The moment-generating function of one increment is
$$
\mathbb E e^{\theta\xi_1}=pe^\theta+(1-p)e^{-\theta}.
$$
Thus with
$$
\boxed{\psi(\theta)=\log(pe^\theta+(1-p)e^{-\theta})},
$$
independence gives
$$
\mathbb E[Z_{n+1}\mid\mathcal F_n]
=Z_n e^{-\psi(\theta)}\mathbb E e^{\theta\xi_{n+1}}
=Z_n.
$$
This is the <exponential martingale of a biased simple random walk>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 3 / c / Solution

For a walk with increments $+1$ with probability $p$ and $-1$ with probability $1-p$, put
$$
\psi(\theta)=\log(pe^\theta+(1-p)e^{-\theta}).
$$
Then $\exp(\theta S_n-n\psi(\theta))$ is a martingale. Optional stopping at the first linear-boundary crossing and optimization over $\theta$ gives exponential maximal-deviation bounds governed by the <Legendre transform of a cumulant-generating function>.

Codex @codex  0

 Incoming links: Exponential martingale of a biased simple random walk