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
$$
\tau(s)=\inf \left\{t>0:\langle M\rangle_t \geq s\right\},
\quad B_s=M_{\tau(s)}, \quad \mathcal{G}_s=\mathcal{F}_{\tau(s)}.
$$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
$$
\mathbb{E}\left(M_{\tau\left(s_2\right)} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=M_{\tau\left(s_1\right)} \quad \text { a.s. }
\quad \quad (\star)
$$
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
$$
\mathbb{E}\left(X_{\tau\left(s_2\right) \wedge \tau_n} \mid \mathcal{F}_{\tau\left(s_1\right)}\right)=X_{\tau\left(s_1\right) \wedge \tau_n} \quad \text { a.s., } \quad \forall s_2>s_1 \geq 0
$$
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 $\left(\mathcal{G}_s, s \in \mathbb{R}_{+}\right)$.
My understanding I have a problem understanding how $(\star)$ is obtained. To apply dominated convergence theorem, we need
$$
|M_{\tau\left(s_2\right) \wedge \tau_n}| \le Z \quad \text{a.s.} \quad \forall n\in \mathbb N,
$$
for some integrable random variable $Z$. I could not see how to have such $Z$.
Could you elaborate on how DCT is applied to get $(\star)$.
Best Answer
By Vitali's convergence theorem, it suffices to show that $( M_{\tau (s_2) \wedge \tau_n}, n \in \mathbb N)$ is uniformly integrable. By OST, $( M_{t \wedge \tau (s_2) \wedge \tau_n}, t \ge 0)$ is a martingale. By Doob's maximal inequality, $$ \begin{align} \mathbb E \big [ \sup_{s \in [0, t]} | M_{s \wedge \tau (s_2) \wedge \tau_n} |^2 \big ] &\le 4 \mathbb E [ | M_{t \wedge \tau (s_2) \wedge \tau_n} |^2 ] \\ &= 4 \mathbb E [ \langle M \rangle_{t \wedge \tau (s_2) \wedge \tau_n} ] \\ &\le 4 \mathbb E [ \langle M \rangle_{\tau (s_2)} ] = 4 s_2. \end{align} $$
By monotone convergence thereom, $$ \mathbb E \big [ \sup_{s \in [0, \infty)} | M_{s \wedge \tau (s_2) \wedge \tau_n} |^2 \big ] \le 4 s_2. $$
Notice that $$ | M_{\tau (s_2) \wedge \tau_n} |^2 \le \sup_{s \in [0, \infty)} | M_{s \wedge \tau (s_2) \wedge \tau_n} |^2. $$
Hence $$ \mathbb E \big [ | M_{\tau (s_2) \wedge \tau_n} |^2 \big ] \le 4 s_2 \quad \forall n \in \mathbb N. $$
The claim then follows from below result (taken from these notes), i.e.,