Consider the assymtric random walk $(S_n)_n$ on $\mathbb{Z}$. Determine $(a_n)_{n \ge 0}$ such that $\exp(\lambda S_n-a_n)_{n \ge 0}$ is a martingale. (We assume that $\lambda \in \mathbb{R}$ such that $\mathbb{E}[\exp(\lambda X_n)] < \infty$.
I tried to check the martingale property, i.e.
$$\mathbb{E}[\exp(\lambda S_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}] = \mathbb{E}[\exp(\lambda S_{n}-a_{n})].$$
So far I got the following:
\begin{align*} \mathbb{E}[\exp(\lambda S_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}] &= \mathbb{E}[\exp(\lambda S_{n}+\lambda X_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}] \\ &= \mathbb{E}[\exp(\lambda S_{n}) \exp(\lambda X_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}] \\ &= \mathbb{E}[\exp(\lambda S_{n}) \mid \mathcal{F}_{n+1}] \cdot \mathbb{E}[\exp(\lambda X_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}] \\ &= \mathbb{E}[\exp(\lambda S_{n})] \cdot \mathbb{E}[\exp(\lambda X_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}] \end{align*}
However, I am not so sure what to do with $\mathbb{E}[\exp(\lambda X_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}]$. Does it suffice to write
$$\mathbb{E}[\exp(\lambda X_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}] = \mathbb{E}[\exp(\lambda X_{n+1}-a_{n+1})] $$
and thus
$$\mathbb{E}[\exp(\lambda S_{n+1}-a_{n+1}) \mid \mathcal{F}_{n+1}] = \mathbb{E}[\exp(\lambda S_{n})] \cdot \mathbb{E}[\exp(\lambda X_{n}-a_{n+1})] = \mathbb{E}[\exp(\lambda S_{n+1}-a_{n+1})] \qquad ?$$
But what should this have to do with the $(a_n)_n$?