# Independence of Brownian Motion with respect to a stopping time

Let $B_t$ be a brownian motion, $B_0=0$, and $\gamma \in \mathbb{R}$. Now, let's build the following stopping time: $$T = \inf \{ t \geq 0 : |B_t + \gamma t| = 1 \}.$$ If $\gamma = 0$, how would you argue that $B$ and $T$ are independents?

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Could you please clarify, what do you mean by the independence here? For example, $$\mathsf P(B_1\geq 1,T\geq 2) = 0 \neq \mathsf P(B_1\geq 1)\mathsf P(T\geq 2)$$ –  Ilya Jan 11 '13 at 21:39
how would you argue that B and T are independent... I would not, they are not. –  Did Jan 12 '13 at 9:44
If $\gamma = 0$ it is really just a simple consequence of symmetry, since $-B_t$ is also a brownian motion. –  mike Jan 15 '13 at 14:30