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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.

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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*}.

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  • $\begingroup$ Thank you so much for your verification! Can we weaken the conditions $(\star)$ or $(\star\star)$? $\endgroup$
    – Analyst
    Commented Mar 20, 2023 at 16:33
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    $\begingroup$ @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 $\endgroup$ Commented Mar 20, 2023 at 16:54

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