In page 45 of the book "Financial Derivatives In Theory and Practice by P.J.Hunt and J.E.Kennedy, it seems to me that the author says the stopped Brownian Motion is not a martingale as follows.
(Quote)
Does the martingale property
$$M(t)=E[M(T)|F(t)]$$
hold if $T$ is a stopping time? In general the answer is no, as can be seen by taking M to be Brownian Motion and $T=\inf\{t>0: M(t)\ge1\}$ (Unquote)
I do not understand why the martingale property does not hold in this case and appreciate any explanation on this.
Best Answer
A stopped martingale is still a martingale, the proof is similar to the one of the Optional Sampling theorem.
For the equality to hold, the martingale needs to be Uniformly Integrable: $$ \lim_{A\to+\infty}\sup_{t\ge0}\mathbb{E}\left[|M_t|\mathbf{1}_{\left\{|M_t|>A\right\}}\right] = 0 $$ This is the weakest assumption, the proof can be found here: Theorem 3.6.
Applications:
Regarding the Brownian Motion, $$\sup_{t\ge0}\mathbb{E}\left[|B_t|\right] = +\infty$$ so the martingale is not Uniformly Integrable.
Now, regarding the stopped martingale $\left(B_{t\wedge T}\right)_{t\ge0}$, it depends on your stopping time. In particular, if the stopping times $S\le T$ are bounded almost-surely, the stopped martingale is Uniformly Integrable and Doob's Optional Stopping theorem applies: $$ \mathbb{E}\left[B_T\;|\;\mathcal{F}_S\right] = B_S $$ for any $S\le T$, $\mathbb{P}$-a.s. bounded stopping times.
Another particular case is when the stopped martingale is bounded in $\mathbb{L}^p$ with $p\in]1,+\infty]$, by Hölder inequality. For example $\left(B_{t\wedge T_a \wedge T_b}\right)_{t\ge 0}$ where $T_a$ and $T_b$ are the reaching times of $a$ and $b$. Then $$ \mathbf{1}_{\left\{|M_t|>A\right\}} = 0 $$ for $A>\max\left\{|a|,|b|\right\}$, and the martingale is Uniformly Integrable.