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

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