I am struggling with this exercise and trying very different approaches.
Let $X_i$ be iid r.v. s.t. $P(X=1)=P(X=-1)=1/2$
Let $S_n = X_1+\ldots+X_n, n\in \mathbb{N}$
Let $\alpha>0$ and let $M_n = (\exp(S_n-\alpha n)), n\in \mathbb{N}$
In particular I had to show for which values of $\alpha$ it holds that the process $M$ is a martingale. In order for it to be a martingale I needed that $E(M_{n+1} \mid \mathcal{F}_n) = M_n = \exp(S_n-\alpha\cdot n)$ where $\mathcal{F}_n$ is the natural filtration of the process.
Since $$\begin{align*} E(M_{n+1} \mid \mathcal{F}_n) &= E(\exp(S_{n+1}-\alpha(n+1)\mid\mathcal{F}_n) \\ &= E(\exp(S_n-\alpha(n+1)\cdot\exp(X_{n+1})\mid\mathcal{F}_n) \\ &= \exp(S_n-\alpha(n+1))\cdot E(\exp(X_{n+1})) \\&= \exp(S_n-\alpha(n+1))\cdot \frac{e+e^{-1}}{2} \end{align*}$$ and thus for it to be equal to $Z_n$ I need $\exp(S_n-\alpha(n+1))\cdot \frac{e+e^{-1}}{2} = \exp(S_n-\alpha(n))$ i.e. $$ \alpha = \log(\frac{e+e^{-1}}{2}).$$
Now, I need to compute $E(\exp(-\alpha T))$ where $T=\inf\{n\geq 1: |S_n|\geq a\}$, via optional stopping theorem and without using the dominated convergence theorem (I understand it is a powerful tool, but the material so far didn't cover it and thus I have to solve the exercise without it). My new idea is the following. Since the martingale is always positive, being an exponential, $|M_n|=M_n$ for every $n$ and thus the $\sup_n(E(|M_n|)) < +\infty$ and by the martingale convergence theorem I can say that $M_n\to M_{+\infty} $. My idea would be that, intuitively, $S_n-\alpha n\to-\infty $ a.s. (although I didn't manage to prove it formally) and thus $M_{+\infty} =0$ a.s. Now, I wanted to say that $E(M_T) = E(M_{T\land n}) = E(M_0)$ and thus show that the expectation I need to compute is $1/E(\exp(S_T))=2/(e^a+e^{-a})$, but in order to do so I need the martingale to be "closed" at infinity, right? Which I don't have since I only have the boundedness of the supremum of the absolute value. Is it provable that $\sup E(M_{T\land n}^2)<+\infty$? I honestly don't think so, because the stopping time is potentially n if the condition is not met and $\sup E(M_{n}^2)=+\infty$. Should I use that $\sup E(|M_{T\land n}|) <+\infty$? And thus conclude that this process is closed at infinity? And would it be $M_{T\land +\infty} \to M_{T\land 0}=M_0$?
Since the random walk $(S_n)_{n \in \mathbb{N}}$ has steps of size 1, it follows trom the definition of the stopping time $T$ that
$$|S_{n \wedge T}| \leq \underbrace{|S_{(n \wedge T)-1}|}_{< a} + \underbrace{|S_{n \wedge T}-S_{(n \wedge T)-1}|}_{\leq 1} \leq a+1.$$
Hence,
$$|M_{n \wedge T}| = \exp \big( S_{n \wedge T}-\alpha (n \wedge T) \big) \stackrel{\alpha >0}{\leq} \exp(S_{n \wedge T}) \leq \exp(a+1)$$
and so
$$\sup_{n \geq 1} \mathbb{E}(M_{n \wedge T}^2) \leq \exp(2(a+1))<\infty.$$
This shows that $(M_{n \wedge T})_{n \geq 1}$ is an $L^2$-bounded martingale.