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Let $S_n=\displaystyle\sum_{i=1}^{n}X_i$ be the simple random walk with $S_0=0$. Now define the stopping time $$\tau=\inf\{n>0| S_n\notin (a,b)\}.$$

I am trying to follow some steps in a book and just need to understand how it ends up with the following inequalities with the goal of showing that $E(\tau)<\infty$.

$$P(S_{b-a}\notin (a,b))\geq2^{-(b-a)}$$ and supposedly iterating this gives $$P(\tau>n(b-a))\leq (1-2^{-(b-a)})^n.$$

Can someone explain how we can derive these inequalities? Thank you!

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First, $a<0<b$ are integers and the first inequality should be $$ \mathsf{P}(x+S_{b-a}\notin (a,b))\ge 2^{-(b-a)}, \quad x\in (a,b). $$ This inequality holds because $\{X_i=1 \text{ for all } 1\le i\le (b-a)\}\subset \{x+S_{b-a}\notin (a,b)\}$ for any $x\in(a,b)$. As for the second inequality, \begin{align} \mathsf{P}(N> 1(b-a))&\le \mathsf{P}(S_{b-a}\in(a,b))\le 1-2^{-(b-a)}, \\[1em] \mathsf{P}(N>2(b-a))&=\mathsf{P}(N>2(b-a),S_{b-a}\in (a,b)) \\ &=\mathsf{E}[\mathsf{P}(N>2(b-a)\mid S_{b-a})1\{S_{b-a}\in (a,b )\}] \\ &\le \mathsf{E}[\mathsf{P}(S_{b-a}+(S_{2(b-a)}-S_{b-a})\in (a,b)\mid S_{b-a})1\{S_{b-a}\in(a,b)\}] \\ &\le (1-2^{-(b-a)})\mathsf{P}(S_{b-a}\in (a,b))\le (1-2^{-(b-a)})^2, \\[1em] &\ldots \end{align}

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