I need a check in the folliwing exercise, which is an application of the optional stopping theorem.
let $T= \inf \{ t \geq 0: |B_t|=a \}$, $a \geq 0$, and $B_t$ a standard Brownian motion. By using the $\mathcal{F_t^+}$-martingale $(B_t^2 – t)_{ t \geq 0}$, show that $$E[T] = a^2$$
Here's what I did:
First I used consider the stopped process $(B_{t \wedge T}^2 – t \wedge T)_{t \geq 0}$. It's again a martingale, and by the optional sampling theorem $$E[B_{t \wedge T}^2 ] =E[ t \wedge T)] $$
(I can apply the Optional sampling theorem since $t \wedge T< t$)
Now, I notice that $t \wedge T \rightarrow_t T$, and so by Monotone Convergence thm: $$\lim_t E[t \wedge T] = E[ \lim_t t \wedge T]= E[T] $$
Moreover I notice that $$B_{t \wedge T}^2 \leq a $$ since if $t \wedge T = T$, then $B_{t \wedge T} = a$, else if $t \wedge T = t$, then $B_{t \wedge T} < a$ (otherwise, if it would be greater than $a$,$B_t$ would had hitten $a$).
So, by DCT: $$ \lim_t E[B_{t \wedge T}^2] =_{(DCT)} E[\lim_t B_{t \wedge T}^2] = E[B_T^2] = a^2$$
and the result follows.
Is everything okay? I want to be sure that all the steps are motivated in the right way!
Best Answer
This is almost perfect, except you do not precise why $T$ is a.s. finite, as pointed out by Xiaohai.
But this is also already in your analysis since
$$E[T] = \lim_{t} E[T \wedge t] = \lim_{t} E[B_{T \wedge t}^2] \le a^2$$
hence the r.v. $T$ being integrable is a.s. finite. From this point your reasoning is OK.