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!

  • $\begingroup$ it seems to be correct $\endgroup$ – Canardini Dec 27 '19 at 19:42
  • $\begingroup$ @Canardini thanks for the check. The " since " steps is okay in your opinion? I'm a bit unsure $\endgroup$ – Vefhug Dec 27 '19 at 19:44
  • $\begingroup$ @Canardini Sorry, but are there other ways to justify that $B_{t \wedge T}$ is bounded? $\endgroup$ – Vefhug Dec 28 '19 at 8:19
  • $\begingroup$ You just have to be careful ,we have $$ |B_{t \wedge T}| \leq a $$ a.s . As stated in the answer, we have to prove that $\tau$ is finite, one can use $$lim_n sup |B_n|=\infty$$ $\endgroup$ – Canardini Dec 28 '19 at 14:55
  • $\begingroup$ @Canardini I've seen the answer, but I'd like t use your hint to prove that $\tau$ is a.s. finite. How can I do? $\endgroup$ – Vefhug Dec 29 '19 at 8:52

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.

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  • 1
    $\begingroup$ Just a confirmation: 1) In your first equality you're already exchanged the limit with the expectation, so you're using Monotone convergence (and don't need $T$ to be a.s. finite), right? $\endgroup$ – Vefhug Dec 29 '19 at 18:03
  • $\begingroup$ 2) So one need $T$ to be a.s. finite because if it wouldn't, when we compute $E[B_T^2]$ we should do $E[B_T^2 | T= \infty] + E[B_T^2 | T < \infty]$, right? $\endgroup$ – Vefhug Dec 29 '19 at 18:03
  • $\begingroup$ you're right on both points, except I would rather write the second line as $E[B_T^2]= E[B_T^2; T <\infty]+E[B_T^2; T =\infty]$ (without conditioning). $\endgroup$ – Olivier Dec 29 '19 at 21:02

I think the proof is flawed in this step:

$$ \lim_t E[B_{t \wedge T}^2] =_{(DCT)} E[\lim_t B_{t \wedge T}^2] = E[B_T^2] = a^2.$$

Without first establishing $T < \infty$ a.s., one can not claim $E[B_T^2] = a^2$ (in fact, $E[B_T^2|T=\infty] < a^2)$.

Full proof:

For $n\ge 0$, $P_0(\{B_n\ge a\ \ i.o.\}) \ge P_0(\{B_n/\sqrt{n}\ge a\ \ i.o.\}) \ge \limsup P_0(\{B_n/\sqrt{n} \ge a\}) = P_0(\{B_1 \ge a\}) > 0.$

Note for any $m > 0$, $\{B_n\ge a\ \ i.o.\} = \{B_n\ge a, n \ge m,\ \ i.o.\} \in \sigma(\mathcal{F}_t^+, t\ge m)$. Hence $\{B_n\ge a\ \ i.o.\} \in \cap_{m \ge 0}\sigma(\mathcal{F}_t^+, t\ge m)\equiv \mathcal{T}.$

By 0-1 law for Brownian motion, we have $P_0(\{B_n\ge a\ \ i.o.\})=1$. Since $B_t$ is continuous, $1 = P_0(\{B_n\ge a\ \ i.o.\}) \le P_0(\{B_n \ge a\ \text{for some }n\}) \le P(T< n\ \text{for some }n) \le P(T < \infty).$ Hence $P(T < \infty) = 1$.

Let $X_t=B_t^2-t, t\ge 0$. Then $X_t$ is a martingale and $T$ is a bounded stopping time. In view of the optional stopping time theorem, $$ 0 = EX_0=EX_T=EB_T^2-ET. $$ Hence $$ET=EB_T^2=E(B_T^2;T<\infty) + E(B_T^2;T=\infty)=E(B_T^2;T<\infty)$$ $$=E(a^2;T<\infty)=a^2P(T<\infty)=a^2.$$

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  • $\begingroup$ Thanks for your answer. First of all, the step where I used (DCT) is okay, right? $\endgroup$ – Vefhug Dec 28 '19 at 8:39
  • $\begingroup$ Yes. But you don't need any DCT in problem. $\endgroup$ – Xiaohai Zhang Dec 28 '19 at 8:40
  • $\begingroup$ Okay, now I'll read again your answer, to be sure to understand everything. $\endgroup$ – Vefhug Dec 28 '19 at 8:42
  • $\begingroup$ Let $X_t=B_t^2-t$. Then $X_t$ is a martingale, and $T < \infty$ a.s. Hence $EX_T=EX_0=0$. Hence $ET=EB_T^2=a^2$. $\endgroup$ – Xiaohai Zhang Dec 28 '19 at 8:46
  • $\begingroup$ $EB_T^2=E(B_T^2; T<\infty)=E(a^2; T<\infty)=a^2$ due to $P(T<\infty)=1$. $\endgroup$ – Xiaohai Zhang Dec 28 '19 at 8:51

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