Why is $E(\tau_{z})=\infty$ with the following proof?

Question

Let $(X_{n})_{n}$ be IID RV's with $P(X_{i}=1)=P(X_{i}=-1)=\frac{1}{2}$ and $S_{n}:=\sum\limits_{i=1}^{n}X_{i}$.

Further define: $\tau:=\tau_{-a}\land\tau_{b}$ where $a,b \in \mathbb N$

Show that for any $z\in \mathbb Z\setminus\{0\}:E[\tau_{z}]=\infty$ where $\tau_{z}$ is the hitting time, i.e. $\tau_{z}=\inf\{n \in \mathbb N: S_{n}=z\}$

I do not understand the following proof:

w.l.o.g. assume $\tau > 0$, and then $\tau_{z} \geq \tau_{-n} \land \tau_{z}, n \in \mathbb N$

$\Rightarrow E[\tau_{z}]\geq \sup\limits_{n}E[\tau_{-n} \land \tau_{z}]=\sup\limits_{n}S_{n}=\infty$

Any ideas why the second last equality holds?

Answer 1

We have $$ \mathbb E[\tau_{-a}\wedge \tau_{b}] = ab $$ from Doob's Optional Stopping theorem, so $$ \mathbb E[\tau_{-n}\wedge \tau_{z}] = nz\xrightarrow[n\to+\infty]{}+\infty. $$