Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 201 / 2

For each deterministic $n$, use the nonnegativity and the martingale property to obtainA nonnegative random variable with zero expectation is zero almost surely. Hence $M_{n+1}=0$ almost surely on $\{M_n=0\}$. Intersecting these probability-one statements over all $n$ makes them hold simultaneously, and induction givesZero is absorbing for a nonnegative martingale. This is the absorption at zero of a nonnegative martingale. On $\{T=\infty \}$ the assertion has no finite-time requirement. No unbounded-time use of the optional stopping theorem is needed.

The time $T_R$ is a stopping time, because $\{T_R\le n\}={\bigcup }_{k\le n}\{M_k\ge R\}\in {\mathcal{F}}_n$. Apply the optional stopping theorem to the bounded time $T_R\wedge n$:On $\{T_R\le n\}$ the stopped value is at least $R$, and elsewhere it is nonnegative. Letting $n\to \infty$ gives the maximal bound for a nonnegative martingale:Stop at a bounded time first, then pass to the increasing event. This avoids assuming uniform integrability of the original process. For $0<R\le 1$, $T_R=0$ and the bound simply says $1\le 1/R$.

The overshoot hypothesis as printed cannot hold for every $R>0$. For $R\le 1$, $T_R=0$ and $M_{T_R}=1$; choosing $R<1/\tau$ contradicts $M_{T_R}\le \tau R$. The intended and sufficient assumption is the overshoot bound for $R\ge 1$, on the event $\{T_R<\infty \}$. We use those thresholds below; when $R\le 1$, the requested stopped-time bound is trivial because $T\wedge T_R=0$.

Fix $R\ge 1$ and set $S=T\wedge T_R$. By part (a), the stopped martingale satisfies $0\le M_{S\wedge n}\le \tau R$: before stopping it is below $R$, at $T$ it is zero, and at $T_R$ the assumed overshoot bound applies. Thus the square of this stopped process is integrable. For $\Delta M_n=M_{n+1}-M_n$,The event $\{S>n\}$ is ${\mathcal{F}}_n$-measurable and contained in $\{T>n\}$. Conditional expectation of the linear term is zero, while the variance assumption makes the remaining conditional expectation nonnegative. The stopped increments here are bounded by the stopping and overshoot bounds, so all conditional expectations are legitimate. ConsequentlySince $X_0=1$, we have ${\sigma }^2\mathbb{E}(S\wedge n)\le \mathbb{E}{M}_{S\wedge n}^2-1$. Bounded-time optional stopping theorem also gives $\mathbb{E}{M}_{S\wedge n}=1$, soApply the monotone convergence theorem to $S\wedge n$. This proves the stronger estimate $(\tau R-1)/{\sigma }^2$ and hence the requested absorption-time bound from conditional variance and overshoot:In particular this stopping time is finite almost surely. For $0<R\le 1$, $S=0$ and $X_n=1$ is a constant submartingale, while the stopping-time expectation is zero.

If $T>n$, either $T_R\le n$, or both stopping times exceed $n$. Therefore the maximal bound for a nonnegative martingale and the Markov inequality give, for $R\ge 1$,The right-hand side is minimized at $R=\sigma \sqrt{n}/\tau$. Whenever the claimed bound is nontrivial, $2\tau /(\sigma \sqrt{n})<1$, this choice has $R>2$ and is allowed under the corrected overshoot hypothesis. Substituting gives the square-root tail bound for martingale absorption:When the displayed upper bound is at least one, the inequality follows from $\mathbb{P}(T>n)\le 1$. This handles every $n$ without using the impossible small-threshold hypothesis.