# Brownian motion stopped at the hitting time of an independent Brownian Motion

While I was working on the exit time of planar BM out of a square I came across the following observation, which I cannot grasp.

I define this exit time as $$\tau = \inf\{t \geq 0: \lvert B(t)\rvert = a \text{ or } \lvert W(t)\rvert = a\}$$ Here $B$ and $W$ are two independent BMs. I further define $$\tau_1 = \inf\{t \geq 0: \lvert B(t)\rvert = a\}$$ $$\tau_2 = \inf\{t \geq 0: \lvert W(t)\rvert = a\}$$ Clearly, $\tau =\tau_1 \wedge \tau_2$. Then, Doob's optional stopping gives me $$E[B_{\tau_1}\mid\mathcal{F}_{\tau_2}] = B_{\tau}$$

To see why this is true note the following facts.

Since both $B$ and $-B$ are martingales, they are also both submartingales. The inequality in this theorem then becomes an equality for our case. Then I apply Doob's optional stopping to obtain $$E[B_{\tau_1\wedge T}\mid\mathcal{F}_{\tau_2}] = B_{\tau\wedge T} \quad \text{ for any } T < \infty$$

Furthermore,$\lvert B_{\tau_1\wedge T}\rvert \leq a$ and $P\{\tau_1 < \infty\} = 1$. The latter implies $B_{\tau_1\wedge T} \rightarrow B_{\tau_1}$ pointwise as $T$ goes to infinity. Clearly, the same arguments hold for $B_{\tau\wedge T}$ as well. Now fix $A \in \mathcal{F}_{\tau_2}$ and apply the dominated convergence theorem. $$\lim_{T\rightarrow\infty}E[B_{\tau_1\wedge T}1_A] = E[B_{\tau_1}1_A]$$ $$\lim_{T\rightarrow\infty}E[B_{\tau\wedge T}1_A] = E[B_{\tau}1_A]$$

But $E[B_{\tau_1\wedge T}1_A] = E[B_{\tau\wedge T}1_A]$. The limit must be unique, hence $E[B_{\tau_1}1_A] = E[B_{\tau}1_A]$. Since $A \in \mathcal{F}_{\tau_2}$ was arbitrary, $E[B_{\tau_1}\mid\mathcal{F}_{\tau_2}] = B_{\tau}$.

$B_{\tau_1}$ and $\tau_2$ are independent. So then, $$E[B_{\tau_1}] = B_{\tau}$$ This last statement makes no sense to me. I feel like going from the last but one step to the last step the way I do is not correct but I wouldn't be able to say why. Can someone point out where I am going wrong?

I thought about this a bit further and I think the problem lies indeed in the last step. $E[B_{\tau_1}\mid\mathcal{F}_{\tau_2}]$ is not the same as $E[B_{\tau_1}\mid\sigma(\tau_2)]$. The latter, even though I have never seen such an expression, should be equal to $E[B_{\tau_1}]$ due to independence but the former is something completely different.

• Source?   – Did Jun 2 '15 at 9:09
• @Did I was just playing around with these notions of exit time for different cases out of curiosity. It is not something that I have seen in a book or something. So there is no source. I just want to reinforce my understanding of these subjects by applying my knowledge to non-textbook (not standard) problems. – Calculon Jun 2 '15 at 9:15
• Then you might want to explain why (you think that) Doob's optional stopping gives $E[B_{\tau_1}\mid\mathcal{F}_{\tau_2}] = B_{\tau_1\wedge\tau_2}$. – Did Jun 2 '15 at 9:19
• @Did For the right-continuous martingale case Doob's optional stopping gives me $E[B_{\tau_1\wedge T}\mid \mathcal{F}_{\tau_2}] = B_{\tau_1\wedge T \wedge \tau_2}$ for any $T < \infty$. I know that $\lvert B_{\tau_1\wedge T}\rvert \leq a$ and $\tau_1,\tau_2 < \infty$ almost surely. Then I can apply the dominated convergence theorem here, which gives $E[B_{\tau_1}\mid \mathcal{F}_{\tau_2}] = B_{\tau_1 \wedge \tau_2}$ almost surely. – Calculon Jun 2 '15 at 13:17
• OK, this settles the debate about $E(B_\sigma\mid\mathcal F_\rho)=B_{\sigma\wedge\rho}$ (thanks!). Re your question, indeed there is no reason to expect that if $X$ and the stopping time $T$ are independent then $E(X\mid\mathcal F_T)=E(X)$. For example, every $X$ is independent of every constant stopping time $T=n$ and then $\mathcal F_T=\mathcal F_n$, but $E(X\mid\mathcal F_n)$ and $E(X)$ can be different. – Did Jun 2 '15 at 19:08