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

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.