Let $M$ be continuous local martingale starting from zero. For $a>0$, let $\tau_a=\inf \left\{t \ge 0: |M_t|>a \right\}$. Show that for every $t \ge 0$ we have: $a^2\mathbb{P}(\tau_a\le t)\le\mathbb{E}<M>_{\tau_a \wedge t}$
So I have few problems with this task, I count on your help. What I've thought: the condition $\left\{ \tau_a \le t \right\}$ is equal to : $\left\{ \tau_a \le t \right\}=\left\{ \sup\limits_{s\le t} |M_s| \ge a \right\}=\left\{ \sup\limits_{s\ge 0} |M_s^{\tau_a \wedge t}| \ge a \right\}$. What is more if I know that $M^{\tau_a \wedge t}$ was a martingale I could use martingale inequality, and got: $a^2\mathbb{P}(\sup\limits_{s\ge 0} |M_s^{\tau_a \wedge t}| \ge a)\le \sup\limits_{s\ge 0}\mathbb{E}(|M_s^{\tau_a \wedge t}|^2)$.
My problem is:
How to show that $M^{\tau_a \wedge t}$ is a martingale? (I know that it is enough to say that it is local martingale and it is bounded, but Im not sure if it is true that it is bounded..)
How do I know the thesis of the problem using above inequality?
It follows from the continuity of the sample paths and the definition of $\tau_a$ that $|M_{t \wedge \tau_a}| \leq a$ for all $t \geq 0$. This means that $M^{\tau_a}$ is, indeed, a bounded process. Moreover, the optional stopping theorem implies that it is a local martingale.
By the definition of the quadratic variation, we know that $(N_s^2 - \langle N \rangle_s)_{s \geq 0}$ is a martingale for any (continuous) square-integrable martingale $(N_s)_{s \geq 0}$. This implies, in particular, $$\mathbb{E}(N_s^2) = \mathbb{E}(\langle N \rangle_s)$$ for all $s \geq 0$. Use this identity with $N_s := M_s^{\tau_a \wedge t}$ and the fact that $s \mapsto \langle M^{\tau_a \wedge t} \rangle_s$ is non-decreasing.