Bound on supremum of local martingale

Question

Let $M$ be a continuous local martingale starting at $0$. How can I prove $$ P(\sup_{s\leq t}M_s>a,\langle M\rangle_t\le b)\leq 4\frac b{a^2} $$ for all $a,b>0$ and $t>0$?

Answer 1

Define a stopping time $\tau$ by $$\tau := \inf\{t>0; \langle M \rangle_t>b\}.$$ Since

\begin{align*} \mathbb{P} \left( \sup_{s \leq t} M_s>a, \langle M \rangle_t \leq b \right) &= \mathbb{P} \left( \sup_{s \leq t} M_s>a, \tau>t \right) \\ &\leq \mathbb{P} \left( \sup_{s \leq t} M_{s \wedge \tau} >a \right) \end{align*}

it follows from Doob's maximal inequality that \begin{align*} a^2 \mathbb{P} \left( \sup_{s \leq t} M_s>a, \langle M \rangle_t \leq b \right) &\leq \mathbb{E} \left( \sup_{s \leq t} M_{s \wedge \tau}^2 \right) \\ &\leq 4 \mathbb{E}\left(\langle M \rangle_{t \wedge \tau} \right) \leq 4b \end{align*}

for any continuous martingale $M$. Dividing both sides yields the desired inequality. Use standard stopping techniques to derive the inequality for continuous local martingales.