Absorption at zero of a nonnegative martingale ($M_n=0(n\ge T_0)$)

Once a nonnegative martingale hits zero, it stays zero almost surely. Conditional expectation on $\{M_n=0\}$ gives $\mathbb{E}[M_{n+1}{1}_{\{M_n=0\}}]=0$; nonnegativity forces the next value to vanish there. A countable intersection makes these assertions simultaneous for all times, proving absorption at the first zero.

Let a nonnegative martingale start at one, let $T$ be its first zero, and let $T_R$ be its first hit of $[R,\infty )$. Suppose $\mathbb{E}[(M_{n+1}-M_n{)}^2\mid {\mathcal{F}}_n]\ge {\sigma }^2{1}_{\{T>n\}}$ and $M_{T_R}\le \tau R$ on finite hits, for thresholds $R\ge 1$. Then $M_{T\wedge T_R\wedge n}^2-{\sigma }^2(T\wedge T_R\wedge n)$ is a submartingale. The bounded stopped values have mean one, so their second moment is at most $\tau R$. Consequently $\mathbb{E}(T\wedge T_R)\le (\tau R-1)/{\sigma }^2$. For $\tau \ge 1$ this implies the weaker bound ${\tau }^{2}R/{\sigma }^2$. The threshold restriction prevents an impossible overshoot requirement below the starting value.

Under the conditional-variance and overshoot assumptions of the absorption-time bound from conditional variance and overshoot, $\mathbb{P}(T>n)\le 2\tau /(\sigma \sqrt{n})$. Bound the probability by $\mathbb{P}(T_R<\infty )+\mathbb{E}(T\wedge T_R)/n$ and optimize $1/R+{\tau }^{2}R/({\sigma }^{2}n)$. If the resulting bound is below one, its minimizing threshold is greater than one; otherwise the probability bound is trivial.