I'm reading about optional stopping theorem from this note. In Optional stopping theorem, version 3. below,
$a, b>0$ and $$ T=\inf \left\{n \in \mathbb{N}: M_n \leq-a \text { or } M_n \geq b\right\}. $$
$M_{T \wedge \infty}$ is the a.s. and $L_2$ limit of the stopped martingale $(M_{T \wedge n})_n$, i.e., $M_{T \wedge n} \to M_{T \wedge \infty}$ a.s. and in $L_2$.
The author then says that $\color{blue}{M_{T \wedge \infty}=M_T}$. I don't see how $T$ is almost surely finite.
Could you explain how $M_T$ is well-define?
Stopped martingale. Let $\left(M_n, n \in \mathbb{N}\right)$ be a martingale and $T$ be a stopping time with respect to a filtration $\left(\mathcal{F}_n, n \in \mathbb{N}\right)$, without any further assumption. Let us also define the stopped process $$ \left(M_{T \wedge n}, n \in \mathbb{N}\right) $$ where $T \wedge n=\min \{T, n\}$ by definition. Then this stopped process is also a martingale with respect to $\left(\mathcal{F}_n, n \in \mathbb{N}\right.$ ) (we skip the proof here, which uses the first version of the optional stopping theorem).
Theorem 14.2. (The martingale convergence theorem: first version) Let $\left(M_n, n \in \mathbb{N}\right.$ ) be a square-integrable martingale (i.e., a martingale such that $\mathbb{E}\left(M_n^2\right)<$ $+\infty$ for all $n \in \mathbb{N})$ with respect to a filtration $\left(\mathcal{F}_n, n \in \mathbb{N}\right)$. Under the additional assumption that $$ \sup _{n \in \mathbb{N}} \mathbb{E}\left(M_n^2\right)<+\infty $$ there exists a limiting random variable $M_{\infty}$ such that
- (i) $M_n \underset{n \rightarrow \infty}{\rightarrow \rightarrow} M_{\infty}$ almost surely.
- (ii) $\lim _{n \rightarrow \infty} \mathbb{E}\left(\left(M_n-M_{\infty}\right)^2\right)=0$ (quadratic convergence).
- (iii) $M_n=\mathbb{E}\left(M_{\infty} \mid \mathcal{F}_n\right)$, for all $n \in \mathbb{N}$ (this last property is referred to as the martingale $M$ being "closed at infinity").
Optional stopping theorem, version 3. Let $\left(M_n, n \in \mathbb{N}\right)$ be a martingale with respect to $\left(\mathcal{F}_n, n \in \mathbb{N}\right)$ such that there exists $c>0$ with $\left|M_{n+1}(\omega)-M_n(\omega)\right| \leq c$ for all $\omega \in \Omega$ and $n \in \mathbb{N}$ (this assumption ensures that the martingale does not make jumps of uncontrolled size: the simple symmetric random walk $S_n$ satisfies in particular this assumption). Let also $a, b>0$ and $$ T=\inf \left\{n \in \mathbb{N}: M_n \leq-a \text { or } M_n \geq b\right\}. $$ Observe that $T$ is a stopping time with respect to $\left(\mathcal{F}_n, n \in \mathbb{N}\right)$ and that $-a-c \leq M_{T \wedge n}(\omega) \leq b+c$ for all $\omega \in \Omega$ and $n \in \mathbb{N}$. In particular, $$ \sup _{n \in \mathbb{N}} \mathbb{E}\left(M_{T \wedge n}^2\right)<+\infty, $$ so the stopped process $\left(M_{T \wedge n}, n \in \mathbb{N}\right)$ satisfies the assumptions of the first version of the martingale convergence theorem. By the conclusion of this theorem, the stopped martingale $\left(M_{T \wedge n}, n \in \mathbb{N}\right)$ is closed, i.e. it admits a limit $\color{blue}{M_{T \wedge \infty}=M_T}$ and $$ \mathbb{E}\left(M_T\right)=\mathbb{E}\left(M_{T \wedge \infty}\right)=\mathbb{E}\left(M_{T \wedge 0}\right)=\mathbb{E}\left(M_0\right). $$