and }\lim\limits_{n\to\infty} \sup S_n=\infty\text{ }\mathbb{P}\text{ }a.s.$.
$\quad$Show:$\tau\lt\infty \text{ }\mathbb{P}\text{ }a.s.$ and $\mathbb{E}(\tau)=\infty.$

$\quad(\text{hint:}\forall{z}\in\mathbb{Z}\setminus\{0\}\text{ and stopping time }\tau_2=\inf\{n\in\mathbb{N}_0|S_n=z\}:\operatorname{\mathbb{E}}(\tau_2)=\infty)$

Part of my idea:

a).$\tau\text{ is a stopping time}:\iff$$\{\tau\le{n}\}\in\mathscr{F}_n\iff\{\exists{k}\le{n}:S_{n-2}(\omega)=-1,S_{n-1}(\omega)=0,S_n(\omega)=1\}$

$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\bigcup_{k=0}^n(\{S_{n-2}(\omega)=-1\}\cap\{S_{n-1}(\omega)=0\}\cap\{S_n(\omega)=1\})\in\mathscr{F}_n$

since each of those 3 part are measurable with respect to $\in\mathscr{F}_n$.

I get stuck on b) and c). I didn't see that the stopping times are equal. And how could it be, when we have $\tau\lt\infty$ but $\mathbb{E}(\tau)=\infty.$

Thanks in advance!

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For $(b)$, call the stopping time defined there $\tau_1$. Clearly $\tau_1 \le \tau$. To prove the other direction, observe that $S_{\tau_1 -1}$ must be $0$ and hence $S_{\tau_1 -2}$ must be $-1$.

For (c), use the hint with $z=-1$. Note there are many random variables that are finite a.s. but have infinite mean.