Is stopped brownian motion not a martingale?

Question

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.

Answer 1

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.

Answer 2

Let $a=1$ or any other positive constant. Then $$E(M(\inf\{t:M(t)\ge a\}))=a$$ since $$P(M(\inf\{t:M(t)\ge a\})=a)=1.$$ And $a\ne M(t)$ with positive probability (actually probability 1).