# Brownian motion: how is DCT for conditional expectation applied on this stopped martingale?

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 $$(\star)$$.

• $\vert M_t\vert\le n$ for all $t\le\tau_n$, a.s., by definition of $\tau_n$. So $Z=n$.
– Will
Commented Mar 19, 2023 at 19:34
• @Will but we take the limit $n \to \infty$... Commented Mar 19, 2023 at 19:35
• Oh you need an upper bound independent of $n$ ok sorry
– Will
Commented Mar 19, 2023 at 19:38
• $\tau_n$ goes to $+\infty$ with $n$. So $X_{\tau(s_1)\wedge\tau_n}$ converges almost surely to $X_{\tau(s_1)}$. And $X$ is a uniformly integrable martingale, so so is $X^{\tau(s_2)}$, hence $X_{\tau(s_2)\wedge\tau_n}$ converges in $L^1$ to $X_{\tau(s_2)}$ as $n\to+\infty$. This is a way to show $(\star)$.
– Will
Commented Mar 19, 2023 at 19:43
• Actually I meant $M$ instead of $X$, but it is true for both.
– Will
Commented Mar 19, 2023 at 20:24

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.,

Example 5.4.3. Let $$(X_i)_{i \in I}$$ be a family of real random variables. If $$\sup _{i \in I} \mathbb E [ |X_i|^r ] < \infty$$ for some $$r>1$$, then $$(X_i)$$ is uniformly integrable.

• Looks correct to me
– Will
Commented Mar 20, 2023 at 20:36
• @Will Thank you so much for your verification! Commented Mar 20, 2023 at 20:37