Let $(M_n)_{n\in\mathbb{N}}$ be a martingale with respect to a filtration $(\mathcal{F}_n)_{n\in\mathbb{N}}$. We can define the bracket of $M_n$ by
$$\langle M\rangle_n=\sum\limits_{k=0}^{n-1}\mathbb{E}\left[(M_{k+1}-M_k)^2|\mathcal{F}_k \right].$$
Morally, $\langle M \rangle_n$ is of the same order as $M_n^2$. This idea can be made precise thanks to the BDG inequality, which says that, for every $p>1$ (actually, any $p>0$ for continuous martingales.), there exists universal constants $c_p$ and $C_p$ such that for every $n\in\mathbb{N}$,
$$c_p\mathbb{E}\left[\langle M \rangle_n^{p/2}\right]\leq \mathbb{E}\left[(M_n^{*})^p \right]\leq C_p\mathbb{E}\left[\langle M \rangle_n^{p/2}\right]$$
where $M_n^*=\underset{1\leq k\leq n}\max|M_k|$.
Therefore, my question is the following one:
Is this possible to compare $\mathbb{P}\left(M_n^*> \lambda\right)$ and $\mathbb{P}(\langle M\rangle_n>\lambda^2)$?
Or in the case where $M$ and its bracket converge almost surely, $\mathbb{P}\left(|M_{\infty}-M_n|> \lambda\right)$ and $\mathbb{P}\left(\langle M\rangle_{\infty}-\langle M\rangle_n>\lambda^2 \right)$?
Best Answer
Regarding your question, the following Lenglart's inequality is useful(cf. J. Jacod, and A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003, Lemma I.3.30, P.35.).
From Lenglart inequality, Since $M^2$ is a càdlàg adapted process which is L-dominated by an predictable increasing process $\langle M\rangle$(i.e. $\mathsf{E}[M^2_T] = \mathsf{E}\langle M \rangle_T$ for every bounded stopping times), then for all stopping times $T$ and all $\epsilon, \eta>0$, \begin{gather*} \mathsf{P}(M^*_T>\epsilon)\le \frac{\eta}{\epsilon^2} + \mathsf{P}(\langle M\rangle_T \ge \eta),\\ \mathsf{P}\Big(\sup_{m\ge n}|M_m-M_n|>\epsilon\Big)\le \frac{\eta}{\epsilon^2} + \mathsf{P}(\langle M\rangle_\infty-\langle M\rangle_n \ge \eta). \end{gather*}