Past exam of the martingales and financial mathematics course of the University of Oxford / 2009 / A / 3 / b

Use ${\mathcal{F}}_0=\{\varnothing ,\Omega \}$; the printed $\{\varnothing ,\mathcal{F}\}$ does not define the intended trivial sigma-algebra. With the simple symmetric random walk convention from part (a), $M_n$ is adapted and integrable because $|S_n|\le n$. Independence givesBy linearity, the pull-out property for the ${\mathcal{F}}_n$-measurable $S_n$, and preservation of measurable variables by conditional expectation,This proves the square-minus-time martingale of a simple symmetric random walk.

A discrete stopping time is a random variable $T:\Omega \to {\mathbb{N}}_0\cup \{\infty \}$ such that $\{T\le n\}\in {\mathcal{F}}_n$ for every $n$. The convention is $\inf \varnothing =\infty$.

For the hitting time in question,Each event in the union is in ${\mathcal{F}}_j\subseteq {\mathcal{F}}_n$ because $S_j$ is adapted. This proves $T_k$ is a stopping time for every positive integer $k$.

The optional stopping theorem applies to $M$ at the bounded stopping time $n\wedge T_k$. As $M_0=0$, it yields the valid stopped second-moment identity for a simple symmetric random walkThe printed bound on $\mathbb{E}{T}_k$ is false. Before hitting the positive level $k$, the walk can travel arbitrarily far downwards, so $|S_{n\wedge T_k}|\le k$ is unavailable.

Here is a valid proof of finiteness of $T_k$. For a positive integer $L$, stop instead at $\tau =\inf \{n:S_n\in \{-L,k\}\}$. The same identity and the bound $|S_{n\wedge \tau }|\le \max (L,k)$ giveThe monotone convergence theorem makes $\mathbb{E}\tau <\infty$, so $\tau <\infty$ almost surely. Applying optional stopping to $S$ at $n\wedge \tau$ and then using dominated convergence gives $\mathbb{E}{S}_{\tau }=0$. Since $S_{\tau }\in \{-L,k\}$, this meansConsequently $\mathbb{P}(T_k<\infty )\ge L/(L+k)$ for every $L$, and letting $L\to \infty$ proves $T_k<\infty$ almost surely.

In fact this is the infinite mean first passage of a simple symmetric random walk. If $\mathbb{E}{T}_k<\infty$, the first displayed identity would bound the second moments of $S_{n\wedge T_k}$ uniformly. Their uniform integrability and their almost sure convergence to $k$ would imply $\mathbb{E}{S}_{n\wedge T_k}\to k$. But the optional stopping theorem applied to $S$ at every bounded $n\wedge T_k$ makes each expectation zero, a contradiction. Hence the correct conclusion is

Intersecting the probability-one hitting events over all positive integers shows that the simple symmetric random walk reaches arbitrarily large positive levels almost surely. The hitting times of levels tending to infinity also tend to infinity, since $|S_n|\le n$. Hence $\mathbb{P}(B_{+})=1$, and in particular $\mathbb{P}(B_{-}\cup B_{+})=1$.

Alternatively, the union conclusion, the equal probabilities from symmetry, and the Kolmogorov zero-one law force both probabilities to equal one. Thus almost every path has arbitrarily large positive and negative excursions. A nearest-neighbor integer path must pass through zero between a positive and a negative excursion. Such crossings occur after every finite time, givingThis is recurrence of the simple symmetric random walk.