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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?

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1 Answer 1

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You need two things:

  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.
  2. The optional stopping theorem holds true for uniformly integrable (right-)continuous martingales.

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.

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  • $\begingroup$ It seems you meant $(X^{\tau_n}_t)_{t\ge0}$ (rather than $(X_t)_{t\ge0}$) is a uniformly integrable martingale. $\endgroup$
    – Akira
    Commented Mar 19, 2023 at 18:36
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    $\begingroup$ Yes you are correct. $\endgroup$
    – Will
    Commented Mar 19, 2023 at 18:43

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