Exponential martingale Brownian tail bound

Fix $0<b<a$, $\theta >0$ and an integer $n\ge 1$. Set $m=2^n$, $t_j=jT\mathrm{/}m$, and use the original adapted process on this grid. Define $J$ as the first index $j\in \{0,\dots ,m\}$ with $B_{t_j}>b$, or $m$ if there is no such index. For $j<m$, the event $\{J\le j\}$ is the finite union ${\bigcup }_{k\le j}\{B_{t_k}>b\}$ and is in ${\mathcal{F}}_{t_j}$; the event for $j=m$ is the whole space. Thus $J$ is a bounded discrete stopping index for the grid filtration. By [F1] and [F2], $E{Z}_{t_J}=1$. This variable is measurable and integrable, being a finite sum of integrable grid values times indicators.

Let $E_n=\{{\max }_{0\le j\le m}{B}_{t_j}>b\}$. On $E_n$ the selected value satisfies $B_{t_J}>b$ and $t_J\le T$, whence $Z_{t_J}\ge \exp (\theta b-{\theta }^{2}T\mathrm{/}2)$. Positivity therefore gives $P(E_n)\le \exp ({\theta }^{2}T\mathrm{/}2-\theta b)\mathrm{.}$ No continuous-time hitting time or finiteness of an unbounded hitting time has entered.

Write $M_T={\sup }_{t\le T}{\widehat{B}}_t$. This is the supremum over the countable union of the nested dyadic grids: for any $t$ in the interval there are grid times tending to it, and continuity gives convergence of the path values. The supremum is finite, since a continuous function on a compact interval is bounded. Measurability also follows from the countable supremum. The normalized grid events ${\widehat{E}}_n=\{{\max }_j{\widehat{B}}_{t_j}>b\}$ increase to $\{M_T>b\}$ and have the same probabilities as $E_n$, since the original and normalized paths agree on the common full event. Consequently [F4] and step 2.1 give $P(M_T>b)\le \exp ({\theta }^{2}T\mathrm{/}2-\theta b)$. Normalizing on another full event gives the same $M_T$ on their full intersection, so its distribution is independent of the choice.

Choose $\theta =b\mathrm{/}T$ in step 3.1, the positive minimizer of the quadratic, to get $P(M_T>b)\le e^{-b^2\mathrm{/}(2T)}$. Since $\{M_T\ge a\}\subseteq \{M_T>b\}$ for every $0<b<a$, take the explicit sequence $b_k=a(1-1\mathrm{/}k)$, $k\ge 2$, and let $k$ tend to infinity in the numerical upper bounds. Continuity of the exponential gives $P(M_T\ge a)\le e^{-a^2\mathrm{/}(2T)}$. This last argument does not assume that a dyadic grid attains the continuous maximum or that $M_T$ has no atoms.

The parameter range is $a,T>0$. At $a=0$ the probability is one and the limiting bound is one; for $a<0$ the probability is also one, but the displayed formula would be less than one and is not asserted. At $T=0$ and $a>0$ the probability is zero and division by $T$ is not used. With $T$ fixed, the bound tends to zero as $a\to \mathrm{\infty }$; with $a>0$ fixed, it tends to one as $T\to \mathrm{\infty }$ and to zero as $T\downarrow 0$. The real exponential martingale has random magnitude; its integrability follows from its Gaussian unit mean, not a deterministic modulus. Only finite-grid optional sampling is used, so no uniform-integrability assertion for an unbounded stopped family is needed. AC has exactly the interface uses in [F5].