Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 2 / a

For $x\ne 0$, the Strong Markov property at the first step gives the discrete mean-value identityAt $x=0$, $v(0)=1$ while the same average is at most one. Thus $v$ is a bounded superharmonic function on ${\mathbb{Z}}^3$. Conditioning on the natural filtration and using the one-step Markov property givesTherefore $(v(X_n))$ is a nonnegative supermartingale.