# Let $M$ be a continuous martingale, $r >0$, and $\tau := \inf \{t \ge 0 : \langle M \rangle_t \ge r \}$. Then $M_\tau$ is square-integrable

Let $$(\Omega, \mathcal F, \mathbb P)$$ be a probability space and $$\mathcal G = (\mathcal G_t, t \ge 0)$$ a filtration. Let $$M$$ be a real-valued continuous martingale w.r.t. $$\mathcal G$$ such that $$(\star) \quad M_0 = 0 \quad \text{and} \quad (\star \star) \quad \lim_{t \to \infty} \langle M \rangle_t=\infty.$$

Fix $$r >0$$ and define a stopping time $$\tau := \inf \{t \ge 0 : \langle M \rangle_t \ge r \}$$. By $$(\star \star)$$, we have $$\tau$$ is finite. I would like to prove that

Theorem $$M_\tau$$ is square-integrable.

Could you have a check on my below attempt?

Proof By OST, $$M^\tau = (M_{t \wedge \tau}, t \ge 0)$$ is a martingale. We define a process $$X$$ by $$X_t := M_t^2 - \langle M \rangle_t$$ for all $$t \ge0$$. Then $$X$$ and thus $$X^\tau$$ are martingales. By $$(\star)$$, we have $$X^\tau_0=0$$. Hence $$\mathbb E [X_{t \wedge \tau}] = \mathbb E [ |M_{t \wedge \tau}|^2 - \langle M \rangle_{t \wedge \tau} ] = 0.$$

We have $$\mathbb E [ \langle M \rangle_{t \wedge \tau} ] \le \mathbb E [ \langle M \rangle_{\tau} ]$$. Because $$M$$ and thus $$\langle M \rangle$$ have continuous sample paths, we get $$\mathbb E [ \langle M \rangle_{\tau} ] = r$$. Hence $$\mathbb E [ |M_{t \wedge \tau}|^2 ] = \mathbb E [ \langle M \rangle_{t \wedge \tau} ] \le r.$$

By Doob's maximal inequality, $$\mathbb E \big [ \sup_{s \in [0, t]} | M_{s \wedge \tau} |^2 \big ] \le 4 \mathbb E [ | M_{t \wedge \tau} |^2 ] \le 4 r.$$

By monotone convergence theorem, $$\mathbb E \big [ \sup_{s \in [0, \infty)} | M_{s \wedge \tau} |^2 \big ] \le 4r.$$

Clearly, $$| M_{\tau} |^2 \le \sup_{s \in [0, \infty)} | M_{s \wedge \tau} |^2.$$

Hence $$\mathbb E [ | M_{\tau} |^2 ] \le4r$$. This completes the proof.

Yes, your proof is correct. One could also use Fatou's lemma for a slightly more direct proof and a tighter bound: After showing $$\mathbb{E}[|M_{t \wedge \tau}|^2] \le r$$, we have \begin{align*} \mathbb{E}[|M_\tau|^2] &= \mathbb{E}\left[\liminf_{t \rightarrow \infty} |M_{t \wedge \tau}|^2\right] \\ &\le \liminf_{t \rightarrow \infty} \mathbb{E}\left[|M_{t \wedge \tau}|^2\right] \\ &\le r\end{align*}.
• Thank you so much for your verification! Can we weaken the conditions $(\star)$ or $(\star\star)$? Mar 20 at 16:33
• @Analyst The condition $M_0 = 0$ can be replaced with $M_0 \in L^2$. The condition $\langle M \rangle_t \rightarrow \infty$ is also unnecessary because $\langle M \rangle_{t \wedge \tau} \le r$ holds regardless of whether $\tau$ is finite or not, but dropping this condition requires showing that $\{\omega: \lim_{t \rightarrow \infty} M_t \text{ exists }\} \subseteq \{\omega : \lim_{t \rightarrow \infty} \langle M \rangle_t < \infty\}$ a.s. See Proposition 4.1.23 in Revuz and Yor's Continuous Martingales and Brownian Motion Mar 20 at 16:54