Independence of $T$ and $B_T$

Question

Let $\{B_t:t\ge0\}$ be a real brownian motion such that $B_0=0$. Let $T=\inf \{t:B_t \notin (-a,a)\}$ with $a>0$. Are $T$ and $B_T$ independent?

I tried the following and I would like your opinion. Let $t>0$, $P(T<t)=P(T<t, B_T=a)+P(T<t, B_T=-a)=2 P(T<t, B_T=a)$.

Answer 1

Let $\mathcal G$ the $\sigma$-algebra generated by $\|B_t\|$. Then $T$ is $\mathcal G$-measurable.

Now, using what you've already tried, see if you can show that $\mathbb E(B_T\cdot 1_A) = 0$ for every $A\in\mathcal G$.

Therefore you can show that $\mathbb E(B_T|\mathcal G) \equiv 0$ hence $B_T$ is independent from $\mathcal G$.

Answer 2

Your approach looks good. You might want to think about why it is that $P(T < t, B_T = -a) = P(T < t, B_T = a)$ (you used this in your last equality). One way to see it is to use the fact that $-B_t$ is also a Brownian motion.

The last thing to note is that $P(B_T = a) = 1/2$, and thus you have shown $$P(T < t, B_T = a) = P(T < t) P(B_T = a)$$ and so the events $\{T < t\}$ and $\{B_T = a\}$ are independent. The same argument shows this also holds with $-a$ in place of $a$. Since $B_T$ can only take the values $a$ and $-a$ (by continuity), this shows that $T$ and $B_T$ are independent.