Construction
Let $(\varepsilon_i)_{i\ge 1}$ be i.i.d. random variables with
$$ \mathbb P(\varepsilon_i=1)=\mathbb P(\varepsilon_i=-1)=\frac12, $$and define the simple random walk
$$ X_n=\sum_{i=1}^n \varepsilon_i, \qquad \mathcal F_n=\sigma(\varepsilon_1,\dots,\varepsilon_n). $$For $\theta\in\mathbb R$, define
$$ \lambda(\theta)=\log \mathbb E[e^{\theta\varepsilon_1}]
\log\left(\frac{e^\theta+e^{-\theta}}2\right). $$
Then
$$ M_n(\theta)=\exp\bigl(\theta X_n-n\lambda(\theta)\bigr) $$is a martingale with respect to $(\mathcal F_n)$.
Verification
Using $X_{n+1}=X_n+\varepsilon_{n+1}$,
$$ M_{n+1}(\theta)
\exp\bigl(\theta X_n+\theta\varepsilon_{n+1}-(n+1)\lambda(\theta)\bigr). $$
Conditioning on $\mathcal F_n$ gives
$$ \mathbb E[M_{n+1}(\theta)\mid \mathcal F_n]
e^{\theta X_n-(n+1)\lambda(\theta)} \mathbb E[e^{\theta\varepsilon_{n+1}}\mid \mathcal F_n]. $$
Since $\varepsilon_{n+1}$ is independent of $\mathcal F_n$,
$$ \mathbb E[e^{\theta\varepsilon_{n+1}}\mid \mathcal F_n]
\mathbb E[e^{\theta\varepsilon_{n+1}}]
e^{\lambda(\theta)}. $$
Therefore,
$$ \mathbb E[M_{n+1}(\theta)\mid \mathcal F_n]
e^{\theta X_n-n\lambda(\theta)}
M_n(\theta). $$