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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-201/3/c/solution