Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 201 / 1

For the simple symmetric random walk, $S_n$ is a martingale and the square-minus-time martingale of a simple symmetric random walk isIndeed, conditioning on ${\mathcal{F}}_n$ and using $\mathbb{E}[X_{n+1}]=0$ and $X_{n+1}^2=1$ gives $\mathbb{E}[S_{n+1}^2\mid {\mathcal{F}}_n]=S_n^2+1$.

Apply the optional sampling theorem for a supermartingale to the bounded stopping time $T\wedge n$:Thus the stopped martingale $(S_{T\wedge n})$ is bounded in $L^2$. The L2 martingale convergence theorem gives convergence in $L^2$, and because $T<\infty$ almost surely its limit is $S_T$. Meanwhile the monotone convergence theorem gives $\mathbb{E}[T\wedge n]\to \mathbb{E}[T]$. Therefore

Finite mean is essential. Let $T=\inf \{n\ge 1:S_n=0\}$ be the first return to zero. The one-dimensional simple symmetric random walk is recurrent, so $T<\infty$ almost surely, but its first-return time has infinite mean. Since $S_T=0$,

The law of the iterated logarithm for a simple symmetric random walk states that almost surelySince $\sqrt{2\log \log n}$ is unbounded, $|S_n|/\sqrt{n}$ exceeds every fixed $c>0$ at some finite time. Hencealmost surely.

Suppose $c>1$ and $\mathbb{E}[T_c]<\infty$. Part (a) would giveBut $T_c\ge 1$ and its defining strict inequality gives $S_{T_c}^2>c^2{T}_c$ almost surely. Taking expectations would yieldwhich is impossible because $c^2>1$. Part (b) shows that $T_c$ is nevertheless finite almost surely. Consequently