# Exponential submartingale inequality

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.

• For context: can you link the paper in which you found this inequality? Commented May 15, 2022 at 17:19
• @JoseAvilez. It's this paper: tandfonline.com/doi/abs/10.1081/SAP-100001183 (But I can't find a "free" link to the paper. I got it by email)
– UBM
Commented May 15, 2022 at 17:33

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 $$X_t$$.