In a paper I am reading I found the following:
"Applying the exponential martingale inequality we derive that $$P\Big(\omega: \sup_{0 \leq t \leq k}[M(t)-1/2 \epsilon \langle M(t) \rangle] \leq 2 \epsilon^{-1} \log k \Big) \leq k^{-2}. \quad (1) "$$ $\epsilon$ and $k$ are positive constants.
My thoughts:
$\{M(t)\}$ is a continuous martingale that vanishes at $t=0,$ so the process $\{M(t)-1/2 \epsilon \langle M(t) \rangle\}$ is a submartingale. The inequality (1) looks like a version of Doob's submartingale inequality but I haven't been able to link it.
Does anybody knows where inequality (1) comes from? I appreciate any idea or reference that can throw some light on it.
Under some mild regularity conditions, the stochastic exponential process $$X_t = \mathcal{E}(\epsilon M)_t = \exp\left( \epsilon M_t - \frac{1}{2} \epsilon^2 \langle M \rangle_t \right)$$ is a non-negative martingale martingale starting at $X_0 = 1$; in such a case $E(X_t) = 1$ for all $t \geq 0$. Then, $$\begin{align*} \mathbb{P} \left( \sup_{0 \leq t \leq k} \left( M_t - \frac{1}{2} \epsilon \langle M \rangle_t \right)\geq 2 \epsilon^{-1} \log k \right) &= \mathbb{P} \left( \sup_{0 \leq t \leq k} \exp \left( \epsilon M_t - \frac{1}{2} \epsilon^2 \langle M \rangle_t \right) \geq k^2 \right) \\ &\leq \frac{E(X_k^+)}{k^2} \\ &= \frac{E(X_k)}{k^2} \\ &= \frac{1}{k^2} \end{align*}$$ where the inequality is Doob's martingale inequality applied to $X_t$.