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.