# Brownian motion: how is $\mathbb E [\sup_{t \ge 0} |\langle M\rangle_{t \wedge \tau_n}|^2] < \infty$ satisfied?

I'm reading a theorem at page $$43$$ of these notes, i.e.,

Proposition 7.12. Let $$M$$ be a continuous local martingale with respect to a filtration $$\left(\mathcal{F}_t, t \in \mathbb{R}_{+}\right)$$ such that $$M_0=\begin{array}{ll} 0 & \text { a.s } \end{array} \text { and } \lim _{t \rightarrow \infty}\langle M\rangle_t=\infty \quad \text { a.s. }$$ Let us also define $$\tau(s)=\inf \left\{t>0:\langle M\rangle_t \geq s\right\}$$ and $$B_s=M_{\tau(s)}, \mathcal{G}_s=\mathcal{F}_{\tau(s)}$$. Then $$B$$ is a standard Brownian motion with respect to $$\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$$.

Proof. As already mentioned, the idea is to use Lévy's theorem, i.e., to show that

• (i) $$B$$ has continuous trajectories.
• (ii) $$B$$ is a local martingale with respect to $$\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$$.
• (iii) $$\langle B\rangle_s=s$$, i.e., $$\left(B_s^2-s, s \in \mathbb{R}_{+}\right)$$is a local martingale with respect to $$\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$$.

Let us verify these three statements.

• (i) As $$M$$ is continuous, $$t \rightarrow\langle M\rangle_t$$ is also continuous. Moreover, if $$\langle M\rangle$$ is constant on some interval, then $$M$$ also is, so the function $$s \mapsto B_s=M_{\tau(s)}$$ is continuous.

• (ii) Let $$\tau_n=\inf \left\{t>0:\left|M_t\right| \geq n\right\}, n \geq 1$$. For each $$n, M^{\tau_n}$$ is a martingale such that $$\mathbb{E}\left(\sup _{t \in[0, T]}\left|M_{t \wedge \tau_n}\right|^2\right)<\infty, \quad \forall T>0,$$ so by the optional stopping theorem (version 2), we have $$\mathbb{E}\left(M_{\tau\left(s_2\right) \wedge \tau_n} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=M_{\tau\left(s_1\right) \wedge \tau_n} \quad \text { a.s., } \quad \forall s_2>s_1 \geq 0 .$$ By the dominated convergence theorem (and some details), this implies that $$\mathbb{E}\left(M_{\tau\left(s_2\right)} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=M_{\tau\left(s_1\right)} \quad \text { a.s. }$$ i.e., $$\mathbb{E}\left(B_{s_2} \mid \mathcal{G}_{s_1}\right)=B_{s_1} \quad \text { a.s. }$$ i.e., $$B$$ is a martingale with respect to $$\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$$.

• (iii) Let $$X_t=M_t^2-\langle M\rangle_t$$. By assumption, $$X^{\tau_n}$$ is a martingale $$\forall n$$, so $$\mathbb{E}\left(X_{\tau\left(s_2\right) \wedge \tau_n} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=X_{\tau\left(s_1\right) \wedge \tau_n} \quad \text { a.s., } \quad \forall s_2>s_1 \geq 0 \quad \quad (\star)$$ Then again by the dominated convergence theorem (and some details), we obtain that $$\mathbb{E}\left(X_{\tau\left(s_2\right)} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=X_{\tau\left(s_1\right)} \quad \text { a.s. }$$ i.e., $$\mathbb{E}\left(M_{\tau\left(s_2\right)}^2-\langle M\rangle_{\tau\left(s_2\right)} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=M_{\tau\left(s_1\right)}^2-\langle M\rangle_{\tau\left(s_1\right)} \quad \text { a.s. }$$ As $$\langle M\rangle_{\tau(s)}=s$$ by definition, we obtain: $$\mathbb{E}\left(B_{s_2}^2-s_2 \mid \mathcal{G}_{s_1}\right)=B_{s_1}^2-s_1 \quad a . s ., \quad \forall s_2>s_1 \geq 0$$ i.e., $$\left(B_s^2-s, s \in \mathbb{R}_{+}\right)$$is a martingale with respect to $$\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$$.

The quoted theorem is

Optional stopping theorem (version 2). Let $$M$$ be a continuous square-integrable martingale such that $$\mathbb{E} \bigg [ \sup _{t \ge 0} |M_t|^2 \bigg ] < \infty.$$ Let $$\tau_1, \tau_2$$ be two stopping times such that $$0 \leq \tau_1 \leq \tau_2 \leq \infty \quad \text { a.s. }$$ Then $$\mathbb{E}\left(M_{\tau_2} \mid \mathcal{F}_{\tau_1}\right)=M_{\tau_1} \quad \text { a.s., } \quad \text { so } \quad \mathbb{E}\left(M_{\tau_2}\right)=\mathbb{E}\left(M_{\tau_1}\right).$$

My understanding It seems to obtain $$(\star)$$, the authors apply Doob's optional stopping theorem (version $$2$$) on the martingale $$X^{\tau_n}$$ and two stopping times $$\tau(s_1), \tau(s_2)$$. We have to show that $$\mathbb{E} \bigg [ \sup _{t \ge 0} |X^{\tau_n}_t|^2 \bigg ] < \infty. \qquad ( \star \star )$$

We have \begin{align} |X^{\tau_n}_t|^2 &= |M_{t \wedge \tau_n}^2-\langle M\rangle_{t \wedge \tau_n}|^2 \\ &\le 2 |M_{t \wedge \tau_n}|^2 + 2 |\langle M\rangle_{t \wedge \tau_n}|^2 \\ &\le 2n^2 + 2 |\langle M\rangle_{t \wedge \tau_n}|^2. \end{align}

I could not see how $$\mathbb E [\sup_{t \ge 0} |\langle M\rangle_{t \wedge \tau_n}|^2] < \infty$$. Could you elaborate on how $$( \star \star )$$ is satisfied?

1. A continuous local martingale $$M$$ such that $$M_0\in L^2$$ is a bounded martingale in $$L^2$$ iff $$\mathbb E[\langle M\rangle_{\infty}]<+\infty$$. In that case, $$M^2-\langle M\rangle$$ is a uniformly integrable martingale.
Here $$(M^{\tau_n}_t)_{t\ge0}$$ is bounded by $$n$$, therefore it is bounded in $$L^2$$, so $$(X^{\tau_n}_t)_{t\ge0}$$ is a uniformly integrable martingale, to which you can apply the optional stopping theorem.
• It seems you meant $(X^{\tau_n}_t)_{t\ge0}$ (rather than $(X_t)_{t\ge0}$) is a uniformly integrable martingale. Commented Mar 19, 2023 at 18:36